A2 - Workbook 1.pdf

A2 CIRCULAR MOTION


CEDAR COLLEGE

1

May/June2014 2014 , ,Question #7 ,#7 qp_42 May/June Question , qp_42 14

17 Q20-

(a) Define the radian. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) A telescope gives a clear view of a distant object when the angular displacement between the edges of the object is at least 9.7 × 10−6 rad. (i)

The Moon is approximately 3.8 × 105 km from Earth. Estimate the minimum diameter of a circular crater on the Moon’s surface that can be seen using the telescope.

diameter = .................................................. km [2] (ii)

Suggest why craters of the same diameter as that calculated in (i) but on the surface of Mars are not visible using this telescope. ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2]

© UCLES 2014

CEDAR COLLEGE

9702/42/M/J/14

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For Examiner’s Use

4 Section A Answer Oct/Nov 2007 , Question #1 , qp_4all the questions in the spaces provided.

2 1

(a) Explain (i)

what is meant by a radian, .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2]

(ii)

why one complete revolution is equivalent to an angular displacement of 2 rad. .................................................................................................................................. ..............................................................................................................................[1]

(b) An elastic cord has an unextended length of 13.0 cm. One end of the cord is attached to a fixed point C. A small mass of weight 5.0 N is hung from the free end of the cord. The cord extends to a length of 14.8 cm, as shown in Fig. 1.1. C

14.8 cm small mass Fig. 1.1 The cord and mass are now made to rotate at constant angular speed in a vertical plane about point C. When the cord is vertical and above C, its length is the unextended length of 13.0 cm, as shown in Fig. 1.2.

© UCLES 2007

CEDAR COLLEGE

9702/04/O/N/07

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For Examiner’s Use

5

13.0 cm

C

C

L

Fig. 1.2 (i)

Show that the angular speed

Fig. 1.3 of the cord and mass is 8.7 rad s–1.

[2] (ii)

The cord and mass rotate so that the cord is vertically below C, as shown in Fig. 1.3. Calculate the length L of the cord, assuming it obeys Hooke’s law.

L = ............................................ cm [4]

© UCLES 2007

CEDAR COLLEGE

9702/04/O/N/07

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4

4 Section A

For Examiner’s Use

May/June 2008 , Question #1 , qp_4 Answer all the questions in the spaces provided. 3 1

(a) (i)

Define the radian. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

(ii)

A small mass is attached to a string. The mass is rotating about a fixed point P at constant speed, as shown in Fig. 1.1. mass rotating at constant speed

P Fig. 1.1 Explain what is meant by the angular speed about point P of the mass. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

5

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© UCLES 2008

9702/04/M/J/08

5 (b) A horizontal flat plate is free to rotate about a vertical axis through its centre, as shown in Fig. 1.2.

For Examiner’s Use

plate M d

Fig. 1.2 A small mass M is placed on the plate, a distance d from the axis of rotation. The speed of rotation of the plate is gradually increased from zero until the mass is seen to slide off the plate. The maximum frictional force F between the plate and the mass is given by the expression F = 0.72W, where W is the weight of the mass M. The distance d is 35 cm. Determine the maximum number of revolutions of the plate per minute for the mass M to remain on the plate. Explain your working.

number = ........................................... [5] (c) The plate in (b) is covered, when stationary, with mud. Suggest and explain whether mud near the edge of the plate or near the centre will first leave the plate as the angular speed of the plate is slowly increased. .......................................................................................................................................... .......................................................................................................................................... ...................................................................................................................................... [2]

© UCLES 2008

CEDAR COLLEGE

9702/04/M/J/08

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4 Section A

For Examiner’s Use

May/June 2010 , Question #1 , qp_41

Answer all the questions in the spaces provided.

4

1

(a) Define the radian. .......................................................................................................................................... .......................................................................................................................................... ...................................................................................................................................... [2] (b) A stone of weight 3.0 N is fixed, using glue, to one end P of a rigid rod CP, as shown in Fig. 1.1. glue

ω P

85 cm

stone, weight 3.0 N

C

Fig. 1.1 The rod is rotated about end C so that the stone moves in a vertical circle of radius 85 cm. The angular speed ω of the rod and stone is gradually increased from zero until the glue snaps. The glue fixing the stone snaps when the tension in it is 18 N. For the position of the stone at which the glue snaps, (i)

on the dotted circle of Fig. 1.1, mark with the letter S the position of the stone,

(ii)

calculate the angular speed ω of the stone.

[1]

angular speed = ................................... rad s–1 [4] © UCLES 2010

CEDAR COLLEGE

9702/41/M/J/10

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4 Section A

For Examiner’s Use

Answer all the questions in the spaces provided. Oct/Nov 2010 , Question #1 , qp_43 5

1

A planet of mass m is in a circular orbit of radius r about the Sun of mass M, as illustrated in Fig. 1.1. planet mass m Sun mass M

r

Fig. 1.1 The magnitude of the angular velocity and the period of revolution of the planet about the Sun are x and T respectively. (a) State (i)

what is meant by angular velocity, .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

(ii)

the relation between x and T. .............................................................................................................................. [1]

(b) Show that, for a planet in a circular orbit of radius r, the period T of the orbit is given by the expression T 2 = cr 3 where c is a constant. Explain your working.

[4] © UCLES 2010

CEDAR COLLEGE

9702/43/O/N/10

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5 (c) Data for the planets Venus and Neptune are given in Fig. 1.2. planet

r / 108 km

T / years

Venus Neptune

1.08 45.0

0.615

For Examiner’s Use

Fig. 1.2 Assume that the orbits of both planets are circular. (i)

Use the expression in (b) to calculate the value of T for Neptune.

T = ....................................... years [2] (ii)

Determine the linear speed of Venus in its orbit.

speed = ..................................... km s–1 [2]

CEDAR COLLEGE © UCLES 2010

9702/43/O/N/10

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4 Section A

For Examiner’s Use

Oct/Nov 2012 , Question #1 , qp_41 Answer all the questions in the spaces provided. 6

1

(a) State Newton’s law of gravitation. .......................................................................................................................................... .......................................................................................................................................... ...................................................................................................................................... [2] (b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M. For this planet, the product GM is 4.00 × 1014 N m2 kg–1, where G is the gravitational constant. The planet may be assumed to be isolated in space. (i)

By considering the gravitational force on the satellite and the centripetal force, show that the kinetic energy EK of the satellite is given by the expression EK =

GMm . 2r

[2] (ii)

The satellite has mass 620 kg and is initially in a circular orbit of radius 7.34 × 106 m, as illustrated in Fig. 1.1.

initial orbit 7.34 × 106 m

7.30 × 106 m new orbit

Fig. 1.1 (not to scale) © UCLES 2012

CEDAR COLLEGE

9702/41/O/N/12

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5 Resistive forces cause the satellite to move into a new orbit of radius 7.30 × 106 m. Determine, for the satellite, the change in 1.

For Examiner’s Use

kinetic energy,

change in kinetic energy = ............................................. J [2] 2.

gravitational potential energy.

change in potential energy = ............................................. J [2] (iii)

Use your answers in (ii) to explain whether the linear speed of the satellite increases, decreases or remains unchanged when the radius of the orbit decreases. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]




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Oct/Nov 2014 , Question #2 , qp_41
 6

7

2

A large bowl is made from part of a hollow sphere. A small spherical ball is placed inside the bowl and is given a horizontal speed. The ball follows a horizontal circular path of constant radius, as shown in Fig. 2.1.

ball 14 cm

Fig. 2.1 The forces acting on the ball are its weight W and the normal reaction force R of the bowl on the ball, as shown in Fig. 2.2. wall of bowl

R

ball

W Fig. 2.2 The normal reaction force R is at an angle θ to the horizontal. (a) (i)

By resolving the reaction force R into two perpendicular components, show that the resultant force F acting on the ball is given by the expression W = F tan θ.

[2] © UCLES 2014

CEDAR COLLEGE

9702/41/O/N/14

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7 (ii)

State the significance of the force F for the motion of the ball in the bowl. ........................................................................................................................................... ...................................................................................................................................... [1]

(b) The ball moves in a circular path of radius 14 cm. For this radius, the angle θ is 28°. Calculate the speed of the ball.

speed = ............................................... m s−1 [3]

CEDAR COLLEGE

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Page 4

= = = =

Mark Scheme GCE A LEVEL – May/June 2014

Syllabus 9702

(b) (i) m R Bqr / v = = R=(640 × 10–3 × 1.6 × 10–19 × 6.2 × 10–2) / (9.6 × 104 ) = = R=6.61 × 10–26 kg = = R=(6.61 × 10–26 ) / (1.66 × 10–27 ) u = = R=40 u

ANSWERS

=

C1 C1 C1 A1

[4]

B1 B1 B1

[3]

(a) angle subtended at the centre of a circle by an arc equal in length to the radius

B1 B1

[2]

(b) (i) arc R distance × angle diameter R 3.8 × 105 × 9.7 × 10–6 = = R 3.7 km

C1

or m constant and q ∝ 1 / r (ii) q / m ∝ 1 / r q / m for A is twice that for B ions in path A have (same mass but) twice the charge (of ions in path B)

Q207 1

Paper 42

(ii) Mars is (much) further from Earth / away (answer must be comparative) angle (at telescope is much) smaller

A1

[2]

B1 B1

[2]

Page 2 8

1

Q21- 1=

= =

Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2014 9702 43 (a) photon energy R hc / λ Page 2 Mark Scheme Syllabus Paper –34 8 R /(6.63 × 3.0f = ×1 10/ T ) / (590 × 10–9 ) C1 (a) (i) either ω = 2π T orAS/A ω× =102πf and C1 GCE LEVEL – May/June 2014 9702 41 –19 R 3.37 × 10 J C1 ω = 2π / 0.30 = 20.9 rad s–1 (accept 2 s.f.) Section A A1 [2] number R (3.2 × 10–3) / (3.37 × 10–19 ) 2 2× 1015) = =(ii) kinetic R bringing 9.5 × 1015 A1 (a) work done unit mass9.4 M1 x0 or v = ωx0 and ½mv2 C1 [3] energy =(allow ½mω from infinity (to the point) A1 = ½ × 0.130 × 20.92 × (1.5 × 10–2)2 = 6.4 × 10–3 J A1 [2] [2]

(b) (i)2 p R h / λ Page Mark Scheme Syllabus C1 –34 –9 =(b) = R (6.63 × 10 ) / (590 × 10 ) B1 rise to induced9702 e.m.f. (b) E(i) as magnet moves, flux is cut –byOctober/November cup / aluminium giving 2007 P = –mφ GCE A/AS LEVEL –27 –1 =

= (inR 1.12 × 10 cup)

kg m s

Paper [1]04

C1 B1

Section A

2 1=

9

2

Q22- 2

2

15 –27 × 1.12 and × 10heating total momentum R 9.5 10currents induced e.m.f. gives rise× to of the cup B1 (c) φ ∝ 1/x C1 –11 –1 = = thermal energy derived R 1.06from × 10oscillations kg m s of magnet so amplitude decreases A1 B1 [3] (a) (i) angle subtended at centre of circle ....................................................................... B1 7 7 –1 or (= 1.05 × 10 J kg ) either at 6R from centre, potential is (6.3 × 10 )/6 arc equal in length B1 –11 to the radius ........................................................................... induced e.m.f. gives currents a magnetic (B1) Nrise to A1 [1] (ii) force × 10 andRat1.06 5R from centre, potential is which (6.3 × generate 107)/5 (= 1.26 × 107field J kg–1) C1 7 the magnet so amplitude decreases (B1) thechange magnetic field opposes the motion of [3] in energy = (1.26 – 1.05) × 10 × 1.3 C1 (ii) arc = rθ and for one=revolution, arc = 2πr ............................................................. M1 6 2.7 × 10 J A1 2 2 = use 2πr/r 2π .................................................................................................. A0 M1 (a) time forθnumber atoms x/0nuclei and x/ activity cmthe or xisotope) C1 (ii) so, either ofof=½mω 0 = 0.75(of 0 is halved so ¼ energy to betoreduced to one half=(of its value) A1 [2] give new 1.6 mJinitial or change in energy potential = (1/5 – 1/6) × (6.3 × 107) (C1) 7 change in energy = (1/5 – 1/6) × (6.3 × 10 ) × 1.3 (C1) 6 or A1 [4] [2] either loss in energy 6.4×–10 1.6 loss = ¾ × 6.4 force giving loss = 4.8 mJ (b) (i) either weight provides/equals centripetal ==2.7 Jthe (A1) (b) (i) or A R λNacceleration of free fall is centripetal acceleration .................................... C1 B1 C1 460 R N × ln 2 2/ (8.1 × 24 × 60 × 60) 9.8 = 0.13 × 8ω ...................................................................................................... M1 (c) the q =number mc∆θ N R 4.6of× atoms 10 -1 A1 [3] (a) M1 ....................................................................................................... A0 ω = 8.7 –3 rad s = 6.2 × 10–3 × 910 × ∆θ C1 [2] in4.8 12×g10 of carbon-12 A1 23 –3 (ii) molecules in 1.0 kg R (6.02 × 10 ) / (18 × 10 ) C1 ∆θ =number 8.5 × 10of–4water K A1 [2] (ii) force in cord = weight + centripetal R force 3.3 × (can 1025 be an equation) .......................... C1 cord (L25–) / 13) ratio in R= (3.3 ×=10 (4.6 ×× 5/1.8 108) or force constant = 5.0/1.8 ................................ C1 (b) (i) force amount 3.2/40 16 
 (a) smooth curve with decreasing gradient, at x2 =.................................................. 0 M1 [1] (L – 13) × 5/1.8 = 5.0 + 5/9.8 × not L ×starting 10-2 × 8.7 R =7.2 (7.3)mol × 10 A1 [2]C1 0.080 A1 endL of line not at g = 0 or horizontal A1 = 17.2 cm ...........................................................................................................[2] A1 (ii) (constant pV = nRT centripetal force ofInternational 5.0 N gives L = 16.6 2014 cm allow 2/4) © Cambridge Examinations C1 p × 210 × 10–6 = 0.080 × 8.31 × 310 (b) straight line ×with gradient M1 [2] p = 9.8 105positive Pa A1 line starts at origin A1 [2] not credit if T in °C not K) (a) (i) pV =(do nRT

[2]

[1]

[2]

[4]

V = (8.31 × 300)/(1.02 × 105) ............................................................................... C1

(iii) either pV = 1/33 × Nm = 0.0244 then 0/2) .......................................................... A1 (c) sinusoidal shapem (if uses Celsius, B1 N = 0.080 × 6.02 × 1023 (= 4.82 × 1022) B1 only positive values and peak / trough –27 height constant –26 and m = 40 × 1.66 × 10 (= 6.64 × 10 ) C1 B1 [3] (ii)4 ‘loops’ volume occupied by one atom = 0.0244 / (6.02 × 1023) =–264.06 2× 10-26 m3 ............ M1 22 9.8 × 103 5 × 210 × 10–6 × C1 14 CEDAR COLLEGE -26= 1/3 × 4.82 × 10 × 6.64 × 10 separation √(4.06 10 ) ................................................................................ A1 ≈ = 1.93 × 10×5 -9 –1 = 3.44 × 10 m ................................................................................... A0 440 ×m10 s 5 × 5.00 A1 T = =(2.40 × 10–4) / 288 = 0.417 M1 [3] 3 (a) initially, pVc/ RMS 5 finally, pV / T = (2.40 × 10–3 × 14.5 × 10–4) / 835 = 0.417

M1

[2]

[2]

Page 2

Mark Scheme GCE A/AS LEVEL – May/June 2008

Syllabus 9702

Paper 04

Section A

3

1

(a) (i) angle (subtended) at centre of circle by an arc equal in length to the radius (of the circle) (ii) angle swept out per unit time / rate of change of angle by the string

(b) friction provides / equals the centripetal force 0.72 W = mdω2 0.72 mg = m × 0.35ω2 ω = 4.49 (rad s–1) n = (ω /2π) × 60 = 43 min–1 (allow 42)

2

4

1

(b) (i) p = !" ρ (b) (i) point S shown below C 5 " × 10 × 0.900 + centripetal force (ii) 1.02 (max) force=/ tension = ×weight ! 2 centripetal force5 = mrω 2 > = 3.4 × 10
2

5

1

LEVEL2 –

5 October/November

T or = 2 × 3.4 ×10 (ii) either ∝ A 830 m sor–1 27.2 (allow 820) RMS = (a) (i) c27.2 + 273.15 + 273.2 Section A 300.4 K

2010

Syllabus 9702

M1 A1

[2]

C1 B1 A1

[5]

M1 A1

[2]

Paper 41 B1 B1 B1 B1 B1

[3] [2]

C1 B1

[1]

C1 C1 C1 C1 A1 A1 [3] [4] Paper 43 C1 A1 C1

[2]

A1

[2]

M1 B1 A1 A1 B1

[1] [2] [2]

(ii) ω × T = 2π (b) (i) ( is the) mean / average square speed

B1 B1

[1] [1]

ρ = Nm/V N explained (b) (ii) centripetal forcewith is provided by the gravitational force 2 2 Nm 2 so, pV = 1/3 or mrω 2 = GMm/r 2 either mr(2π/T) = GMm/r and pV = NkT with k explained r 3 × 4π2 = GM × T 2 so mean kinetic energy / <EK> = ½m = 3/2 kT GM/4π2 is a constant (c) T 2 = cr 3

B1 B1 B1 M1 B1 A1 B1 A1 A0

[4]

(c) (i) pV = nRT 2.1 × 107 × 7.8 × 10–3 = n × 8.3 × 290 (c) (i) either T 2 = (45/1.08)3 × 0.6152 or T 2 = 0.30 × 453 n = 68 mol T = 165 years

C1 C1 A1 A1

[2] [2]

(ii) mean kinetic energy = 3/2 kT© UCLES 2008 × 365 × 24 × 3600) (ii) speed = (2π × 1.08 × 108) / (0.615 = 3/2 × 1.38 × 10–23 × 290 = 35 km s–1 = 6.0 × 10–21 J

C1 C1 A1 A1

[2] [2]

C1 (1) C1 (1) A1

[3]


 (c) (a) c(i) rate of change of angle / angular temperature (alone) displacement RMS depends (ii) 11.6 Kouton swept by radius so no effect

CEDAR COLLEGE

2

[2]

B1 C1

(c) either centripetal force increases as r increases or centripetal force larger at edge so flies off at edge first (F = mrω2 so edge first – treat as special case and allow one mark) Page 2 Mark Scheme: Teachers’ version Syllabus GCE AS/A LEVEL – May/June 2010 9702 (a) molecule(s) rebound from wall of vessel / hits walls change in momentum gives rise to impulse / force Section A either (many impulses) averaged to give constant force / pressure the molecules are in of random circle motion (a) or angle (subtended) at centre (by) arc equal in length to radius

2GCE

B1 B1

realisation that/ total internal energy is the total kineticspheres energy (a) (iii) atoms / molecules particles behave as elastic (identical) –21 23 × 68 × 6.02 × 10 energy = 6.0 × 10 volume of atoms 5/ molecules negligible compared to volume of containing vessel = 2.46 × 10 J

15

[4]

Page 2

Mark Scheme GCE AS/A LEVEL – October/November 2012

Syllabus 9702

Paper 41

Section A

6

1

(a) force is proportional to the product of the masses and inversely proportional to the square of the separation either point masses or separation >> size of masses

M1 A1

[2]

(b) (i) gravitational force provides the centripetal force mv2/r = GMm/r2 and EK = ½mv2 hence EK = GMm/2r

B1 M1 A0

[2]

∆EK = ½ × 4.00 × 1014 × 620 × ({7.30 × 106}–1 – {7.34 × 106}–1) = 9.26 × 107 J (ignore any sign in answer) that EK evaluated separately for each r) (allow 1.0 × 108 J if evidence Page 2 Mark Scheme Syllabus Cambridge International AS/A Level – October/November 2014 9702 2. ∆EP = 4.00 × 1014 × 620 × ({7.30 × 106}–1 – {7.34 × 106}–1) = 1.85 × 108 J (ignore any sign Section A in answer) (allow 1.8 or 1.9 × 108 J) 1 (a) g = GM / R2 6 –1 6 –1 6 2 23 ) ×– 10 (7.34 × 10× ) 10or) ∆E is N positive (iii) =either × 10 × 6.4 ) / (3.4 = K3.7 kg–1 / EK increased (6.67(7.30 × 10–11 speed has increased (ii) 1.

2

3

7

(b) ∆EP = mg∆h (a) (i) sum of kinetic atoms / molecules / particles because ∆hpotential ! R (or energy 1800 mand ! 3.4 × 106energy m) g isof constant to×random ∆EPreference = 2.4 × 3.7 1800 = 1.6 × 104 J (ii) no intermolecular forces (use of g = 9.8 m s–2 max. 1 for explanation) no potential energy internal energy is kinetic energy (of random motion) of molecules (reference to random motion here then allow back credit to (i) if M1 scored) (c) gravitational potential energy = (–)GMm / x v2 = 2GM / x x = 4D = 4 × 6.8 × 106 (b) kinetic energy ∝ thermodynamic temperature either temperature in Celsius, not kelvin so incorrect v2 = (2 × 6.67 × 10–11 × 6.4 × 1023 ) / (4 × 6.8 × 106 ) or temperature 6in kelvin is not doubled = 3.14 × 10 v = 1.8 × 103 m s–1 (use of 3.5 Dofgiving 1.9 × 10is3 m s–1same , allow max. 3) (a) temperature the spheres the no (net) transfer of energy between the spheres

2 
 (a) (i) F = R cosθ W = R =sinθ (b) (i) power m × c × ∆θ where m is mass per second dividing, W×=4.2 F tanθ 3800 = m × (42 – 18) to final line not shown) g s–1 m(max. 1= if38derivation (ii) some provides the centripetal force (ii) thermal energy is lost to the surroundings so rate is an overestimate

4

3

(b) either F = mv2 / r and W = mg / tan θ or v2 = rg (a) straight line through origin v2 = (14 × 10–2 × 9.8) / tan 28° to displacement shows acceleration proportional negative = 2.58gradient shows acceleration and displacement in opposite directions v = 1.6 m s–1

(a) obeys the equation pV / T = constant (accept pV = nRT)

C1 A1 [2] Paper 41 C1 A1 [2] C1 M1 A1 A1

[2] [2]

M1 B1 A1 C1

[2]

A1 B1 B1 B1

[3] [3]

C1 C1 C1 B1 B1

[2]

A1

[4]

B1 B1

[2]

M1 M1 C1 A0 C1 A1

[2] [3]

B1 M1 A1

[1] [2]

C1 M1 C1 A1 M1 A1 A1

[4] [3]

B1

[1]

© Cambridge International Examinations 2012

(b) (i) pV = nRT CEDAR COLLEGE 7

5.0 × 10 × 3.0 × 10–4 = n × 8.31 × 296 giving n = 6.1 mol

(ii) pressure ∝ amount of substance

16 C1 A1

[2]

A2 GRAVITATION


CEDAR COLLEGE

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1

1

4 May/June 2002 , Question #1 , qp_4 Answer all the questions in the spaces provided.

For Examiner’s Use

(a) The Earth may be considered to be a uniform sphere of radius 6.38 × 106 m. Its mass is assumed to be concentrated at its centre. Given that the gravitational field strength at the Earth’s surface is 9.81 N kg–1, show that the mass of the Earth is 5.99 × 1024 kg.

[2] (b) A satellite is placed in geostationary orbit around the Earth. (i)

Calculate the angular speed of the satellite in its orbit.

angular speed = ........................................ rad s–1 (ii)

[3]

Using the data in (a), determine the radius of the orbit.

radius = ........................................ m

CEDAR COLLEGE

9702/4 M/J/02

[3]

18

8 4

If an object is projected vertically upwards from the surface of a planet at a fast enough speed, it can escape the planet’s gravitational field. This means that the object can arrive at 2 infinity Oct/Nov 2002 , Question #4 ,energy. qp_4 The speed that is just enough for this to happen is where it has zero kinetic known as the escape speed.

For Examiner’s Use

(a) (i)

By equating the kinetic energy of the object at the planet’s surface to its total gain of potential energy in going to infinity, show that the escape speed v is given by v2 =

2GM , R

where R is the radius of the planet and M is its mass.

(ii)

Hence show that v2 = 2Rg, where g is the acceleration of free fall at the planet’s surface.

[3]




9702/4 O/N/02

CEDAR COLLEGE

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4

For Examiner’s Use

Answer all the questions in the spaces provided. May/June 2003 , Question #1 , qp_4

3 1

(a) Define gravitational potential. .......................................................................................................................................... ..................................................................................................................................... [2] (b) Explain why values of gravitational potential near to an isolated mass are all negative. .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [3] (c) The Earth may be assumed to be an isolated sphere of radius 6.4 × 103 km with its mass of 6.0 × 1024 kg concentrated at its centre. An object is projected vertically from the surface of the Earth so that it reaches an altitude of 1.3 × 104 km. Calculate, for this object, (i)

the change in gravitational potential,

change in potential = ……………………………………. J kg–1 (ii)

the speed of projection from the Earth’s surface, assuming air resistance is negligible.

speed = ……………………………………. m s–1 [5] 9702/4/M/J03

CEDAR COLLEGE

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5

For Examiner’s Use

(d) Suggest why the equation v 2 = u 2 + 2as is not appropriate for the calculation in (c)(ii). .......................................................................................................................................... ..................................................................................................................................... [1]

9702/4/M/J03

CEDAR COLLEGE

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21

4

4 1

Answer all the questions in the spaces provided. Oct/Nov 2003 , Question #1 , qp_4 (a) (i)

For Examiner’s Use

On Fig. 1.1, draw lines to represent the gravitational field outside an isolated uniform sphere.

Fig. 1.1 (ii)

A second sphere has the same mass but a smaller radius. Suggest what difference, if any, there is between the patterns of field lines for the two spheres. ................................................................................................................................... ................................................................................................................................... [3]

(b) The Earth may be considered to be a uniform sphere of radius 6380 km with its mass of 5.98 ⌅ 1024 kg concentrated at its centre, as illustrated in Fig. 1.2.

Fig. 1.2 A mass of 1.00 kg on the Equator rotates about the axis of the Earth with a period of 1.00 day (8.64 ⌅ 104 s).

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For Examiner’s Use

5 Calculate, to three significant figures, (i)

the gravitational force FG of attraction between the mass and the Earth,

FG = ………….…………………………. N (ii)

the centripetal force FC on the 1.00 kg mass,

FC = …………………….………………. N (iii)

the difference in magnitude of the forces.

difference = …………………………………….. N [6] (c) By reference to your answers in (b), suggest, with a reason, a value for the acceleration of free fall at the Equator. .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [2] 9702/4/O/N03

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5 3

May/June 2004 , Question #3 , qp_4

For Examiner’s Use

6

A binary star consists of two stars that orbit about a fixed point C, as shown in Fig. 3.1.

R2 C

M2

M1 R1

Fig. 3.1 The star of mass M1 has a circular orbit of radius R1 and the star of mass M2 has a circular orbit of radius R2. Both stars have the same angular speed , about C. (a) State the formula, in terms of G, M1, M2, R1, R2 and (i)

for

the gravitational force between the two stars, ...................................................................................................................................

(ii) the centripetal force on the star of mass M1. ................................................................................................................................... [2] (b) The stars orbit each other in a time of 1.26 speed for each star.

108 s (4.0 years). Calculate the angular

angular speed = ................................... rad s–1 [2]

24

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7 (c) (i)

Show that the ratio of the masses of the stars is given by the expression M1 R = 2. M2 R1

[2] (ii)

The ratio

M1 is equal to 3.0 and the separation of the stars is 3.2 M2

1011 m.

Calculate the radii R1 and R2.

R1 = ........................................ m R2 = ........................................ m [2] (d) (i)

By equating the expressions you have given in (a) and using the data calculated in (b) and (c), determine the mass of one of the stars.

mass of star = ......................................... kg (ii)

State whether the answer in (i) is for the more massive or for the less massive star. ................................................................................................................................... [4]

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For Examiner’s Use

4

6 1

Answer all the questions in the spaces provided. May/June 2005 , Question #1 , qp_4 The orbit of the Earth, mass 6.0 × 1024 kg, may be assumed to be a circle of radius 1.5 × 1011 m with the Sun at its centre, as illustrated in Fig. 1.1. Earth, mass 6.0 x 1024 kg Sun 1.5 x 1011 m

Fig. 1.1 The time taken for one orbit is 3.2 × 107 s. (a) Calculate (i)

the magnitude of the angular velocity of the Earth about the Sun,

angular velocity = ............................... rad s–1 [2] (ii)

the magnitude of the centripetal force acting on the Earth.

force = ....................................... N [2]

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For Examiner’s Use

5 (b) (i)

State the origin of the centripetal force calculated in (a)(ii). ................................................................................................................................... ...............................................................................................................................[1]

(ii)

Determine the mass of the Sun.

mass = ..................................... kg [3]

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4 Answer all the questions in the spaces provided. Oct/Nov 2006 , Question #1 , qp_4

7 1

For Examiner’s Use

The definitions of electric potential and of gravitational potential at a point have some similarity. (a) State one similarity between these two definitions. .......................................................................................................................................... ..................................................................................................................................... [1] (b) Explain why values of gravitational potential are always negative whereas values of electric potential may be positive or negative. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .................................................................................................................................... [4]

2

A mercury-in-glass thermometer is to be used to measure the temperature of some oil. The oil has mass 32.0 g and specific heat capacity 1.40 J g–1 K–1. The actual temperature of the oil is 54.0 °C. The bulb of the thermometer has mass 12.0 g and an average specific heat capacity of 0.180 J g–1 K–1. Before immersing the bulb in the oil, the thermometer reads 19.0 °C. The thermometer bulb is placed in the oil and the steady reading on the thermometer is taken. (a) Determine (i)

the steady temperature recorded on the thermometer,

temperature = ………………………… °C [3] © UCLES 2006

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8 4

Oct/Nov 2006 , Question #4 , qp_4

For Examiner’s Use

8

A rocket is launched from the surface of the Earth. Fig. 4.1 gives data for the speed of the rocket at two heights above the Earth’s surface, after the rocket engine has been switched off. speed / m s–1

height / m h1 = 19.9

106

v1 = 5370

h2 = 22.7

106

v2 = 5090 Fig. 4.1

The Earth may be assumed to be a uniform sphere of radius R = 6.38 106 m, with its mass M concentrated at its centre. The rocket, after the engine has been switched off, has mass m. (a) Write down an expression in terms of (i)

G, M, m, h1, h2 and R for the change in gravitational potential energy of the rocket, .............................................................................................................................. [1]

(ii)

m, v1and v2 for the change in kinetic energy of the rocket. .............................................................................................................................. [1]

(b) Using the expressions in (a), determine a value for the mass M of the Earth.

M = ………………………… kg [3]

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4 Section A Answer all the questions in the spaces provided.

9

May/June 2007 , Question #1 , qp_4 1

(a) Explain what is meant by a gravitational field. .......................................................................................................................................... ..................................................................................................................................... [1] (b) A spherical planet has mass M and radius R. The planet may be considered to have all its mass concentrated at its centre. A rocket is launched from the surface of the planet such that the rocket moves radially away from the planet. The rocket engines are stopped when the rocket is at a height R above the surface of the planet, as shown in Fig. 1.1.

2R

R

planet

R

Fig. 1.1 The mass of the rocket, after its engines have been stopped, is m. (i)

Show that, for the rocket to travel from a height R to a height 2R above the planet’s surface, the change ΔEP in the magnitude of the gravitational potential energy of the rocket is given by the expression ΔEP =

GMm . 6R

[2]

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5 (ii)

During the ascent from a height R to a height 2R, the speed of the rocket changes from 7600 m s–1 to 7320 m s–1. Show that, in SI units, the change ΔEK in the kinetic energy of the rocket is given by the expression

For Examiner’s Use

ΔEK = (2.09 × 106)m.

[1] (c) The planet has a radius of 3.40 × 106 m. (i)

Use the expressions in (b) to determine a value for the mass M of the planet.

M = …………………………… kg [2] (ii)

State one assumption made in the determination in (i). .................................................................................................................................. ............................................................................................................................. [1]

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4 Section A

10 1

For Examiner’s Use

Answer all the questions in the spaces provided. Oct/Nov 2008 , Question #1 , qp_4 A spherical planet has mass M and radius R. The planet may be assumed to be isolated in space and to have its mass concentrated at its centre. The planet spins on its axis with angular speed , as illustrated in Fig. 1.1.

mass m

R equator of planet pole of planet Fig. 1.1 A small object of mass m rests on the equator of the planet. The surface of the planet exerts a normal reaction force on the mass. (a) State formulae, in terms of M, m, R and , for (i)

the gravitational force between the planet and the object, .............................................................................................................................. [1]

(ii)

the centripetal force required for circular motion of the small mass, .............................................................................................................................. [1]

(iii)

the normal reaction exerted by the planet on the mass. .............................................................................................................................. [1]

(b) (i)

Explain why the normal reaction on the mass will have different values at the equator and at the poles. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

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5 (ii)

The radius of the planet is 6.4 106 m. It completes one revolution in 8.6 Calculate the magnitude of the centripetal acceleration at

104 s.

For Examiner’s Use

1. the equator,

acceleration = .........................................m s–2 [2] 2. one of the poles.

acceleration = .........................................m s–2 [1] (c) Suggest two factors that could, in the case of a real planet, cause variations in the acceleration of free fall at its surface. 1. ...................................................................................................................................... .......................................................................................................................................... 2. ...................................................................................................................................... .......................................................................................................................................... [2]

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4 Section A Answer all the questions in the spaces provided.

11

May/June 2009 , Question #1 , qp_4 1

For Examiner’s Use

(a) Define gravitational field strength. .......................................................................................................................................... .................................................................................................................................... [1] (b) A spherical planet has diameter 1.2 × 104 km. The gravitational field strength at the surface of the planet is 8.6 N kg–1. The planet may be assumed to be isolated in space and to have its mass concentrated at its centre. Calculate the mass of the planet.

mass = .......................................... kg [3] (c) The gravitational potential at a point X above the surface of the planet in (b) is – 5.3 × 107 J kg–1. For point Y above the surface of the planet, the gravitational potential is – 6.8 × 107 J kg–1. (i)

State, with a reason, whether point X or point Y is nearer to the planet. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................ [2]

(ii)

A rock falls radially from rest towards the planet from one point to the other. Calculate the final speed of the rock.

speed = ...................................... m s–1 [2] © UCLES 2009

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4 Section A

12 1

For Examiner’s Use

Answer the questions in the spaces provided. Oct/Nov 2009 , Question #1 all , qp_41 (a) State Newton’s law of gravitation. .......................................................................................................................................... .......................................................................................................................................... .................................................................................................................................... [2] (b) The Earth may be considered to be a uniform sphere of radius R equal to 6.4 × 106 m. A satellite is in a geostationary orbit. (i)

Describe what is meant by a geostationary orbit. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................ [3]

(ii)

Show that the radius x of the geostationary orbit is given by the expression gR 2 = x 3ω 2 where g is the acceleration of free fall at the Earth’s surface and ω is the angular speed of the satellite about the centre of the Earth.

[3] (iii)

Determine the radius x of the geostationary orbit.

radius = ........................................... m [3] © UCLES 2009

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4 Section A

13 1

For Examiner’s Use

Answer all the questions in the spaces provided.

May/June 2010 , Question #1 , qp_42

(a) Define gravitational potential at a point. .......................................................................................................................................... .......................................................................................................................................... .................................................................................................................................... [2] (b) The Earth may be considered to be an isolated sphere of radius R with its mass concentrated at its centre. The variation of the gravitational potential φ with distance x from the centre of the Earth is shown in Fig. 1.1. distance x 0

0

R

2R

3R

4R

5R

–2.0 / 107 J kg–1 –4.0

–6.0

–8.0 Fig. 1.1 The radius R of the Earth is 6.4 × 106 m. (i)

By considering the gravitational potential at the Earth’s surface, determine a value for the mass of the Earth.

mass = ......................................... kg [3] © UCLES 2010

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5 (ii)

A meteorite is at rest at infinity. The meteorite travels from infinity towards the Earth.

For Examiner’s Use

Calculate the speed of the meteorite when it is at a distance of 2R above the Earth’s surface. Explain your working.

speed = ..................................... m s–1 [4] (iii)

In practice, the Earth is not an isolated sphere because it is orbited by the Moon, as illustrated in Fig. 1.2. initial path of meteorite Moon Earth

Fig. 1.2 (not to scale) The initial path of the meteorite is also shown. Suggest two changes to the motion of the meteorite caused by the Moon. 1. .............................................................................................................................. .................................................................................................................................. 2. .............................................................................................................................. .................................................................................................................................. [2]

37

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4 Section A

14 1

For Examiner’s Use

Answer all the questions in the spaces provided. Oct/Nov 2011 , Question #1 , qp_41 (a) A moon is in a circular orbit of radius r about a planet. The angular speed of the moon in its orbit is ω. The planet and its moon may be considered to be point masses that are isolated in space. Show that r and ω are related by the expression r 3ω 2 = constant. Explain your working.

[3] (b) Phobos and Deimos are moons that are in circular orbits about the planet Mars. Data for Phobos and Deimos are shown in Fig. 1.1.

moon

radius of orbit /m

Phobos Deimos

9.39 × 106 1.99 × 107

period of rotation about Mars / hours 7.65

Fig. 1.1

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5 (i)

Use data from Fig. 1.1 to determine

For Examiner’s Use

1. the mass of Mars,

mass = ............................................ kg [3] 2. the period of Deimos in its orbit about Mars.

period = ...................................... hours [3] (ii)

The period of rotation of Mars about its axis is 24.6 hours. Deimos is in an equatorial orbit, orbiting in the same direction as the spin of Mars about its axis. Use your answer in (i) to comment on the orbit of Deimos. .................................................................................................................................. .............................................................................................................................. [1]

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4 Section A

For Examiner’s Use

Oct/Nov 2012 , Question #1 , qp_41 15 Oct/Nov 2012 , Question #1 all , qp_41 Answer the questions in the spaces provided. Q16-

1

(a) State Newton’s law of gravitation. .......................................................................................................................................... .......................................................................................................................................... ...................................................................................................................................... [2] (b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M. For this planet, the product GM is 4.00 × 1014 N m2 kg–1, where G is the gravitational constant. The planet may be assumed to be isolated in space. (i)

By considering the gravitational force on the satellite and the centripetal force, show that the kinetic energy EK of the satellite is given by the expression EK =

GMm . 2r

[2] (ii)

The satellite has mass 620 kg and is initially in a circular orbit of radius 7.34 × 106 m, as illustrated in Fig. 1.1.

initial orbit 7.34 × 106 m

7.30 × 106 m new orbit

Fig. 1.1 (not to scale) © UCLES 2012

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5 Resistive forces cause the satellite to move into a new orbit of radius 7.30 × 106 m. Determine, for the satellite, the change in 1.

For Examiner’s Use

kinetic energy,

change in kinetic energy = ............................................. J [2] 2.

gravitational potential energy.

change in potential energy = ............................................. J [2] (iii)

Use your answers in (ii) to explain whether the linear speed of the satellite increases, decreases or remains unchanged when the radius of the orbit decreases. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

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4 Section A

16 Q18-1

For Examiner’s Use

Oct/Nov 2013 , Question #1 , qp_43 Answer the questions in the spaces provided. May/June 2013 , Question #1all , qp_42

(a) Explain what is meant by a geostationary orbit. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [3] (b) A satellite of mass m is in a circular orbit about a planet. The mass M of the planet may be considered to be concentrated at its centre. Show that the radius R of the orbit of the satellite is given by the expression R3 =

!

GMT 2 4π2

"

where T is the period of the orbit of the satellite and G is the gravitational constant. Explain your working.

[4] (c) The Earth has mass 6.0 × 1024 kg. Use the expression given in (b) to determine the radius of the geostationary orbit about the Earth.

radius = ............................................. m [3]

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4 Section A May/June 2014 , Question #1 , qp_42 May/June 2014 , Question #1 , qp_42

17

Answer all the questions in the spaces provided.

Q19-1

The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet. (a) A stone, travelling at speed v, is in a circular orbit of radius r about the planet, as illustrated in Fig. 1.1. stone

v planet

r

Fig. 1.1 Show that the speed v is given by the expression v =

! GM r "

where G is the gravitational constant. Explain your working.

[2]

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5 (b) A second stone, initially at rest at infinity, travels towards the planet, as illustrated in Fig. 1.2. stone V0

planet x

Fig. 1.2 (not to scale) The stone does not hit the surface of the planet. (i)

Determine, in terms of the gravitational constant G and the mass M of the planet, the speed V0 of the stone at a distance x from the centre of the planet. Explain your working. You may assume that the gravitational attraction on the stone is due only to the planet.

[3] (ii)

Use your answer in (i) and the expression in (a) to explain whether this stone could enter a circular orbit about the planet. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2]

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4

Oct/Nov 2014 , Question #1 , qp_41

18 Q21-

Section A

Oct/Nov 2014 , Question #1 all , qp_41 Answer the questions in the spaces provided. 1

An isolated spherical planet has a diameter of 6.8 × 106 m. Its mass of 6.4 × 1023 kg may be assumed to be a point mass at the centre of the planet. (a) Show that the gravitational field strength at the surface of the planet is 3.7 N kg−1.

[2] (b) A stone of mass 2.4 kg is raised from the surface of the planet through a vertical height of 1800 m. Use the value of field strength given in (a) to determine the change in gravitational potential energy of the stone. Explain your working.

change in energy = ..................................................... J [3] (c) A rock, initially at rest at infinity, moves towards the planet. At point P, its height above the surface of the planet is 3.5 D, where D is the diameter of the planet, as shown in Fig. 1.1. D

3.5 D

P

planet

path of rock

Fig. 1.1

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5 Calculate the speed of the rock at point P, assuming that the change in gravitational potential energy is all transferred to kinetic energy.

speed = ............................................... m s−1 [4]

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Oct/Nov 2014 , Question #2 , qp_43

19

Oct/Nov 2014 , Question #2 , qp_43 8

Q22- 2

(a) On the axes of Fig. 2.1, sketch the variation with distance from a point mass of the gravitational field strength due to the mass.

gravitational field strength

0

0

distance Fig. 2.1 [2]

(b) On the axes of Fig. 2.2, sketch the variation with speed of the magnitude of the force on a charged particle moving at right-angles to a uniform magnetic field.

force

0

0

speed Fig. 2.2 [2]

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9 (c) On the axes of Fig. 2.3, sketch the variation with time of the power dissipated in a resistor by a sinusoidal alternating current during two cycles of the current.

power

0

0

time Fig. 2.3 [3]

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4

May/June 2016 , Question #1 , qp_41 Answer#1 all, the questions in the spaces provided. May/June 2016 , Question qp_41

20

Q23- 1

(a) By reference to the definition of gravitational potential, explain why gravitational potential is a negative quantity. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) Two stars A and B have their surfaces separated by a distance of 1.4 × 1012 m, as illustrated in Fig. 1.1. 1.4 1012 m star A

star B P x Fig. 1.1

Point P lies on the line joining the centres of the two stars. The distance x of point P from the surface of star A may be varied. The variation with distance x of the gravitational potential φ at point P is shown in Fig. 1.2. x / 1012 m –2

0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

–4 –6 –8 / 108 J kg–1 –10 –12 –14 –16 Fig. 1.2

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5 A rock of mass 180 kg moves along the line joining the centres of the two stars, from star A towards star B. (i)

Use data from Fig. 1.2 to calculate the change in kinetic energy of the rock when it moves from the point where x = 0.1 × 1012 m to the point where x = 1.2 × 1012 m. State whether this change is an increase or a decrease.

change = ............................................................. J .................................................................................. [3] (ii)

At a point where x = 0.1 × 1012 m, the speed of the rock is v. Determine the minimum speed v such that the rock reaches the point where x = 1.2 × 1012 m.

minimum speed = ............................................... m s−1 [3] [Total: 8]

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4 Section A

For Examiner’s Use

Answer all the questions in the spaces provided. 21 Oct/Nov 2011 , Question #1 , qp_43 1

The planet Mars may be considered to be an isolated sphere of diameter 6.79 × 106 m with its mass of 6.42 × 1023 kg concentrated at its centre. A rock of mass 1.40 kg rests on the surface of Mars. For this rock, (a) (i)

determine its weight,

weight = ............................................ N [3] (ii)

show that its gravitational potential energy is –1.77 × 107 J.

[2] (b) Use the information in (a)(ii) to determine the speed at which the rock must leave the surface of Mars so that it will escape the gravitational attraction of the planet.

speed = ....................................... m s–1 [3]

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ANSWERS ANSWERS ANSWERS ANSWERS

Q1Q11 Q1-

Q1-

Q22Q2-

Q2Q2Page 2 Page 2

3
 Q31 (a)

Mark Scheme A/AS LEVEL EXAMINATIONS - JUNE 2003 Mark Scheme A/AS LEVEL EXAMINATIONS - JUNE 2003

Syllabus 9702 Syllabus 9702

Paper 04 Paper 04

work done in bringing/moving unit mass ......................................M1 Scheme Syllabus from infinity to the Mark point.......................................................... ...... A1 Paper[2] A/AS LEVEL EXAMINATIONS JUNE 2003 9702 Q3- 1 (a) (use 1 kgininbringing/moving the definition – max 1/2) ......................................M1 04 workof done unit mass Page 2 Mark Scheme Syllabus Paper from infinity to the point.......................................................... ...... A1 [2] A/AS LEVEL EXAMINATIONS - JUNE 2003 9702 04 (b) potential as–unit being zero........................ ............. B1 (use done of 1 at kg in the defined definition maxmass 1/2) Q3- 1 (a) work ininfinity bringing/moving ......................................M1 forces are always attractive.......................................................... B1 from infinity to the point.......................................................... ...... A1 [2] so work got out in moving to point...................... .......................... B1 [3] potential at infinity defined as being zero........................ ............. B1 Q3-(b) 1 (a) bringing/moving unit mass ......................................M1 (use ofwork 1 kgdone in theindefinition – max 1/2) (max is at infinity – allow 1/3) forcespotential are always attractive.......................................................... B1 from infinity to the point.......................................................... ...... A1 [2] so work got out in defined moving to point...................... B1 [3] (use 1 kg in the definition max 1/2) .......................... (b) potential at of infinity as being–zero........................ ............. B1 (c) (i) φ = -GM/R (max potential is at infinity – allow 1/3) forces are always attractive.......................................................... B1 -11 6 -1 change = 6.67 x 10 x 6.0 x point...................... 1024 as x({6.4 x 10 } - .......................... {1.94 x 107}-1) .....C2 (b) potential at infinity defined being zero........................ ............. B1[3] so work got out in moving to B1 7 -1 J kg (ignore sign) .........................................A1 change = 4.19 x 10 (c) (i) φ = -GM/R forces are always attractive.......................................................... B1 (max potential is at -11 infinity – allow 1/3) change 6.67 got x 10out xin6.0 x 1024tox({6.4 x 106}-1- {1.94 x.......................... 107}-1) .....C2 B1 so=work moving point...................... [3] 7 -1 2 J kg (ignore sign) .........................................A1 change = 4.19 x 10 ½mv = m φ ................................................................................ C1 (max potential is at infinity – allow 1/3) (ii) (c) (i) φ = -GM/R 7 -11 = 8.38 v2 = 2 x=4.19 change 6.67x x10 10 x 6.0xx10 10724 x({6.4 x 106}-1- {1.94 x 107}-1) .....C2 2 -1 7 -1 (c) φ===mm -GM/R C1 [5] v½mv = 9150 sφ ................................................................................ A1 (ii) (i) .........................................A1 change 4.19 x.............................................................................. 10 J kg -11(ignore sign) 7 7 x 1024 x({6.4 x 106}-1- {1.94 x 107}-1) .....C2 6.67 10 x x106.0 x 4.19 x=10 = x8.38 v2 = 2change -1= 4.19 x 107 J kg-1 (ignore sign) .........................................A1 2 change [5] (d) (ii) acceleration not constant.......................................................... C1 B1 [1] v = 9150 .............................................................................. A1 ½mv = mmφsis ................................................................................ 2 7 7 v = 2 x 4.19 2 x 10 = 8.38 x 10 ½mv ................................................................................ 2 (a) (d) notφconstant.......................................................... B1 C1[5] [1] (ii) vacceleration = 9150 m s=-1ism .............................................................................. A1 v2 = 2 x 4.19 x 107 = 8.38 x 107 -1 2 (d) (a) [5] v = 9150 m sconstant.......................................................... .............................................................................. A1[1] acceleration is not (-1 for each error or omission) ........................................ B1 B2 [2] Page 2

(d) 2 (a) (b) CEDAR COLLEGE 2 (a) (b) CEDAR COLLEGE (b) CEDAR COLLEGE (b) (c) CEDAR COLLEGE CEDAR COLLEGE 3 (a) (c)

acceleration is not constant.......................................................... B1 [1] for each error or omission) B2 heat lost (-1 by liquid gold = 0.95m x 129 x ........................................ ∆T.................................. C1 [2] heat gained (silver) = 0.05m x 235 x (1340 – 300) + 0.05m x 105 000..C1, C1 52 heat lost(-1 by=for liquid gold = 0.95m x 129 x........................................ ∆T..................................B2 C1 error or omission) 122.5m∆T 17each 470m [2] heat gained (silver) = 0.05m x 235 x (1340 – 300) + 0.05m x 105 000..C1, C1 52 ∆T = 143 K.......................................................................................C1 for each errorxor129 omission) ........................................ B2 [2] 122.5m∆T 17(-1 470m heat lost by=liquid gold = 0.95m x ∆T.................................. C1 temperature = 143 + 1340 = 1483 K................................................A1 52 [5] heat (silver) = 0.05m x 235 x (1340 – 300) + 0.05m x 105 000..C1, C1 ∆T =gained 143 K.......................................................................................C1 heat lost by liquid gold = 0.95m x 129 x ∆T.................................. C1 122.5m∆T = 17 470m e.g. thermocouple/resistance thermometer .................................. B1 [1] 52 [5] temperature = 143 + 1340 = 1483 K................................................A1 heat gained (silver) = 0.05m x 235 x (1340 – 300) + 0.05m x 105 000..C1, C1 52 ∆T = 143 K.......................................................................................C1 122.5m∆T = 17 470m fe.g. natural frequency (system) .................................. ................................. B1 thermocouple/resistance thermometer B1 [1] 0 is at [5] temperature = 143 + 1340of= spring 1483 K................................................A1 = 143 K.......................................................................................C1 this is ∆T at the driver frequency ....................................................... B1 [2]

Page 2

4

1 (a)

Mark Scheme A/AS LEVEL EXAMINATIONS - NOVEMBER 2003

(i)

1 (b) (a) (i) Page 2

(b)

(ii)1

Q42

1 (a) (ii)2 (i) (a)

(b) (b) (c)

(ii) (i) (i)

(ii) (ii)

2 (a) 5

Q5- 3

(a)

(c) (b) (b)

(i) (iii) (ii)

Paper 02

acceleration (allow a definition of acceleration) ............................ B1

Page 2

(ii)

Syllabus 9702

Mark Scheme

Syllabus

the velocity is LEVEL decreasing or force/acceleration A/AS EXAMINATIONS - JUNE 2004is in negative 9702 direction – accept ‘body is decelerating’/‘slowing down’ ............... B1 charge is quantised/enabled electron charge to be e.g. separation of dots becomes constant/does notmeasured continue to increase (must make a reference B1 Mark Scheme to-19the diagram) ........................ Syllabus (approximately) n x (1.6 x- NOVEMBER 10 C) all are A/AS LEVEL-19EXAMINATIONS 2003 9702 C (allow 2 sig. fig. only so e = 1.6=x132 10 cm........................................................................ distance B1 summing charges and dividing ten, without explanation scores 1/2 Total lines.................................................................................... B1 atradial constant speed, distance travelled in 0.1 s = 25 cm (value of the) square mean pointing inwards ........................................................................... B1 (allow ± 1 cm)............................................................................... C1 of the speeds (velocities) of the atoms/particles/molecules distance = 132 + (4 x 25) no difference closer near surface of smaller sphere ......A1 B1 = 232OR cmlines ...................................................................... 1

< c 22> 2 .............................................................................. C1 p F=G =ρGMm/R s = ut + ½at 3 -11 24 x 5.98 10m )/(6380 x 103)2 X 10 -2 2 ==½(6.67 1.6 t25(allow g =xx10 > = 3 xx 9.8 2 x x10 /2.4 = 2.5 105 s ............................................ C1 photographs scores 2 marks = 2 1.0 x 10 or(‘bald’ answer increases by factor of 4 only) ........... A1 new F2C = =∝2π/T C1 6 of body’s resistance/inertia 5 measure to changes in Tmass: = {(1.0 2x 10 ) / (2.5 x 3 10 )} x 300 4 2 FC = (4π x 6380 x 10 )/8.64 x 10 ) velocity/motion ............................................................................. B1 = 1200 K = 0.0337 N............................................................................... A1 weight: effect of gravitational field on mass or force of gravity ..... B1 Total 2 any further comment e.g. mass constant, weight varies/ (force) GM 1M2/(R1 + R2) FG - F= 9.77 N............................................................................ A1 C==mg/scalar 2 2 weight and ..................................................... B1 (force) = M1R1 ω or M2Rvector 2ω

Paper 04

[2] B1

[1]

Paper

M104 A1 [2] [3] M1 A1

[2] [3] [4]

C1 C1 A1 [3] C1 [3] C1 A1

[3] [8]

B1 [6] B1 [3][2]

because acceleration (of free fall) is (resultant) force per unit e.g.= where gravitational field strength changes 2π/(1.26 x 108) or 2π/T ω mass ....................................................................................... B1 C1

-8 surrounding (change) fluid body…. 1 each, max 2 ................... B2 rad s-1 = 4.99 in x 10 acceleration = 9.77 m s-2 .............................................................. B1 -8 allow 2 s.f.: 1.59π x 10 scores 1/2 3 (a) force x perpendicular distance .....................................................M1 2 (a) (i) a,ω and x identified ………(-1 each error or omission) ................. B2 (of the force) from taking the pivot .......................................................... (c) (i) reference to either moments (about C) or same (centripetal)A1 (ii) force (-)ve because a and x in opposite directions (b) noRresultant (in any 2 direction) 2 .............................................. B1 = directed M2R2 force or M = M2Rposition/centre................................ M OR towards B1 1 1a 1R1 ω mean 2ω no resultant moment (about any point)......................................... B1 hence M1/M2 = R2/R1 11 (b) (i) springs are + x) and k(e x 3.2 x 1011 mk(e = 2.4 x 10 m – x) .................................... C1 (ii) Rforces 2 = 3/4 in (c) (i) correct direction B1 11 in both................................................................ 10 = k(e + x) – k(e – x) ......................................................M1 = (3.2 x 10 ) – R = 8.0 x 10 m (allow vice versa) Rresultant 1 2 2kx ............................................................................ if values are=both wrong but have ratio of four to three, then allow A0 (ii)1 moment = 150 x 0.3 = 45 N m (1 sig. fig. -1) ................................ A1 1/2 (ii) F = ma ....................................................................................... B1 (ii)2 torque = 45 .................................................................................... N 2m i.e. same 2 is (i) .................................................... A1 a -2kx/m A0 (d) (i) M2 == {(R 1 + R2) x R1 x ω } I G (any subject for equation) 11 2 10 -8 2 -11 (-)ve sign explained...................................................................... B1 ) x 8.0 x 10 x (4.99 x 10 ) /(6.67 x 10 ) = (3.2 x 10 (ii)3 45 = 0.12 x T................................................................................ C1 29 = 3.06 x 10 kg Tω=2 375 N ..................................................................................... A1 (iii) = 2k/m ..................................................................................... C1 (ii) less massive (only award this mark if reasonable attempt at (i)) 2 29 x 120)/0.90 .................................................................. C1 (2πf) = (2 (9.17 x 10 kg for more massive star) 4 (a) (i)1 amplitude 0.4(0) mm ................................................................. A1 f = 2.6 Hz=..................................................................................... A1 Total 4 (a) e.g. amplitude is not constant or wave is damped wavelength = 7.5 x 10-2bymattractive forces (c) (i)2 atomdo held position notinallow 'displacement constant' (1atom sig. oscillates, fig. -1 unless already penalised) ........................................ A1 should be (-)cos, (not sin) Page 1 Scheme Syllabus not just two forces ORMark 3D not 1D A LEVEL JUNE 2005 9702 C1 (i)3 period not = 0.225 ms ........................................................................ (b) T force = 0.60 s proportional to x frequency = 1/T = points, 4400 any two = relevant 1 each, max 2 ........................................A1 B2 -1 Hz........................................................... ω = 2π/T (i) angular speed =10.5 2π/Trad s (allow 10.4 → 10.6) C1 Q6- 1 (a) 
 = 2π/(3.2 × 107) 3 (a) (i)4 C1 vpV/T = fλ= constant............................................................................ (c) same period= 1.96 -2 -7 rad s-1 × 10 A1 30 x................................................................... 300)/(1.1 x 105 x 540).................................C1 C1 T ==4400 (6.5 x 7.5 106 x 10 displacement-1always less 985mKs.................................................................................... A1 ............................................................................... A1 ==330 2 reducing appropriately 2 amplitude (ii) force = mrω or force = mv /r and v = rω C1 (if uses °C,rdallow 1/3 marks for clear formula) nd -7 2 and 3 24 marks, the first quarter period for=26.0 × 10 × 1.5 × ignore 1011 × (1.96 × 10 ) Total 22 CEDAR 3 COLLEGE (b) (i) = 3.46 10 N A1 ∆U = q +× w © University of Cambridge Examinations Syndicate 2003 symbols identified correctlyLocal ..........................................................M1 (b) (i) gravitation/gravity/gravitational field (strength) B1 directions correct.......................................................................... A1

A1 [2] [2] [2]

[2] B1 B1 [3] [2] A0 [2] A1 [1] A1 [2] [2]

C1 C1 [2] A1 [4] B1 [4]

[3] [12] B1 B1 [2] Paper 4 C1 A1 [2] [2] B1 [2] M1 [6] [3] A1 [3]

[2]53[7] [1] [2]

GCE A/AS LEVEL or - OCT/NOV 2006 9702 B1 (a) e.g. amplitude is not constant wave is damped (c)Page 2 atomdo held in position by attractive forces Mark Scheme Syllabus not allow 'displacement constant' atomshould oscillates, GCE A/AS(not LEVEL - OCT/NOV 2006 9702 B1 [2] be (-)cos, sin) Scheme Syllabus Paper 1 Page (a) 1either ratio work done to mass/charge notofjust two forces ORMark 3D not 1D A LEVEL -from JUNE 2005 9702 4 or work force done moving unit mass/charge infinity not proportional to x (b) or bothThave = 0.60 s potential C1 zero at infinity of two workrelevant done to mass/charge any points, 1 each, 2 ........................................ B2 A1 [2] 6 1 (a) either ratio ω = 2π/T = =10.5 s-1 (allow 10.4max → 10.6) [2] (i)or work angular 2π/Trad C1 Q6- 1 (a) 
 donespeed moving unit mass/charge from infinity 7 (b) gravitational forces are (always×attractive) = 2π/(3.2 10 ) both have potential at infinity 3 (a) or pV/Tzero =can constant............................................................................ C1 electricsame forces be attractive or -7 repulsive -1 (c) period B1 6 5 = 1.96 × 10 rad s A1 x 30 x 300)/(1.1 x 10 x 540)................................. C1 [2] T = (6.5 x 10 for gravitational, work got out as masses come together displacement M1 (b) gravitational forces arealways (alwaysless attractive) = 985 K .................................................................................... A1 [3] /mass moves from infinity 2 reducing 2 repulsive electric forces can attractive amplitude appropriately (ii) = uses mrω orbeforce =1/3 mvor /rifand v sign, =clear rω work C1 A1 [3] forforce electric, work done on charges same got out if opposite sign as charges (if °C, allow marks for formula) nd rd for gravitational, work out as 11 masses come -7 together 3 24got marks, the first quarter period come for together =26.0and × 10 × 1.5 × ignore 10 ×/mass (1.96 × 10 )2 infinity moves from Total 22 3 (b) for (i) electric, = 3.46 ×w 10 Ncharges if same sign, work got out if opposite sign as A1 [2] [7] ∆U =q+ done charges 2 (a) (i) idea work of heat lost on (by oil) = heat gained (by thermometer) identified come together 32symbols x 1.4 x (54 – t) = 12 correctly x 0.18 x (t..........................................................M1 – 19) (b) (i) gravitation/gravity/gravitational field (strength) B1 A1 [1] [2] t =directions 52.4°C correct..........................................................................

4

04 Paper 04 B1 B1 B1 B1

B1 B1 B1 B1 B1

C1 B1 C1 A1 2 (a) (i) idea of heat lost (by oil) = heat gained (by thermometer) C1 2 (54 3 x2 0.18 x (t – 19) 32 x 1.4 x – t) = 12 C1 (ii) q is zero ....................................................................................... B1 = r =ω0.030 or (=1.6/327) = 0.0049 C1 (ii)(ii)F = either GMm/x ratioor (=GM 1.6/54) A1 22 -11 24 11 2 t = 52.4°C A1 w is positive ∆U =×wMand UInternational B1 3.46 × 10 = University (6.67OR × 10 × 6.0 ×increases 10 )/(1.5.................................... × 10 ) C1 © of Cambridge Examinations 2004 Syllabus Page 2 Mark Scheme Paper 30 (b) thermistor (allow energy ‘resistance B1 ∆Uthermometer is× rise kinetic of thermometer’) atoms ............................................M1 M =either 1.95 10(= in kg A1 [3] GCE A/AS LEVEL - OCT/NOV 2006 9702 04 (ii) ratio 1.6/54) =capacity 0.030 or (=1.6/327) = 0.0049 A1 because small mass/thermal B1 [4] and mean kinetic energy ∝ T ....................................................... A1

2 point Mark Scheme Syllabus (allow one of the last‘resistance two marks if states ‘U increases so T rises’) (b) thermistor thermometer (allow thermometer’) (c)Page boiling temperature is constant GCE A/AS LEVEL OCT/NOV 2006 9702 any two named gas laws M1 obeys the law pV/T = constant or because small mass/thermal capacity either of work done to mass/charge furtherratio comment or work unit mass/charge from infinity e.g. ofmoving bulb would affect of boiling atheating alldone values of p, V and T only rate A1 [2] (c) boiling point temperature is at constant or both have zero potential infinity correct assumptions of kinetic theory of ideal gas (B1) or two 2© University of Cambridge Local Examinations Syndicate 2003 Q7-7 13 (a) of work done to mass/charge further comment (a) either use ofratio a = – ω x clear third correct assumption (B1) 2 (b) or gravitational forces attractive) work moving mass/charge from infinity e.g. heating of bulb are would affect only rate of boiling either ωdone = √(2k/m) orunit ω(always = (2k/m) electric forces canpotential be attractive or repulsive or both have zero at infinity = 2 πf ω (b) (i)use speed B1 [1] formean gravitational, work got out as masses come together 3 (a) of a = square – ω 2x clear f = (1/2 π)√(2 x 300)/0.240) 2 (b) gravitational forces are attractive) /mass moves from infinity either ω = √(2k/m) or ω(always = (2k/m) 2 =mean 7.96 ≈kinetic 8 Hz energy > M1 (ii) = ½m = 3/2 kT A0 [2] = 7.96 ≈ 8ofHz 2 (a) for (i) idea heat lost on (bycharges oil) = heat gained (by work thermometer) done if same sign, got out if opposite sign as charges (ii)electric, 8 Hzwork 32 x 1.4 x -27 (54 – t) = 12 x 0.18 x (t – 19) together (b) (i) ½ × resonance C1 (c) (i)come × 10 × (1.1 ×104)2 = 3/2 × 1.38 × 10-23 ×T t 6.6 = 52.4°C (c) (increase amount 4 of) damping T = 1.9 × 10 K A1 [2] 2 (a) (i) idea of heat lostm(by oil) = heat gained (by thermometer) without altering (k or) …(some indirect reference is acceptable) (ii) 8either Hz (ii) ratio (= 1.6/54) = 0.030 or (=1.6/327) = 0.0049 32 suggestion x 1.4 x (54 – t) = 12 x 0.18 x (t – 19) sensible (ii) Nott =all52.4°C atoms of) have same speed/kinetic energy B1 [1] (c) (increase amount damping –1 –1 8 4 (b) thermistor thermometer (allow ‘resistance thermometer’) Q8(a) without (i) GMm {(R + h ) – (R + h ) } 1 2 indirect reference is acceptable) altering (k or) 2 mass/thermal 2 m …(some because small capacity ½m {vratio (ii) either 1 – v(= 2 }1.6/54) = 0.030 or (=1.6/327) = 0.0049 sensible suggestion 3 (a) (thermal) energy/heat required to convert unit mass/1 kg of –11 6 –1 (c) boiling point temperature is constant –1 (allow (b) thermistor 2M x 6.67 x {(R 10 {(26.28 x 10 ) )–1–} (29.08 thermometer’) x 106)–1} = 53702 – 50902 thermometer ‘resistance Q8- 4 (b) (a) (i) GMm + h ) – (R + h solid to liquid M1 1 2 –19 6 further comment 2 mass/thermal M x 4.888 x 10 2.929 x 10 because small capacity ½m {v –bulb v2=2}would 1of24 with no change in temperature/at melting point A1 [2] e.g. heating affect only rate of boiling M = 6.00 x 10 kg

Q7- 21 (a) (a)

[1

[4

[3

[1

[2

Paper B1 M1 04 B1 A1 [2 M1 [1 B1 C1 B1 A1 B1 B1 B1 C1 C1 B1 B1 B1 B1 A0 [4 B1 C1 B1 [4 B1 [1 B1 B1 A0 C1 B1 [1 C1 B1 B1 A1 [3 B1 C1 B1 B1 [1 A1 C1 B1 [3 A1 B1 B1 B1 B1 [2 B1 B1 [2 A1 B1 M1 B1 B1 B1 C1 B1 B1 [2 A1 A1 [3

–11 is dimensionally 6 –1 unsound, then 6 –1 (c) boiling point temperature constant M1 (If equation in (a) 0/3 marks2in (b), if 2dimensionally sound but xof6.67 10 {(26.28 is x 10 ) – (29.08 x 10 -3 ) } = 53703 – 5090 3 (b) (a) (i)2M useenergy a = x–required ω 2–19 x clear C1B1 C1 (b) to warm 6ice = 24 × 10 × 2.1 × 10 × 15 (= 756 J) incorrect, treat as e.c.f.) further comment 2 x 10 M x 4.888 10 = or 2.929 C1 either ω =x√(2k/m) = (2k/m) B1 e.g. heating of24bulb would affect of 24 boiling A1 required toωmelt iceonly at 0rate °C = × 10-3 × 330 × 103 (= 7920 J) C1 Mω =energy 6.00 x 10 kg A1 = 2 (induced) πf C1 5 (a) (i) total e.m.f proportional/equal to rate of change of flux (linkage) B1 energy J A1sound [3]but (If equation in 2(a)=is8700 dimensionally unsound, then 0/3 marks in (b), if dimensionally f = of (1/2 xx 300)/0.240) B1 3 (a) use a(allow =π)√(2 – ω‘induced clear voltage, induced p.d.) C1 incorrect, treat as e.c.f.) 2 flux is cust as the disc moves M1 = 7.96 ≈ 8 Hz A0 -3 3 either ω = √(2k/m) or ω = (2k/m) B1 C1 (ii) energy lost by warm water = 200 × 10 × 4.2 × 10 × (28 - T) hence inducing an e.m.f A0 = 2 πf ω C1 5 (a) (i) (induced) e.m.f proportional/equal to rate of change of flux (linkage) Page(i)2 200resonance SyllabusC1 Paper B1B1 × 4.2 × (28 - T) = 24Mark × 4.2Scheme × T + 8676 (b) (allow ‘induced voltage, induced p.d.) f (ii) = (1/2 π)√(2 x 300)/0.240) B1 LEVELof–cutting May/June 2007 discGCE is notA/AS uniform/rate not same/speed of disc not9702 same (over T =field 16in°C A1 whole [3] 04 flux is cust as the disc moves M1 = 7.96 ≈ 8 Hz A0 (ii)[allow 8 Hz B1 disc) 2 marks if ∆ Te.m.f calculated] hence inducing an Q9- 1 Page so e.m.f.’s different parts of disc (a) (region of different space) a force B1Paper M1 [1]A0 2 [allow Mark Syllabus 2 markswhere if damping (24 axinmass 4.2 x experiences T)Scheme omitted] (b) resonance B1 (c) (i) (increase amount of) B1 lead to eddy currents A0 A/AS LEVEL May/June (ii) field disc is uniform/rate of–-cutting not same/speed same (over whole04 B1 [allow 1inmark fornot 224 x 4.2 xindirect (28 T) = 8676, T - 19 °C]of disc not9702 without altering (kGCE or) m …(some reference is 2007 acceptable) (ii) 8 Hz B1 disc) sensible suggestion B1 (b) eddy currents dissipate thermal energy (b) (i) potential energy = (–)GMm / x in disc C1 Q9- 1 (a) (region so different e.m.f.’s in different parts of disc M1 of space) where a mass experiences a force B1 [1] energy from oscillation of–1disc B1 ∆EPderived =amount GMm/2R – GMm/3R M1 –1 of) damping B1 4 (c) (a) (increase (i) GMm {(R + h ) – (R + h ) } B1 lead to eddy currents A0 1 2 energy of disc2 depends on amplitude of oscillations B1 = GMm/6R A0 [2] 2 m …(some indirect reference is acceptable) without altering B1 ½m {v1 –(kv2or) } B1 suggestion B1 (b) eddy currents dissipate thermal (b) (i)sensible potential energy / x in disc 6 –1 C1 2 = (–)GMm 2 energy –11 6) –1 2 2 E = ½m (7600 – 7320 M1 K6.67 (b) (ii) 2M x x 10 {(26.28 x 10 ) – (29.08 x 10 ) } = 5370 – 5090 B1 energy from oscillation of disc B1 ∆EPderived = GMm/2R – GMm/3R M1 –196)m –1 (2.09 A0 [1] 4 (a) (i) GMm {(R += h12.929 ) –on (R + h6 2)–1} of oscillations B1 © UCLES 2006 M x=4.888 x× 1010 xamplitude 10 C1 energy of disc depends B1 = GMm/6R A0 [2] 224 2 ½mx{v10 v2 } B1 M = 6.00 A1 1 – kg (If equation in6(a) is 2dimensionally unsound, then 0/3 marks in (b), if dimensionally sound but 2 –11 6 –11 6 –1 E = ½m – 7320 M1 K6.67 (c) (ii) (i) ×treat =6 (6.67 × x10 M)/(6 × 3.4x × C1 (b) 2M x2.09 x10 10(7600 {(26.28 106))–1 – (29.08 1010 ) )} = 53702 – 50902 B1 incorrect, as e.c.f.) –19 )m 6 23 = (2.09 × 10 A0 [1] © UCLES 2006 M xM 4.888 x 10× 10= 2.929 x 10 C1 = 6.39 kg A1 [2] 24 54 B1A1 CEDAR x 10 kg 5 COLLEGE (a) M (i)= 6.00 (induced) e.m.f proportional/equal to rate of change of flux (linkage) ‘induced voltage, induced p.d.) (If equation (a) is dimensionally unsound, then 0/3 marks in (b), if dimensionally sound but (ii) e.g.(allow no©in energy dissipated due to friction with atmosphere/air University of Cambridge International Examinations 2005 6 (c) (i)incorrect, 2.09 ×treat 10 = (6.67 ×disc 10–11 M)/(6 × 3.4 × 106) C1 flux is as the moves M1 as e.c.f.) rocket is cust outside atmosphere 23 hence inducing an e.m.f A0 M = 6.39 × 10 kg A1 not influenced by another planet etc. B1 [1][2]

[4 [2 [1

[1

[2

[3

[3 [2

[3

[2

(allow voltage, induced p.d.) f = (1/2 π)√(2‘induced x 300)/0.240) flux≈ is cust as the disc moves = 7.96 8 Hz hence inducing an e.m.f Page(i)2 Mark Scheme Syllabus Paper (b) resonance LEVELof–cutting May/June 2007 (ii) field in discGCE is notA/AS uniform/rate not same/speed of disc not9702 same (over whole04 (ii) 8 Hz disc) Q9-9 1 (a) (region so e.m.f.’sainmass different parts of disc of different space) where experiences a force B1 (c) (increase damping leadamount to eddyof) currents without altering (k or) m …(some indirect reference is acceptable) suggestion (b) eddy currents dissipate thermal energy (b) (i)sensible potential energy = (–)GMm / x in disc C1 energy from oscillation of disc ∆EPderived = GMm/2R – GMm/3R M1 4 (a) (i) GMm {(Rdepends + h1)–1 –on (Ramplitude + h2)–1} of oscillations energy of disc = GMm/6R A0 2 2 ½m {v1 – v2 } 2

M = 6.39 × 1023 kg

B1 [1] C1 A1

A1

(induced) e.m.f proportional/equal to rate of change of flux (linkage) ‘induced voltage, induced (ii) e.g.(allow no energy dissipated due to p.d.) friction with atmosphere/air flux is cust as the disc moves rocket is outside atmosphere inducing e.m.f planet etc. nothence influenced byan another (ii)

B1Paper Mark Scheme Syllabus GCE A/AS LEVEL – October/November 2008 of disc not 9702 field in disc is not uniform/rate of cutting not same/speed same (over whole 04

disc) bonds between molecules are broken/weakened (a) (on melting,) Section A so different e.m.f.’s in different of disc or molecules further apart/are able parts to slide over one another lead to eddy currents energy unchanged so no temperature change 10 1 (a)kinetic Q10(i) F = GMm / R2 potential energy increased/changed energy required (b) eddy currents dissipate thermal energyso in disc energy (ii) F derived = mRω2from oscillation of disc energy of disc depends on amplitude of oscillations

(b) thermal energy/heat required to convert unit mass of solid to liquid 2 (iii)noreaction = GMm / R2at–its mRω (allow e.c.f.) with changeforce in temperature/ normal boiling point

[

[

4 2 (2π / {8.6 1. acceleration 6.4 × 106 ×melts (ii)(ii)more energy (than=calculated) ice × 10 }) –2 = 0.034 m s the accepted value so, (calculated) L is lower than 2. acceleration = 0

(a) field strength = potential gradient (c)correct e.g. ‘radius’ of directions planet varies sign OR discussed density of planet not constant planet spinning 2 2 (b) area isnearby 21.2 cm ± 0.4/ cm planets stars (if outside 0.4 cm2 comments, but within ±10.8 cm2each, , allowmaximum 1 mark) 2) (any±sensible mark 2 –2 1.0 cm represents (1.0 × 10 × 2.5 × 103 =) 25 V potential difference = 530 V 2 (a) (Thermal) energy / heat required to convert unit mass of solid to liquid Page 2 Mark Scheme: Teachers’ version Syllabus at its normal melting point / without any change in temperature GCE A/AS LEVEL – May/June 2009 9702 (c) ½mv2 = qV (reference to 1 kg or to ice → water scores max 1 mark) ½ × 9.1 × 10–31 × v2 = 1.6 × 10–19 × 530 Section A v = 1.37 × 107 ms–1

heat gains from the atmosphere (b) (i) To make (a) force per unit allowance mass (ratiofor idea essential) (d) (i) d = 0 (ii) e.g. constant rate of production of droplets from funnel (ii) decreases then increases of water collected per minute in beaker (b) g acceleration = constant GM / R2 mass 7 2 suggestion, –11 some quantitative analysis (e.g. minimum at 4.0 cm) (any sensible 1 mark) 8.6 × (0.6 × 10 ) = M × 6.67 × 10 24 (any suggestion that acceleration becomes zero or that there is a M = 4.6 × 10 kg scores 0/2) in 5 minutes = 64.7 – ½ × 16.6 = 56.4 g (iii)deceleration mass melted by heater 56.4 × 10–3 × L = 18 CEDAR COLLEGE = 320potential kJ kg–1 decreases as© distance UCLES 2007 (c) (i) L either from planet decreases –1 ,closer scorestomax (Use of m = 64.7, giving L = 278 kJ X kgis or potential zero at infinity and zero1 mark –1 , scores max 2 marks) use of m = 48.1, giving L = 374 kJ kg or potential α –1/r and Y more negative so point Y is closer to planet.

[

B1 B1 B1 B1

M1 A0 [1]

[

B1 M1 A0

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[1] [3] B1 B1[1] B1

B1 M1 A1B1

[2][1]

© UCLES 2006

(c)(b)(i) (i)thermal byexpression water = 0.16 x 100 varies either energy value oflost R in Rω×2 4.2 = or 67.2 kJmRω2 no longer parallel to GMm / R2 / normal to surface 67.2 = 0.205smaller × L as object approaches a pole / is zero at pole becomes L = 328 kJ kg–1

Q11- 1

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B1

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B1 M1 [1] B1 A0 B1 B1 B1 B1 B1 [2] B1

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–11 = ½m – 7320 M1 (b) (ii) 2M E x K6.67 x 10(7600 {(26.28 x 106))–1 – (29.08 x 106)–1} = 53702 – 50902 6 –19 6 (2.09x ×1010 )m A0 © UCLES 2006 M x=4.888 = 2.929 x 10 M = 6.00 x 1024 kg (If equation in6(a) is dimensionally unsound, then 0/3 marks in (b), if dimensionally sound but –11 6 (c) (i)incorrect, 2.09 ×treat 10 as = (6.67 C1 e.c.f.) × 10 M)/(6 × 3.4 × 10 )

5 (a)

B1 M1 A0 A0 B1

C1B1 C1B1 A1

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M1C1 A1A1 A1

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C2 B2 [2] C1 A1 [4] M1 Paper A1 [2] 04 C1 A1 [2]

B1 B1 [1] [1] B1 [1] B1 C1 B1B1 C1 [2][1]

A1 C1 C1 55 A1

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(a) (Thermal) energy / heat required to convert unit mass of solid to liquid Page 2 Mark Scheme: Teachers’ version Syllabus at its normal melting point / without any change in temperature GCE A/AS LEVEL – May/June 2009 9702 (reference to 1 kg or to ice → water scores max 1 mark)

M1 Paper A1 [2] 04

Section A 11 1 Q11-

heat gains from the atmosphere (b) (i) To make (a) force per unit allowance mass (ratiofor idea essential)

B1 B1

[1] [1]

(ii) e.g. constant rate of production of droplets from funnel (b) g = constant GM / R2 mass of water collected per minute in beaker –11 (any 1 10 mark) = M × 6.67 × 8.6 × (0.6sensible × 107 )2 suggestion,

C1 B1 C1

[1]

A1

[3]

24

M = 4.6 × 10 kg (iii) mass melted by heater in 5 minutes = 64.7 – ½ × 16.6 = 56.4 g 56.4 × 10–3 × L = 18 = 320potential kJ kg–1 decreases as distance from planet decreases (c) (i) L either –1 ,closer scorestomax (Use m = 64.7,zero giving L = 278and kJ X kgis or of potential at infinity zero1 mark –1 , scores max 2 marks) use = 48.1, αgiving L = 374 kJ kg or of m potential –1/r and Y more negative so point Y is closer to planet. 3

2 (a) (ii) acceleration / force (directly) proportional to displacement idea of ∆φ = ½v 7 towards 2 fixed point and (6.8 either directed – 5.3) × 10 = ½v –1 displacement in opposite directions & Page 2 v or= acceleration Scheme: Teachers’ version Syllabus 5.5 × Mark 103 ms

12 1 Q12-

3

[3]

M1 A1

[2]

M1C1 PaperA1 A1 41

[2] [2]

B1 (b) (i) maximum / minimum 8ismm A long cloth / 14 mm below cloth (a) either the half-life of theheight source/Section veryabove or decay constant is very small 2 (a) F or ∝ Mm …..…(words or explained symbols) ................................................M1 A1 (ii) 1. a/ R = 11 mm half-life >> 40 days either M and m are point masses = 2πf constant << 0.02 day–1 C1 B1 or 2. ω decay or R >> diameter = 2π × 4.5 of masses …(do not allow ‘size’) ....................................... A1 [2] = 28.3 rad s–1 (do not allow 1 s.f.) A1 6 C1 (b) number of helium atoms = 3.5 × 10 × 40 × 24 × 3600 (b) (i) equatorial orbit .................................................................................................... B1 13 = 1.21 × 10 period 24 hours / same angular speed ............................................................... B1 n = N2008 / ..................................................... NA C1 either pV =toNkT pV direction = nRT and © rotation UCLES from west east /orsame of B1 [3] 5 13 –23 nd rd × Vof =the1.21 10marks × 1.38 10 ×overhead’ 290 1.5 × 10one (allow last × two for × ‘always if 2 or 3 marks not scored) A1 V = 3.2 × 10–13 m3 (if uses T/°C or n = 1 or n = 4, then 1 mark max for calculation of number of (ii) gravitational force provides centripetal force atoms) / gives rise to centripetal acceleration ….(in ‘words’) ........................................ B1

[1]

GCE A/AS LEVEL – October/November 2009

2

C1 C1 A1

9702

GM / x2 = xω2 ....................................................................................................M1 g = GM / R2 .......................................................................................................M1 2 (a) increasing separation molecules / breaking bonds between molecules to give gR = x3ω2 of ............................................................................................ A0

B1 [3]

(allow atoms/molecules, overcome forces) against atmosphere (iii) doing ω = work 2π / (24 × 3600) = 7.27 × (during 10-5 rad expansion) s-1 ........................................................ C1

B1

6 2

3

[1] [1] [2]

[3]

[2]

-5 2

9.81 × (6.4 × 10 ) = x × (7.27 × 10 ) ............................................................. C1 x3 = 7.6 × 1022 (b) (i)x =1 4.2either bubbles produced at a constant rate / mass evaporates/lostA1at [3] × 107 m ................................................................................................. constant rate 1 mark but once only in the Paper) (use of g = 10 m s-2, loses

2

or find mass loss more than once and this rate should be constant [Total: 11] or temperature of liquid remains constant B1 to allow/cancel out/eliminate/compensate for heat losses (to atmosphere) B1 (do not allow ‘prevent’/‘stop’)

(a)Page either pV =Scheme: nRT andTeachers’ n = N / Nversion ..................................................... C1 A 3 pV = NkT orMark Syllabus Paper clear correct substitution e.g. (ii) use of power × time = mass × specific latent heat GCE AS/A LEVEL – May/June 2010 9702 41 C1 5 3 -6 -23 2.5 × 10 × 10 ×× 10 = (13.6 N × 1.38 × 10 × 290 ...............................................M1 (70× –4.5 50) × 5 60 = – 6.5) × L C1 23 A0B1 [2] = 2.8 10 work Q13- 4 (a)N ability A1 Lto×=do 845 J.......................................................................................................... g–1 (allow markoffor calculation of n = 0.467 as a 1result the position/shape, etc. of mol) an object B1 [2]

[1] [1]

2

-10 3 / r GMm 1 ∆E=gpe(1.2 =× 10 (b)(b)(i)(i) volume ) × 2.8 × 1023 or 4 πr3 × 2.8 × 1023 .............................. C1C1 = (6.67 × 10–11 × {2 × 1.663× 10–27}2-7) / (3.8 × 10–15) C1 -7 3 3 2.53 × 10 m ..................................... A1A1 [2][3] = 4.8 = × 10 1.93m× 10–49 J 3 3 (ii) either 4.5 × 10= cm cm3 or ratio of volumes is about 10-4 ................ B1C1 2 ∆E Qq />> 4πε0.48 © UCLES 2009 epe 0r justified because is negligible ........................................... B1C1 [2] = volume (1.6 × 10of–19molecules )2 / (4π × 8.85 × 10–12 × 3.8 × 10–15) –14 56 [3] CEDAR COLLEGE = 6.06 × 10 J A1 [Total: 6]

(ii) idea that 2EK = ∆Eepe – ∆Egpe EK = 3.03 × 10–14 J

B1

[3]

from2 infinity to the point Page Mark Scheme: Teachers’ version GCE AS/A LEVEL – May/June 2010

Syllabus 9702

7 0.1 × 10A ) (b) (i) at R, φ = 6.3 × 107 J kg–1 (allow ±Section φ = GM / R 7 Q14-13 1 (a) work unit× mass 6.3done × 10moving = (6.67 10–11 × M) / (6.4 × 106) 24 from the point Minfinity = 6.0to × 10 kg (allow 5.95 → 6.14) Maximum of 2/3 for any value chosen for φ not at R 7 –1 × 10 ) kg–1×(allow (b)(ii)(i) change at R, φ in= potential 6.3 × 107=J 2.1 107 J ±kg0.1 (allow ± 0.1 × 107) φ =inGM / R energy = gain in kinetic energy loss potential 6 2107 = (6.67 × 10–11 6.3 × M) // (6.4 3R × 10 ) ½ mv = φ m or ½ mv 2 =×GM 24 × 10 ½Mv 2= =6.0 2.1 × 107kg (allow 5.95 → 6.14) –1 any value chosen for φ not at R of32/3 vMaximum = 6.5 × 10 m sfor ………..……(allow 6.3 → 6.6) (answer 7.9 × 103 m s–1, based on x = 2R, allow max 3 marks) (ii) change in potential = 2.1 × 107 J kg–1 (allow ± 0.1 × 107) lossspeed in potential energy = gain inwould kineticbe energy (iii) e.g. / velocity / acceleration greater 2 2 = φ m or ½ mv = GM / 3R ½ mv deviates / bends from straight path 2 ½ v sensible = 2.1 ×ideas, 107 1 each, max 2) (any v = 6.5 × 103 m s–1 ………..……(allow 6.3 → 6.6) (answer 7.9 × 103 m s–1, based on x = 2R, allow max 3 marks) 2 (a) (i) reduction in energy (of the oscillations) in /amplitude energy of oscillations (iii) reduction e.g. speed velocity / /acceleration would be greater due to force (always) motion / resistive forces deviates / bends fromopposing straight path two of the above, 2 max (any sensible ideas, 1max each, 2) Page 2any Mark Scheme: Teachers’ version Syllabus GCE AS/A LEVEL – October/November 2011 9702 (ii) amplitude is decreasing (very) gradually / oscillations would (for long time) /many oscillations 2 (a) (i) continue reduction in aenergy (of the oscillations) Section A light damping reduction in amplitude / energy of oscillations 14 due to force (always) opposing motion /force resistive forces force provides the centripetal Q15- 1 (a) gravitational 2two of the 2 above, max 2 any Page 2 Mark Scheme: Teachers’ version Syllabus GMm/r = mrω (must be in terms of ω) 2 (b) (i)r 3ωfrequency = 1 / 0.3 GCEGM AS/A – October/November 2011 9702 = GM and is aLEVEL constant = decreasing 3.3 Hz (ii) amplitude is (very) gradually / oscillations would allow points taken from time axis giving f = 3.45 continue (for a long time) /many oscillations Section A Hz (b) (i) light 1. damping for Phobos, ω 2= 2π/(7.65 × 3600) –4 v ×=centripetal (ii) gravitational energy = ½ mv 1 (a) force provides the Q15=and 2.28 10ωa rad s–1force –2 2 2 2 2 6× 30.065 × (2π –4 2 = ½ /0.3) (1.5 ) ×M = mrω be in×terms GMm/r (9.39 ) × (2.28 10 ) of×=ω) 6.67× ×1010–11 × 10(must 2 23 (b) (i)r 3ωfrequency = 1 / 0.3 = 3.2 mJ =M GM= and is a kg constant 6.46GM × 10 = 3.3 Hz allow points taken from time axis giving f = 3.45 Hz 2. (9.39 × 106)3 × (2.28 × 10–4)2 = (1.99 × 107)3 × ω2 (c) (b) amplitude not decrease linearly (i) 1. reduces 2π/(7.65 × 3600) ωfor= Phobos, 7.30exponentially × 10ω–52=rad s–1/ does –4 v = ωa (ii)willenergy = ½cm mv =and so be not be 0.7 –5 2.28 ×× 10 10 )2rad s–1 T = 2π/ω 6= 3 2π/(7.30 –2 2 = ½ × 0.065 × (2π /0.3) ) ×M 4 ) ×s(2.28 × 10–4)2 ×= (1.5 6.67× ×1010–11 (9.39 × 10 = 8.6 × 10 23 = 3.2 mJ M==23.6 6.46hours × 10 kg

A1 [2] Paper 42 B1 M1 C1 A1 [3] [2] A1

B1 C1 B1 C1 C1 A1 [3] A1 [4] C1 B1 B1 C1 B1 [2] A1

[4]

(B1) (B1) B1 (B1) B1 [2] [2] Paper 41 M1 (B1) A1 [2] (B1) (B1) B1 [2] Paper B1 41 [3] B1 A1 [1] M1 A1 C1 [2] C1 B1 M1C1 B1 A0A1 B1 [2][3] [3] A1 [1]

C1 M1C1 C1 C1 [2] A1 M1C1 A0A1 [2][3] A1 [3]

6 3 ) × (2.28 × 10–4)2 = (1.99 × 107)3 × ω2 C1 2. (9.39 × 10‘geostationary’ (ii) either almost –1/ does not decrease linearly (c) amplitude reduces exponentially M1C1 ω satellite = 7.30 ×would 10–5 rad or takes a long time to cross the sky B1 so will be not be 0.7 cm A1 T = 2π/ω = 2π/(7.30 × 10–5) 4 = 8.6 × 10 s = random 23.6 hours A1 2 (a) e.g. moving in (rapid) motion of molecules/atoms/particles no intermolecular forces of attraction/repulsion (ii) volume either almost ‘geostationary’ negligible compared to volume of of molecules/atoms/particles or satellite would take a long time to cross the sky B1 container © UCLES 2010 time of collision negligible to time between collisions (1 each, max 2) B2 2 (a) e.g. moving in random (rapid) motion of molecules/atoms/particles no intermolecular forces of attraction/repulsion negligible compared to volume of B1 of ofmolecules/atoms/particles (b) (i) 1.volume number (gas) molecules container © UCLES of collision to time collisions 2.timemean squarenegligible speed/velocity (ofbetween gas2010 molecules) B1 (1 each, max 2) B2 57 CEDAR COLLEGE (ii) either pV = NkT or pV = nRT and links n and k M1 and <EK> = ½m B1 (b) (i) 1. number of (gas) molecules 3 clear algebra leading to <EK> = kT A1

[1] [2] [3]

[1]

[2]

[1] [1] [2]

[1] [2]

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Mark Scheme GCE AS/A LEVEL – October/November 2012

Syllabus 9702

Paper 41

Section A 15 1 Q16-

(a) force is proportional to the product of the masses and inversely proportional to the square of the separation either point masses or separation >> size of masses

M1 A1

[2]

(b) (i) gravitational force provides the centripetal force mv2/r = GMm/r2 and EK = ½mv2 hence EK = GMm/2r

B1 M1 A0

[2]

∆EK = ½ × 4.00 × 1014 × 620 × ({7.30 × 106}–1 – {7.34 × 106}–1) = 9.26 × 107 J (ignore any sign in answer) (allow 1.0 × 108 J if evidence that EK evaluated separately for each r)

C1 A1

[2]

∆EP = 4.00 × 1014 × 620 × ({7.30 × 106}–1 – {7.34 × 106}–1) = 1.85 × 108 J (ignore any sign in answer) (allow 1.8 or 1.9 × 108 J)

C1 A1

[2]

(ii) 1.

2.

2 161

Q17-

(iii) either (7.30 × 106)–1 – (7.34 × 106)–1 or ∆EK is positive / EK increased speed has increased Page 2 Mark Scheme Syllabus GCE AS/A LEVEL – May/June 2013 9702 (a) (i) sum of potential energy and kinetic energy of atoms / molecules / particles Section A reference to random

M1 A1 Paper 42 M1 A1

[2]

[2]

3 orbit / above equator Mark Scheme Syllabus Paper (a) Page equatorial B1 (ii) no intermolecular forces B1 GCE west AS/A LEVEL – October/November 2012 9702 43 satellite from B1 no moves potential energy to east / same direction as Earth spins B1 periodinternal is 24 hours / same period as spinning of Earth B1 [3] energy is kinetic energy (ofofrandom oftest) molecules 3
 (a) (i) 1 (tangent to line gives) direction force onmotion) a (small mass marks scored) B1B1 [3][1] (allow mark for ‘appears to be stationary/overhead’ if none of above (reference to random motion here then allow back credit to (i) if M1 scored)

(ii) (tangent to line gives) direction of force on a (small test) charge charge is positive (b) force provides/is the centripetal force (b) gravitational kinetic 2energy ∝2 thermodynamic temperature 2 2 GMm/R = mRω or in GMm/R = not mv kelvin /R either temperature Celsius, so incorrect ωor= temperature 2π /T or v = 2πR / T or clear substitution in kelvin is not doubled (b) similarity: clear working to give R3 = (GMT2 / 4π2) e.g. radial fields lines normal to surface 3 (a) temperature of the spheres iswith the same greater separation of lines increased from sphere 3 2 distance (c) Rno =(net) 6.67 transfer × 10–11 ×of6.0 × 1024 × (24 × the 3600) / 4π2 2 energy between spheres field strength ∝ 1 / (distance to centre of sphere) =(allow 7.57 ×any 1022 sensible answer) R = 4.2 × 107 m scores 2/3 marks) out = 3600 × 105mmisand (b) (missing (i) difference: power m ×gives c × ∆θ1.8 where mass per second 3800 = m × 4.2 × (42 – 18) towards sphere e.g. gravitational force (always) –1 g s m = 38 electric force direction depends on sign of charge on sphere / towards or 2 (a) (i) away 1. pV = nRT from sphere 1.80 × 10–3 energy ×field/force 2.60 × × 8.31 × 297 (ii) e.g. some thermal is 105 lost to nthe surroundings gravitational is = attractive so rate is an overestimate n = 0.19 mol electric field/force is attractive or repulsive (allow any sensible comparison) 2. ∆q = mc∆T 4 (a) straight 95.0 line through = 0.190 origin × 12.5 × ∆T shows acceleration proportional to displacement ∆T = 40force K = 1.67 × 10–27 × 9.81 (c) gravitational –26 negative(allow gradient 2 marks for correct with clear logic shown) N = 1.6 × 10 answer –19 shows acceleration and displacement in opposite × 270 / (1.8 × 10–2) directions electric force = 1.6 × 10 –15 (ii) p/T = constant = 2.4 × 10 N 5 electric force very=much greater (2.6 × 10 ) / 297 p / (297 + 40)than gravitational force 5 p = 2.95 × 10 Pa

M1 A1 [2] B1 B1 M1 M1 B1 [2] A1 [4]

B1 C1 B1 [2] C1 B1 A1 [3] C1 C1B1 A1 [3] B1 C1 M1 (B1) A1 A1 (B1)[2][2] [3] M1 B1 A1 [2] A1 M1A1 A1C1 [4] A1 M1 B1 [4] A0 [1]

© Cambridge International Examinations 2012

4 (a) force on proton is normal to velocity and field provides centripetal force (forJcircular (b) change in internal energy is 120 / 25 J motion) CEDAR COLLEGE internal energy decreases / ∆U is negative / kinetic energy of molecules decreases (b) magnetic force = Bqv so temperature lower centripetal force = mrω2 or mv2/r

M1 A1 [2] B1 58 M1 A1 B1[3] B1

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Mark Scheme GCE A LEVEL – May/June 2014

Syllabus 9702

Paper 42

Section A

17 1

(a) gravitational force provides/is the centripetal force GMm / r2 R mv2 / r v R √(GM / r) allow gravitational field strength provides/is the centripetal acceleration GM / r2 R v2 / r (b) (i) kinetic energy increase/change R loss / change in (gravitational) energy ½mV0 2 R GMm / x V0 2 R 2GM / x V0 R √(2GM / x)

B1 M1 A0

[2]

(B1) (M1) potential B1 C1 A1

[3]

M1 A1

[2]

(max. 2 for use of r not x) (ii) V0 is (always) greater than v (for x = r) so stone could not enter into orbit (expressions in (a) and (b)(i) must be dimensionally correct)

2 (a) use of kelvin temperatures B1 Page 2 both values of (V / T) correct (11.87), Mark Scheme SyllabusM1 Paper V / T is constant so pressure is constant [2] Cambridge International AS/A Level – October/November 2014 9702 41 (allow use of n R1. Do not allow other values of n.) Section A (b) (i) work done R p∆V 181 (a) g = GM / R2 R 4.2 × 105 × (3.87 – 3.49) × 103 × 10–6 C1 C1 –11 = = = == =(6.67 × 10 R ×160 A1 A1[2] [2] 6.4J × 1023 ) / (3.4 × 106 )2 = 3.7 N kg–1

=

3

2

(do not allow use of V instead of ∆V) (b) ∆EP = mg∆h (ii) increase / change in internal energy of system R (or 1800 m ! 3.4 × 106 R m)heating g is constant because ∆h ! N work done on system ∆EP = 2.4 × 3.7 × 1800 R 565 – 160 4 = 1.6 × 10 J = = = R 405 J (use of g = 9.8 m s–2 max. 1 for explanation)

B1 C1 C1

(c) internal energy R sum of kinetic energy and potential energy / EK N EP (c) gravitational potential energy = (–)GMm / x no intermolecular forces v2 = 2GM / x no potential energy (so ∆U R ∆EK) x = 4D = 4 × 6.8 × 106

B1 C1 M1 C1 A1 [3] C1

2 = (2 × 6.67 × 10–11 × 6.4 × 1023 ) / (4 × 6.8 × 106 ) (a) vresonance = 3.14 × 106 v = 1.8 × 103 m s–1 (b) (use Pt R of mc3.5 ∆θD giving 1.9 × 10 3 m s–1, allow max. 3) 750 × 2 × 60 R 0.28 × c × (98 – 25) c R 4400 J kg–1 K–1 (a) (i) F = R cosθ (useW of =∆θRRsinθ 73 N 273 max. 1 / 3) (usedividing, of t R 2 sWnot s max. 2 / 3) = 120 F tanθ (max. 1 if derivation to final line not shown)

A1

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[1]

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[4]

[3] M1 M1 A0 [2]

© Cambridge International Examinations 2014

(ii) provides the centripetal force

(b) either F = mv2 / r and W = mg CEDAR COLLEGE or v2 = rg / tan θ v2 = (14 × 10–2 × 9.8) / tan 28° = 2.58 v = 1.6 m s–1

B1

[1]

59 C1 C1 A1

[3]

192

3

(c) q = mc∆θ 4.8 × 10–3 = 6.2 × 10–3 × 910 × ∆θ ∆θ = 8.5 × 10–4 K

C1 A1

[2]

(a) smooth curve with decreasing gradient, not starting at x = 0 end of line not at g = 0 or horizontal

M1 A1

[2]

(b) straight line with positive gradient line starts at origin

M1 A1

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(c) sinusoidal shape only positive values and peak / trough height constant 4 ‘loops’

B1 B1 B1

[3]

(a) initially, pV / T = (2.40 × 105 × 5.00 × 10–4) / 288 = 0.417 finally, pV / T = (2.40 × 105 × 14.5 × 10–4) / 835 = 0.417 ideal gas because pV / T is constant (allow 2 marks for two determinations of V / T and then 1 mark for V / T and p constant, so ideal)

M1 M1 A1

[3]

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Mark Scheme Cambridge International AS/A Level – May/June 2016

Syllabus 9702

(a) (gravitational) potential at infinity defined as/is zero (gravitational) force attractive so work got out/done as object moves from infinity (so potential is negative)

(b) (i) ∆E = m∆φ © Cambridge International Examinations 2014 = 180 × (14 – 10) × 108 = 7.2 × 1010 J increase (ii) energy required = 180 × (10 – 4.4) × 108 or energy per unit mass = (10 – 4.4) × 108

2

Paper 41 B1

B1

[2]

C1 A1 B1

[3]

C1

½ × 180 × v2 = 180 × (10 – 4.4) × 108 or ½ × v2 = (10 – 4.4) × 108

C1

v = 3.3 × 104 m s–1

A1

[3]

(a) e.g. time of collisions negligible compared to time between collisions no intermolecular forces (except during collisions) random motion (of molecules) large numbers of molecules

(total) volume of molecules negligible compared to volume of containing vessel or average/mean separation large compared with size of molecules 60 CEDAR COLLEGE any two

B2

[2]

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Mark Scheme: Teachers’ version GCE AS/A LEVEL – October/November 2011

Syllabus 9702

Paper 43

Section A 21 1

(a) (i) weight = GMm/r 2 = (6.67 × 10–11 × 6.42 × 1023 × 1.40)/(½ × 6.79 × 106)2 = 5.20 N (ii) potential energy = –GMm/r = –(6.67 × 10–11 × 6.42 × 1023 × 1.40)/(½ × 6.79 × 106) = –1.77 × 107 J

(b) either

or

½mv 2 = 1.77 × 107 v 2 = (1.77 × 107 × 2)/1.40 v = 5.03 × 103 m s–1 ½mv 2 = GMm/r v 2 = (2 × 6.67 x 10–11 × 6.42 × 1023)/(6.79 × 106/2) v = 5.02 × 103 m s–1

(c) (i) ½ × 2 × 1.66 × 10–27 × (5.03 × 103)2 =

3 × 1.38 × 10–23 × T 2

T = 2030 K

[3]

C1 M1 A0

[2]

C1 C1 A1 (C1) (C1) (A1)

[3]

C1 A1

[2]

M1 A1 (M1) (A1)

[2]

(a) temperature scale calibrated assuming linear change of property with temperature neither property varies linearly with temperature

B1 B1

[2]

(b) (i) does not depend on the property of a substance

B1

[1]

B1

[1]

A1

[1]

A1

[1]

(ii) either because there is a range of speeds some molecules have a higher speed or some escape from point above planet surface so initial potential energy is higher

2

C1 C1 A1

(ii) temperature at which atoms have minimum/zero energy

(c) (i) 323.15 K (ii) 30.00 K

© University of Cambridge International Examinations 2011 CEDAR COLLEGE

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A2 OSCILLATION


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May/June 2002 , Question #3 , qp_4 1




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May/June 2002 , Question #4 qp_4 2




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Oct/Nov 2002 , Question #3 , qp_4 3


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May/June 2003 , Question #3 , qp_4 4

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Oct/June 2003 , Question #2 , qp_4 5

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May/June 2004 , Question #4 , qp_4 For Examiner’s Use

9

6

A student claims that the motion of the mass may be represented by the equation y = y0sin t. (a) Give two reasons why the use of this equation is inappropriate. 1. ..................................................................................................................................... .......................................................................................................................................... 2. ..................................................................................................................................... .................................................................................................................................... [2] (b) Determine the angular frequency

of the oscillations.

angular frequency = .................................. rad s–1 [2] (c) The mass is a lump of plasticine. The plasticine is now flattened so that its surface area is increased. The mass of the lump remains constant and the large surface area is horizontal. The plasticine is displaced downwards by 1.5 cm and then released. On Fig. 4.2, sketch a graph to show the subsequent oscillations of the plasticine. [3]


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[Turn over

Oct/Nov 2004 , Question #3 , qp_4 6

7 3

The vibrations of a mass of 150 g are simple harmonic. Fig. 3.1 shows the variation with displacement x of the kinetic energy Ek of the mass.

For Examiner’s Use

16

Ek / mJ 12

8

4

0 -6

-4

-2

0

2

4

x / cm

6

Fig. 3.1 (a) On Fig. 3.1, draw lines to represent the variation with displacement x of (i)

the potential energy of the vibrating mass (label this line P),

(ii)

the total energy of the vibrations (label this line T). [2]

(b) Calculate the angular frequency of the vibrations of the mass.

angular frequency = ......................................... rad s–1 [3]

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For Examiner’s Use

7 (c) The oscillations are now subject to damping. (i)

Explain what is meant by damping. ................................................................................................................................... ................................................................................................................................... ...............................................................................................................................[2]

(ii)

The mass loses 20% of its vibrational energy. Use Fig. 3.1 to determine the new amplitude of oscillation. Explain your working.

amplitude = ............................................... cm [2]




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May/June 2005 , Question #4 , qp_4

4

For Examiner’s Use

8

8

A tube, closed at one end, has a constant area of cross-section A. Some lead shot is placed in the tube so that the tube floats vertically in a liquid of density ρ, as shown in Fig. 4.1. tube, area of cross-section A

liquid, density

lead shot Fig. 4.1 The total mass of the tube and its contents is M. When the tube is given a small vertical displacement and then released, the vertical acceleration a of the tube is related to its vertical displacement y by the expression a=–

Aρg y, M

where g is the acceleration of free fall. (a) Define simple harmonic motion. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] (b) Show that the tube is performing simple harmonic motion with a frequency f given by f=

1 2π

! AMρg .

[3] © UCLES 2005

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For Examiner’s Use

9 (c) Fig. 4.2 shows the variation with time t of the vertical displacement y of the tube in another liquid.

y / cm

3 2 1 0

0

0.2

0.4

0.6

0.8

1.0

1.2

–1

1.4 t/s

–2 –3 Fig. 4.2 (i)

The tube has an external diameter of 2.4 cm and is floating in a liquid of density 950 kg m–3. Assuming the equation in (b), calculate the mass of the tube and its contents.

mass = ..................................... kg [3] (ii)

State what feature of Fig. 4.2 indicates that the oscillations are damped. ................................................................................................................................... ...............................................................................................................................[1]

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[Turn over

Oct/Nov 2005 , Question #4 , qp_4

4

For Examiner’s Use

8

9

The centre of the cone of a loudspeaker is oscillating with simple harmonic motion of frequency 1400 Hz and amplitude 0.080 mm. (a) Calculate, to two significant figures, (i)

the angular frequency

of the oscillations,

= ………………………………. rad s–1 [2] (ii)

the maximum acceleration, in m s–2, of the centre of the cone.

acceleration = ……………………………….. m s–2 [2] (b) On the axes of Fig. 4.1, sketch a graph to show the variation with displacement x of the acceleration a of the centre of the cone. a

0

0

x

[2] Fig. 4.1

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 79

For Examiner’s Use

9 (c) (i)

State the value of the displacement x at which the speed of the centre of the cone is a maximum. x = ……………………………… mm [1]

(ii) Calculate, in m s–1, this maximum speed.

speed = ……………………………. m s–1 [2]

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May/June 2006 , Question #4 , qp_4

4

For Examiner’s Use

8

10

A piston moves vertically up and down in a cylinder, as illustrated in Fig. 4.1.

cylinder pivot

piston

pivot P wheel

Fig. 4.1 The piston is connected to a wheel by means of a rod that is pivoted at the piston and at the wheel. As the piston moves up and down, the wheel is made to rotate. (a) (i)

State the number of oscillations made by the piston during one complete rotation of the wheel. number = ………………………. [1]

(ii)

The wheel makes 2400 revolutions per minute. Determine the frequency of oscillation of the piston.

frequency = ………………………. Hz [1]

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For Examiner’s Use

9 (b) The amplitude of the oscillations of the piston is 42 mm. Assuming that these oscillations are simple harmonic, calculate the maximum values for the piston of (i)

the linear speed,

speed = …………………………. m s–1 [2] (ii) the acceleration.

acceleration = …………………………. m s–2 [2] (c) On Fig. 4.1, mark a position of the pivot P for the piston to have (i)

maximum speed (mark this position S),

[1]

(ii)

maximum acceleration (mark this position A).

[1]

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Oct/Nov 2006 , Question #3 , qp_4

3

For Examiner’s Use

6

11

Two vertical springs, each having spring constant k, support a mass. The lower spring is attached to an oscillator as shown in Fig. 3.1.

mass

oscillator

Fig. 3.1 The oscillator is switched off. The mass is displaced vertically and then released so that it vibrates. During these vibrations, the springs are always extended. The vertical acceleration a of the mass m is given by the expression ma = –2kx, where x is the vertical displacement of the mass from its equilibrium position. (a) Show that, for a mass of 240 g and springs with spring constant 3.0 N cm–1, the frequency of vibration of the mass is approximately 8 Hz.

[4]

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83

For Examiner’s Use

7 (b) The oscillator is switched on and the frequency f of vibration is gradually increased. The amplitude of vibration of the oscillator is constant. Fig. 3.2 shows the variation with f of the amplitude A of vibration of the mass.

A

0

0

f Fig. 3.2

State (i)

the name of the phenomenon illustrated in Fig. 3.2, .............................................................................................................................. [1]

(ii)

the frequency f0 at which maximum amplitude occurs. frequency = ………………………… Hz [1]

(c) Suggest and explain how the apparatus in Fig. 3.1 could be modified to make the peak on Fig. 3.2 flatter, without significantly changing the frequency f0 at which the peak occurs. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [3]

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Oct/Nov 2007 , Question #3 , qp_4

3

For Examiner’s Use

8

12

A spring is hung from a fixed point. A mass of 130 g is hung from the free end of the spring, as shown in Fig. 3.1.

spring

mass 130 g Fig. 3.1 The mass is pulled downwards from its equilibrium position through a small distance d and is released. The mass undergoes simple harmonic motion. Fig. 3.2 shows the variation with displacement x from the equilibrium position of the kinetic energy of the mass. 3.0 kinetic energy / mJ

2.0

1.0

–1.0

–0.5

0

0

Fig. 3.2

x / cm

+1.0


 85

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+0.5

9702/04/O/N/07

For Examiner’s Use

9 (a) Use Fig. 3.2 to (i)

determine the distance d through which the mass was displaced initially, d = ............................................ cm [1]

(ii)

show that the frequency of oscillation of the mass is approximately 4.0 Hz.

[6] (b) (i) (ii)

On Fig. 3.2, draw a line to represent the total energy of the oscillating mass.

[1]

After many oscillations, damping reduces the total energy of the mass to 1.0 mJ. For the oscillations with reduced energy, 1.

state the frequency, frequency = .............................................Hz

2.

using the graph, or otherwise, state the amplitude. amplitude = ............................................ cm [2]




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Oct/Nov 2008 , Question #3 , qp_4 13 3

8

The needle of a sewing machine is made to oscillate vertically through a total distance of 22 mm, as shown in Fig. 3.1.

22 mm

For Examiner’s Use

needle at its maximum height

8.0 mm

cloth Fig. 3.1

The oscillations are simple harmonic with a frequency of 4.5 Hz. The cloth that is being sewn is positioned 8.0 mm below the point of the needle when the needle is at its maximum height. (a) State what is meant by simple harmonic motion. .......................................................................................................................................... .......................................................................................................................................... ...................................................................................................................................... [2] (b) The displacement y of the point of the needle may be represented by the equation y = a cos t. (i)

Suggest the position of the point of the needle at time t = 0. .............................................................................................................................. [1]

(ii)

Determine the values of 1.

a, a = .......................................... mm [1]

2.

.

= ..................................... rad s–1 [2] © UCLES 2008

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 87

9 (c) Calculate, for the point of the needle, (i)

For Examiner’s Use

its maximum speed,

speed = ........................................ m s–1 [2] (ii)

its speed as it moves downwards through the cloth.

speed = ........................................ m s–1 [3]




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May/June 2009 , Question #4 , qp_4 8

14 4

A vertical peg is attached to the edge of a horizontal disc of radius r, as shown in Fig. 4.1.

For Examiner’s Use

peg disc r

Fig. 4.1 The disc rotates at constant angular speed ω. A horizontal beam of parallel light produces a shadow of the peg on a screen, as shown in Fig. 4.2. screen

peg R

Q

θ

parallel beam of light

P

r

S

ω

Fig. 4.2 (plan view) At time zero, the peg is at P, producing a shadow on the screen at S. At time t, the disc has rotated through angle θ. The peg is now at R, producing a shadow at Q. (a) Determine, (i)

in terms of ω and t, the angle θ, ............................................................................................................................ [1]

(ii)

in terms of ω, t and r, the distance SQ. ............................................................................................................................ [1]




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9 (b) Use your answer to (a)(ii) to show that the shadow on the screen performs simple harmonic motion.

For Examiner’s Use

.......................................................................................................................................... .......................................................................................................................................... .................................................................................................................................... [2] (c) The disc has radius r of 12 cm and is rotating with angular speed ω of 4.7 rad s–1. Determine, for the shadow on the screen, (i)

the frequency of oscillation,

frequency = ......................................... Hz [2] (ii)

its maximum speed.

speed = .................................... cm s–1 [2]




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[Turn over

Oct/Nov 2009 , Question #4 , qp_41 15 4

8

The variation with time t of the displacement x of the cone of a loudspeaker is shown in Fig. 4.1.

For Examiner’s Use

0.3 x / mm 0.2 0.1 0

0

0.2

0.4

0.6

0.8

– 0.1

1.0

1.2

1.4 1.6 t / ms

– 0.2 – 0.3 Fig. 4.1 (a) Use Fig. 4.1 to determine, for these oscillations, (i)

the amplitude, amplitude = ........................................ mm [1]

(ii)

the frequency.

frequency = .......................................... Hz [2] (b) State two times at which (i)

the speed of the cone is maximum, time ............................... ms and time ............................... ms [1]

(ii)

the acceleration of the cone is maximum. time ............................... ms and time ............................... ms [1]




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9 (c) The effective mass of the cone is 2.5 g.

For Examiner’s Use

Use your answers in (a) to determine the maximum kinetic energy of the cone.

kinetic energy = ............................................ J [3] (d) The loudspeaker must be designed so that resonance of the cone is avoided. (i)

State what is meant by resonance. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................ [2]

(ii)

State and briefly explain one other situation in which resonance should be avoided. .................................................................................................................................. .................................................................................................................................. ..................................................................................................................................

................................................................................................................. [2] 


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May/June 2010 , Question #3 , qp_41 7

16 3

(a) State what is meant by (i)

For Examiner’s Use

oscillations, .................................................................................................................................. .............................................................................................................................. [1]

(ii)

free oscillations, .................................................................................................................................. .............................................................................................................................. [1]

(iii)

simple harmonic motion. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

(b) Two inclined planes RA and LA each have the same constant gradient. They meet at their lower edges, as shown in Fig. 3.1. ball L

R

A Fig. 3.1 A small ball moves from rest down plane RA and then rises up plane LA. It then moves down plane LA and rises up plane RA to its original height. The motion repeats itself. State and explain whether the motion of the ball is simple harmonic. .......................................................................................................................................... .......................................................................................................................................... ...................................................................................................................................... [2]

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May/June 2010 , Question #2 , qp_42 6

17 2

A long strip of springy steel is clamped at one end so that the strip is vertical. A mass of 65 g is attached to the free end of the strip, as shown in Fig. 2.1.

For Examiner’s Use

mass 65 g

springy steel

clamp

Fig. 2.1 The mass is pulled to one side and then released. The variation with time t of the horizontal displacement of the mass is shown in Fig. 2.2. 2 displacement / cm 1

0

0.1

0

0.2

0.3

0.4

0.5

0.6 t/s

0.7

–1

–2 Fig. 2.2 The mass undergoes damped simple harmonic motion. (a) (i)

Explain what is meant by damping. .................................................................................................................................. ..................................................................................................................................

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............................................................................................................................ [2]

94

Fig. 2.1 The mass is pulled to one side and then released. The variation with time t of the horizontal displacement of the mass is shown in Fig. 2.2. 2 displacement / cm 1

0

0.1

0

0.2

0.3

0.4

0.5

0.6 t/s

0.7

–1

–2 Fig. 2.2 The mass undergoes damped simple harmonic motion. (a) (i)

Explain what is meant by damping. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................ [2]

© UCLES 2010

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7 (ii)

Suggest, with a reason, whether the damping is light, critical or heavy. ..................................................................................................................................

For Examiner’s Use

.................................................................................................................................. ............................................................................................................................ [2] (b) (i)

Use Fig. 2.2 to determine the frequency of vibration of the mass.

frequency = ......................................... Hz [1] (ii)

Hence show that the initial energy stored in the steel strip before the mass is released is approximately 3.2 mJ.

[2] (c) After eight complete oscillations of the mass, the amplitude of vibration is reduced from 1.5 cm to 1.1 cm. State and explain whether, after a further eight complete oscillations, the amplitude will be 0.7 cm. .......................................................................................................................................... .......................................................................................................................................... .................................................................................................................................... [2]

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Oct/Nov 2010 , Question #3 , qp_41 8

18 3

A student sets up the apparatus illustrated in Fig. 3.1 in order to investigate the oscillations of a metal cube suspended on a spring. pulley

For Examiner’s Use

variable-frequency oscillator thread

spring

metal cube Fig. 3.1 The amplitude of the vibrations produced by the oscillator is constant. The variation with frequency of the amplitude of the oscillations of the metal cube is shown in Fig. 3.2. 20

15 amplitude / mm 10

5

0

2

4

6

8 10 frequency / Hz

Fig. 3.2 (a) (i)

State the phenomenon illustrated in Fig. 3.2. .............................................................................................................................. [1]

(ii)

For the maximum amplitude of vibration, state the magnitudes of the amplitude and the frequency. amplitude = ............................................. mm frequency = ............................................... Hz [1]

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9 (b) The oscillations of the metal cube of mass 150 g may be assumed to be simple harmonic. Use your answers in (a)(ii) to determine, for the metal cube, (i)

For Examiner’s Use

its maximum acceleration,

acceleration = ...................................... m s–2 [3] (ii)

the maximum resultant force on the cube.

force = .......................................... N [2] (c) Some very light feathers are attached to the top surface of the cube so that the feathers extend outwards, beyond the vertical sides of the cube. The investigation is now repeated. On Fig. 3.2, draw a line to show the new variation with frequency of the amplitude of vibration for frequencies between 2 Hz and 10 Hz. [2]

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98

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Oct/Nov 2010 , Question #3 , qp_43 8

19 3

A cylinder and piston, used in a car engine, are illustrated in Fig. 3.1.

For Examiner’s Use

cylinder C

D

A

B piston

Fig. 3.1 The vertical motion of the piston in the cylinder is assumed to be simple harmonic. The top surface of the piston is at AB when it is at its lowest position; it is at CD when at its highest position, as marked in Fig. 3.1. (a) The displacement d of the piston may be represented by the equation d = – 4.0 cos(220t ) where d is measured in centimetres. (i)

State the distance between the lowest position AB and the highest position CD of the top surface of the piston. distance = .......................................... cm [1]

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9 (ii)

Determine the number of oscillations made per second by the piston.

For Examiner’s Use

number = ................................................ [2] (iii)

On Fig. 3.1, draw a line to represent the top surface of the piston in the position where the speed of the piston is maximum. [1]

(iv)

Calculate the maximum speed of the piston.

speed = ..................................... cm s–1 [2]

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[Turn over

10 (b) The engine of a car has several cylinders. Three of these cylinders are shown in Fig. 3.2. Y

X

C

D

A

B

For Examiner’s Use

Z

Fig. 3.2 X is the same cylinder and piston as in Fig. 3.1. Y and Z are two further cylinders, with the lowest and the highest positions of the top surface of each piston indicated. The pistons in the cylinders each have the same frequency of oscillation, but they are not in phase. At a particular instant in time, the position of the top of the piston in cylinder X is as shown. (i)

In cylinder Y, the oscillations of the piston lead those of the piston in cylinder X by a phase angle of 120° ( 23 p rad). Complete the diagram of cylinder Y, for this instant, by drawing 1.

a line to show the top surface of the piston,

[1]

2.

an arrow to show the direction of movement of the piston.

[1]

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101

11 (ii)

In cylinder Z, the oscillations of the piston lead those of the piston in cylinder X by a phase angle of 240° ( 43 p rad). Complete the diagram of cylinder Z, for this instant, by drawing

For Examiner’s Use

(iii)

1.

a line to show the top surface of the piston,

[1]

2.

an arrow to show the direction of movement of the piston.

[1]

For the piston in cylinder Y, calculate its speed for this instant.

speed = ..................................... cm s–1 [2]

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May/June 2011 , Question #3 , qp_41 6

20 3

(a) Define simple harmonic motion. ..........................................................................................................................................

For Examiner’s Use

.......................................................................................................................................... ...................................................................................................................................... [2] (b) A tube, sealed at one end, has a total mass m and a uniform area of cross-section A. The tube floats upright in a liquid of density ρ with length L submerged, as shown in Fig. 3.1a. tube

liquid density ρ

L

L+x x

Fig. 3.1a

Fig. 3.1b

The tube is displaced vertically and then released. The tube oscillates vertically in the liquid. At one time, the displacement is x, as shown in Fig. 3.1b. Theory shows that the acceleration a of the tube is given by the expression a=–

CEDAR COLLEGE

A ρg x. m




103

7 (i)

Explain how it can be deduced from the expression that the tube is moving with simple harmonic motion.

For Examiner’s Use

.................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2] (ii)

The tube, of area of cross-section 4.5 cm2, is floating in water of density 1.0 × 103 kg m–3. Calculate the mass of the tube that would give rise to oscillations of frequency 1.5 Hz.

12 mass = ............................................. g [4] (d) (i)

Name the phenomenon illustrated on your completed graph of Fig. 5.3.

For Examiner’s Use

............................................................................................................................ [1] (ii)

State one situation where the phenomenon named in (i) is useful. .................................................................................................................................. ............................................................................................................................ [1]

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104

Oct/Nov 2011 , Question #3 , qp_41 8

21 3

A bar magnet is suspended from the free end of a helical spring, as illustrated in Fig. 3.1.

For Examiner’s Use

helical spring magnet coil

Fig. 3.1 One pole of the magnet is situated in a coil of wire. The coil is connected in series with a switch and a resistor. The switch is open. The magnet is displaced vertically and then released. As the magnet passes through its rest position, a timer is started. The variation with time t of the vertical displacement y of the magnet from its rest position is shown in Fig. 3.2. 2.0 y / cm 1.0

0

0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0 10.0 t /s

–1.0

–2.0 Fig. 3.2 At time t = 4.0 s, the switch is closed.

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 105

9 (a) Use Fig. 3.2 to (i)

state the evidence for the magnet to be undergoing free oscillations during the period t = 0 to t = 4.0 s,

For Examiner’s Use

.................................................................................................................................. .............................................................................................................................. [1] (ii)

state, with a reason, whether the damping after time t = 4.0 s is light, critical or heavy, .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

(iii)

determine the natural frequency of vibration of the magnet on the spring.

frequency = ........................................... Hz [2] (b) (i)

State Faraday’s law of electromagnetic induction. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

(ii)

Explain why, after time t = 4.0 s, the amplitude of vibration of the magnet is seen to decrease. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [4]

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106 [Turn over

May/June 2012 , Question #4 , qp_41 9

22 4

A small metal ball is suspended from a fixed point by means of a string, as shown in Fig. 4.1.

For Examiner’s Use

string

ball x Fig. 4.1 The ball is pulled a small distance to one side and then released. The variation with time t of the horizontal displacement x of the ball is shown in Fig. 4.2. 6 x / cm 4 2 0

0

0.2

0.4

0.6

0.8

t /s

1.0

–2 –4 –6 Fig. 4.2 The motion of the ball is simple harmonic. (a) Use data from Fig. 4.2 to determine the horizontal acceleration of the ball for a displacement x of 2.0 cm.

acceleration = ....................................... m s–2 [3] © UCLES 2012 CEDAR

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107

10 (b) The maximum kinetic energy of the ball is EK. On the axes of Fig. 4.3, sketch a graph to show the variation with time t of the kinetic energy of the ball for the first 1.0 s of its motion. kinetic energy

For Examiner’s Use

EK

0

0

0.2

0.4

0.6

0.8

t /s

1.0

Fig. 4.3 [3]

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108

May/June 2012 , Question #2 , qp_42 6

23 2

A ball of mass 37 g is held between two fixed points A and B by two stretched helical springs, as shown in Fig. 2.1.

For Examiner’s Use

ball mass 37 g A

B

Fig. 2.1 The ball oscillates along the line AB with simple harmonic motion of frequency 3.5 Hz and amplitude 2.8 cm. (a) Show that the total energy of the oscillations is 7.0 mJ.

[2] (b) At two points in the oscillation of the ball, its kinetic energy is equal to the potential energy stored in the springs. Calculate the magnitude of the displacement at which this occurs.

displacement = ............................................ cm [3]

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109

7 (c) On the axes of Fig. 2.2 and using your answers in (a) and (b), sketch a graph to show the variation with displacement x of (i)

the total energy of the system (label this line T),

[1]

(ii)

the kinetic energy of the ball (label this line K),

[2]

(iii)

the potential energy stored in the springs (label this line P).

[2]

For Examiner’s Use

8

6 energy / mJ 4

2

–3

–2

–1

0

0

1

2

x / cm

3

Fig. 2.2 (d) The arrangement in Fig. 2.1 is now rotated through 90° so that the line AB is vertical and the ball oscillates in a vertical plane. Suggest one form of energy, other than those in (c), that must be taken into consideration when plotting new graphs to show energy changes with displacement. ...................................................................................................................................... [1]

110

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May/June 2013 , Question #3 , qp_42 7

24 3

A mass of 78 g is suspended from a fixed point by means of a spring, as illustrated in Fig. 3.1.

For Examiner’s Use

spring

mass 78 g

Fig. 3.1 The stationary mass is pulled vertically downwards through a distance of 2.1 cm and then released. The mass is observed to perform simple harmonic motion with a period of 0.69 s. (a) The mass is released at time t = 0. For the oscillations of the mass, (i)

calculate the angular frequency ω,

ω = ...................................... rad s–1 [2] (ii)

determine numerical equations for the variation with time t of 1. the displacement x in cm, .................................................................................................................................. ............................................................................................................................. [2] 2. the speed v in m s–1. .................................................................................................................................. ............................................................................................................................. [2]

© UCLES 2013

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111

8 (b) Calculate the total energy of oscillation of the mass.

energy = ............................................... J [2]

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For Examiner’s Use




112

Oct/Nov 2014 , Question #4 , qp_41 10

25 4

(a) State what is meant by simple harmonic motion. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) A small ball rests at point P on a curved track of radius r, as shown in Fig. 4.1. curved track, radius r

x

P Fig. 4.1 The ball is moved a small distance to one side and is then released. The horizontal displacement x of the ball is related to its acceleration a towards P by the expression a = −

gx r

where g is the acceleration of free fall. (i)

Show that the ball undergoes simple harmonic motion. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2]

(ii)

The radius r of curvature of the track is 28 cm. Determine the time interval τ between the ball passing point P and then returning to point P.

τ = ..................................................... s [3] © UCLES 2014

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 113

11 (c) The variation with time t of the displacement x of the ball in (b) is shown in Fig. 4.2.

x

0

0

2t

3t

4t

t

Fig. 4.2 Some moisture now forms on the track, causing the ball to come to rest after approximately 15 oscillations. On the axes of Fig. 4.2, sketch the variation with time t of the displacement x of the ball for the first two periods after the moisture has formed. Assume the moisture forms at time t = 0. [3]




© UCLES 2014

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May/June 2015 , Question #4 , qp_41 11

26 4

(a) For an oscillating body, state what is meant by (i)

forced frequency, ........................................................................................................................................... ...................................................................................................................................... [1]

(ii)

natural frequency of vibration, ........................................................................................................................................... ...................................................................................................................................... [1]

(iii)

resonance. ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2]

(b) State and explain one situation where resonance is useful. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (c) In some situations, resonance should be avoided. State one such situation and suggest how the effects of resonance are reduced. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2]




115

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May/June 2016 , Question #3 , qp_41 7

27 3

(a) State, by reference to displacement, what is meant by simple harmonic motion. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) A mass is undergoing oscillations in a vertical plane. The variation with displacement x of the acceleration a of the mass is shown in Fig. 3.1.

a

0

0

x

Fig. 3.1 State two reasons why the motion of the mass is not simple harmonic. 1. .............................................................................................................................................. ................................................................................................................................................... 2. .............................................................................................................................................. ................................................................................................................................................... [2]




116

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8 (c) A block of wood is floating in a liquid, as shown in Fig. 3.2. oscillation of block

block

liquid

Fig. 3.2 The block is displaced vertically and then released. The variation with time t of the displacement y of the block from its equilibrium position is shown in Fig. 3.3. 2.0 y / cm 1.5 1.0 0.5 0

0

0.4

0.8

1.2

1.6

2.0

2.4

t/s

–0.5 –1.0 –1.5 –2.0 Fig. 3.3 Use data from Fig. 3.3 to determine (i)

the angular frequency ω of the oscillations,

ω = ............................................. rad s−1 [2] © UCLES 2016

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9 (ii)

the maximum vertical acceleration of the block.

maximum acceleration = ............................................... m s−2 [2] (iii)

The block has mass 120 g. The oscillations of the block are damped. Calculate the loss in energy of the oscillations of the block during the first three complete periods of its oscillations.

energy loss = ...................................................... J [3] [Total: 11]




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May/June 2016 , Question #4 , qp_42 8

28 4

A metal block hangs vertically from one end of a spring. The other end of the spring is tied to a thread that passes over a pulley and is attached to a vibrator, as shown in Fig. 4.1. pulley

vibrator

spring

block Fig. 4.1 (a) The vibrator is switched off. The metal block of mass 120 g is displaced vertically and then released. The variation with time t of the displacement y of the block from its equilibrium position is shown in Fig. 4.2. 3 y / cm

2 1 0

0

0.2

0.4

–1

0.6

0.8

1.0 /s

–2 –3 Fig. 4.2 For the vibrations of the block, calculate (i)

the angular frequency ω,

ω = ............................................. rad s−1 [2] © UCLES 2016

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 119

9 (ii)

the energy of the vibrations.

energy = ...................................................... J [2] (b) The vibrator is now switched on. The frequency of vibration is varied from 0.7f to 1.3f where f is the frequency of vibration of the block in (a). For the block, complete Fig. 4.3 to show the variation with frequency of the amplitude of vibration. Label this line A. [3]

amplitude

0 0.7

1.3 frequency Fig. 4.3

(c) Some light feathers are now attached to the block in (b) to increase air resistance. The frequency of vibration is once again varied from 0.7f to 1.3f. The new amplitude of vibration is measured for each frequency. On Fig. 4.3, draw a line to show the variation with frequency of the amplitude of vibration. Label this line B. [2] [Total: 9]




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120

May/June 2016 , Question #3 , qp_43 7

29 3

(a) State, by reference to displacement, what is meant by simple harmonic motion. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) A mass is undergoing oscillations in a vertical plane. The variation with displacement x of the acceleration a of the mass is shown in Fig. 3.1.

a

0

0

x

Fig. 3.1 State two reasons why the motion of the mass is not simple harmonic. 1. .............................................................................................................................................. ................................................................................................................................................... 2. .............................................................................................................................................. ................................................................................................................................................... [2]

121

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8 (c) A block of wood is floating in a liquid, as shown in Fig. 3.2. oscillation of block

block

liquid

Fig. 3.2 The block is displaced vertically and then released. The variation with time t of the displacement y of the block from its equilibrium position is shown in Fig. 3.3. 2.0 y / cm 1.5 1.0 0.5 0

0

0.4

0.8

1.2

1.6

2.0

2.4

t/s

–0.5 –1.0 –1.5 –2.0 Fig. 3.3 Use data from Fig. 3.3 to determine (i)

the angular frequency

of the oscillations,

= ............................................. rad s−1 [2] © UCLES 2016

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122




9 (ii)

the maximum vertical acceleration of the block.

maximum acceleration = ............................................... m s−2 [2] (iii)

The block has mass 120 g. The oscillations of the block are damped. Calculate the loss in energy of the oscillations of the block during the first three complete periods of its oscillations.

energy loss = ...................................................... J [3] [Total: 11]

123

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ANSWERS 1

2

3




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124

∆T = 143 K.......................................................................................C1 temperature = 143 Mark + 1340 = 1483 K................................................A1 Scheme Syllabus Paper[5]

Page 2

A/AS LEVEL EXAMINATIONS - NOVEMBER 2003

(c) 4 1 3 (a)

(i)

(ii) (b) (b)

(i)

(c)

(ii)

(iii) 4 (c) (a)

5 2 (a) (b)

(i) (i) (ii)

(b)

(i)

(ii)

(iii)

(c)

04

[6]

because acceleration fall) is .............................................M1 (resultant) force per unit field causes forces on(of thefree electrons mass ....................................................................................... and the nucleus in opposite directions ......................................... B1 A1 -2 .............................................................. acceleration = 9.77 m s (field causes) electrons (to be) stripped off the atom.................... B1 B1

[2] [3]

a,ω x 0identified ………(-1 each error or omission) ................. B2 r2 .....................................................................................C1 E = and Q/4!ε 20 x 103 x 102 = Q/(4! x 8.85 x 10-12 x 0.212 ....................................C1 (-)ve because a and x in opposite directions charge = 9.8 x 10-6 C .......................................................................A1 OR a directed towards mean position/centre................................ B1

[3] [3]

forces in springs are k(e + x) and k(e – x) .................................... C1 resultant = k(e + x) – k(e – x) ......................................................M1 © University of Cambridge Local Examinations Syndicate 2003 = 2kx ............................................................................ A0

[2]

F = ma ....................................................................................... B1 a = -2kx/m .................................................................................... A0 (-)ve sign explained...................................................................... B1

[2]

ω2 = 2k/m ..................................................................................... C1 (2πf)2 = (2 x 120)/0.90 .................................................................. C1 f = 2.6 Hz ..................................................................................... A1

[3]

pV/T = constant............................................................................ C1 T = (6.5 x 106 x 30 x 300)/(1.1 x 105 x 540)................................. C1 = 985 K .................................................................................... A1 (if uses °C, allow 1/3 marks for clear formula) (i)

(ii)

[1]

radial f0 is atlines.................................................................................... natural frequency of spring (system) ................................. B1 B1 pointing ........................................................................... this is at inwards the driver frequency ....................................................... B1 (allow 1 mark for recognition that this is resonance) no difference OR lines closer near surface of smaller sphere ...... B1 line: amplitude less at all frequencies ......................................... B1 2 FG = peak GMm/R .............................................................................. C1 flatter .......................................................................... B1 -11 24 3 2 x 5.98 x 10 )/(6380 x 10 ) = peak (6.67 at X 10 f0 or slightly below f0 ................................................ B1 = 9.80 N ................................................................................. A1 (aluminium) sheet cuts the magnetic flux/field.............................. B1 F mRω2 .................................................................................... C = currents/e.m.f. (so) induced in the (metal) sheet .......................... C1 B1 ω = 2π/T ...................................................................................... C1 these currents dissipate energy ...................................................M1 2 F x 6380 x 103for )/8.64 104)2 less energy available thexoscillations ...................................... A1 C = (4π 0.0337 N............................................................................... so =amplitude smaller .................................................................... A1 A0 (‘current opposes motion of sheet’ scores one of the last two F G - FC = 9.77 N............................................................................ A1 marks)

atom held in position by attractive forces atom oscillates, not just two forces OR 3D not 1D force not proportional to x any two relevant points, 1 each, max 2 ........................................ B2

3 (a)

3 (b)

9702

e.g. thermocouple/resistance thermometer .................................. B1

∆U = q + w symbols identified correctly ..........................................................M1 directions correct.......................................................................... A1 q is zero ....................................................................................... B1 w is positive OR ∆U = w and U increases .................................... B1 ∆U is rise in kinetic energy of atoms ............................................M1 and mean kinetic energy ∝ T ....................................................... A1 (allow one of the last two marks if states ‘U increases so T rises’)

[2]

[3]

[3]




[4]

[2]

[3]

[2]

[4]

125

CEDAR COLLEGE

© University of Cambridge Local Examinations Syndicate 2003

θ 1 + =R20.182 sig. fig.) = {(R )2 x R1 x ω 2} I G (n.b. (any 3subject for equation) (d)(b) (i)(i) M2tan 11 2 = (3.2 x 10 ) x 8.0 x 1010 x (4.99 x 10-8)2/(6.67 x 10-11) 29 (ii) percentage kg = (0.002/0.180) x 100 = 3.06 x 10error (ii) less massive (only award this mark if reasonable attempt at (i)) = 1.1massive (%) star) (9.17 x 1029 kg for more 6 Total (allow 0.002/0.182 allow 1! 4 sig.isfig.) 4 (a) e.g. amplitude is notand constant or wave damped do not allow 'displacement constant' /R 2 (a) (i) grav. pot. energy = GM 1M2sin) should be (-)cos, (not energy = {6.67 x 10-11 x 197 x 4 x (1.66 x 10-27)2}/9.6 x 10-15 -47 (b) T = 0.60 s = 1.51 x 10 J ω = 2π/T = 10.5 rad s-1 (allow 10.4 → 10.6) (ii) elec. pot. energy = Q1Q2/4π ε 0R energy = {79 x 2 x (1.6 x 10-19)2}/4π x 8.85 x 10-12 x 9.6 x 10-15 (c) same period -12 = 3.79 x 10less J displacement always amplitude reducing appropriately (Forfor the2nd substitution, -1 each error orfirst omission to period max 2 in (i) and in (ii)) and 3rd marks, ignore the quarter Total (b) electric potential energy >> gravitational potential energy (c)

either 6 MeV = 9.6 x 10-13 J or 3.79 x 10-12 J = 24 MeV not enough energy to get close to the nucleus

1C1 C1 1A1 B1 [4] 1 [3] [12] B1 1 B1 [2] 1 1C1 [3] A1 [2] 1 1 B1 1M1 [3] A1 [3]

1

[7] [1]

1 1

[2]

7

3

(a)

(b)

(c)

(i)

© University Cambridge reasonable shapeofas ‘inverse’ International of k.e. line Examinations 2004

1

(ii)

straight line, parallel to x-axis at 15 mJ

1

[2]

either

(max) kinetic energy (= ½ mv2) = ½ m ω 2a02 15 x 10-3 = ½ x 0.15 x ω 2 x (5.0 x 10-2)2 ω = 8.9(4) rad s-1

1 1 1

or

(k.e. = ½ mv2), v = 0.44(7) m s-1 ω = v/a = (0.447)/(5.0 x 10-2) ω = 8.9(4) rad s-1

1 1 1

[3]

either loss of energy (from the system) or amplitude decreases or additional force acting (on the mass) either continuous/gradual loss or force always opposing motion

1 1

[2]

either (now has 80% of its) p.e./k.e. = 12 mJ or loss in k.e. = 3 mJ new amplitude = 4.5 cm (allow ± 0.1 cm)

1 1

[2]

(i)

(ii)

© University of Cambridge International Examinations 2005

CEDAR COLLEGE

126

2

(a)

obeys the law pV = constant × T ………………………..………………..

M1

Page 2 at all values of p, V and T Mark Scheme Syllabus ………………………………………………. A LEVEL - JUNE 2005 9702

(b) 8 4 (a) (c)

(b) 3

(a)

= (2.9 × 105 × 3.1 × 10–2) / (8.31 × 290) …..………………..………... = 3.73 mol ………………………………………………………………. acceleration proportional to displacement (from a fixed point) M1 2 3 .4 290 or a = - ω x with a, ω and x explained at new pressure, n n = 3.73 × × and directed towards a fixed point A1 2 .9 300 or negative sign explained = 4.23 mol ….………………………………………. change = 0.50 mol ……………………………………………………….… for s.h.m., a = (-)ω=2x0.50 / 0.012 = 42 (must round up for mark) ……. B1 number of strokes identifies ω2 as Aρg/M and therefore s.h.m. (may be implied) B1 correct 2πf= ωstatement, words or symbols …..…………………………...….. B1 n

1

4

[2]

C1 A1

[2]

[2] C1 C1 A1

[3]

B1

[1]

Apg

(b)(i) whence = p∆V √ A0 f= ………………………………………………………………….. = 1.03 2×π105 ×M(2.96 × 10–2 – 1.87 × 10–5) = (–) 3050 J …………………..……………….…………..…………… C1 (c) (i) T = 0.60 s or f= 1.7 Hz -2 2 4 C1 (2π√M)/√(π {1.2 × 10 } × 950 × 9.81) (ii) q 0.60 == 4.05 × 10 J ×…………………………………………………………. M = 0.0384 kg A1 (iii) ∆U = 4.05 × 104 – 3050 = 37500 J …no e.c.f. from (a)………………… 2 sig.fig. only peak once height/amplitude B1 (ii) penalise decreasing (c) 5 (a)

Paper A1 4

number of molecules = NA ………………………………………………. energy = 37500 / (6.02 ×gradient 1023) [- sign not required] field strength = potential B1 –20 × 10but Jnot(accept 1 sig.fig.) …………………………..…. [allow E==6.2 ∆V/∆x E = V/d]

9 (a) (i) ωNo =field 2πffor………………………………………………………....………….. (b) x B1 –1 = 8800 rad s ………………………………………………………….. discontinuity at x = r (vertical line required) B1

= (–)ω2x0 ……………………………..………………………………… 2 –3 = (8800) × 0.080 10changing 6 (a) (i) flux/field in core must ×be M1 –2 = 6200 ms …………………….……………………………………. so that an e.m.f./current is induced in the secondary A1 (ii) a0

C1 [3] A1

[2]

B1 [3] A1 [1]

[1] [1]

C1 [1] A1

[2]

C1 A1 [3]

[2]

C1 A1 [2]

[2]

straight line through origin with negative gradient …….…………….... (ii) power = VI M1 end points of line correctly labelled …………………………………….. A1 output power is constant so if VS increases, IS decreases

M1 A1 [2]

[2]

(c) (i) zero displacement ………………………………………………………… B1 (b) (i) same shape and phase as IP graph (ii) v = ωx0 ……………………………………………………………………. (ii) same frequency M1 = 8800 × 0.080 × 10–3 –1 correct phase Fig. 6.3 A1 = 0.70 m sw.r.t.…………………………………………………………….

B1 [1] C1

[1]

[2] A1

[2]

(b)

(iii) ½π rad or 90° © University of Cambridge International Examinations 2005

7 (a)

curve levelling out (at 1.4 µg) correct shape judged by masses at nT½ [for second mark, values must be marked on y-axis)

(b) (i) N0 = (1.4 × 10-6 × 6.02 × 1023)/56 = 1.5 × 1016 (ii) A = λN λ = ln2/(2.6 × 3600) (= 7.4 × 10-5 s-1) A = 1.11 × 1012 Bq (c)

1/10 of original mass of Manganese remains 0.10 = exp(-ln2 × t/2.6) t = 8.63 hours [use of 1/9, giving answer 8.24 hrs scores 1 mark]

CEDAR COLLEGE © University of Cambridge International Examinations 2005

B1

[1]

M1 A1

[2]

C1 A1

[2]

C1 C1 A1

[3]

C1 A1

[2]

127

the gradient varies with temperature (b)Page 20402± 20 Ω corresponds to 15.0Mark ± 0.2Scheme °C T / K = T / °C + 273.15GCE (allow 273.2) A/AS LEVEL - OCT/NOV 2006 temperature is 288.2 K 10 (a) (i)either 4 1 (a) 1.0ratio of work done to mass/charge

Syllabus 9702

or work done moving unit mass/charge from infinity have zero potential at infinity (ii)or both 40 Hz (b)

A1

[2]

C1 C1 A1

Paper 04 [3]

B1

[1]

B1

B1 [1]

C1

B1 B1

A1

B1 [2]

C1

B1

A1

[2] C1

gravitational forces are (always attractive)

(b) (i)electric speed = 2π fa be attractive or repulsive forces can -3 = 2π 40 ×got42out × 10 for gravitational, ×work as masses come together -1 /mass moves from infinity = 10.6 m s for electric, work done on charges if same sign, work got out if opposite sign as charges together = 4π2 f2 a (ii)come acceleration 2

[1]

[4]

-3

= (80π) × 42 × 10 2 (a)

(i)

idea of heat lost (by oil)-2= heat gained (by thermometer) = 2650 32 x 1.4 x (54 – t) =m 12sx 0.18 x (t – 19) t = 52.4°C

(c) (i) S marked correctly (on ‘horizontal line through centre of wheel) (ii)

(c)

11 3 (a)

(b)

(c)

4 (a)

(b)

5 (a)

B1

thermistor thermometer (allow ‘resistance thermometer’) because small mass/thermal capacity

© temperature University ofisCambridge International Examinations 2006 boiling point constant further comment e.g. heating of bulb would affect only rate of boiling

[3]

A1

[1]

[2] B1 B1

[2]

M1 A1

[2]

use of a = – ω 2x clear either ω = √(2k/m) or ω 2 = (2k/m) ω = 2 πf f = (1/2 π)√(2 x 300)/0.240) = 7.96 ≈ 8 Hz

C1 B1 C1 B1 A0

[4]

(i)

resonance

B1

[1]

(ii)

8 Hz

B1

[1]

B1 B1 B1

[3]

B1 B1

[2]

2M x 6.67 x 10–11 {(26.28 x 106)–1 – (29.08 x 106)–1} = 53702 – 50902 M x 4.888 x 10–19 = 2.929 x 106 M = 6.00 x 1024 kg (If equation in (a) is dimensionally unsound, then 0/3 marks in (b), if dimensionally sound but incorrect, treat as e.c.f.)

B1 C1 A1

[3]

(i)

B1

(increase amount of) damping without altering (k or) m …(some indirect reference is acceptable) sensible suggestion (i)

(ii)

(b)

A1

B1

either ratio (= 1.6/54) = 0.030 or (=1.6/327) = 0.0049

(ii) A marked correctly (on ‘vertical line’ through centre of wheel) (b)

C1

GMm {(R + h1)–1 – (R + h2)–1} ½m {v12 – v22}

(induced) e.m.f proportional/equal to rate of change of flux (linkage) (allow ‘induced voltage, induced p.d.) flux is cust as the disc moves hence inducing an e.m.f

M1 A0

[2]

field in disc is not uniform/rate of cutting not same/speed of disc not same (over whole disc) so different e.m.f.’s in different parts of disc lead to eddy currents

B1 M1 A0

[2]

B1 B1 B1

[3]

eddy currents dissipate thermal energy in disc energy derived from oscillation of disc energy of disc depends on amplitude of oscillations

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© UCLES 2006

128

2. acceleration = 0

A1

Page Syllabus Paper (c) e.g.3 ‘radius’ of planet varies Mark Scheme density of planet notLEVEL constant GCE A/AS – October/November 2007 9702 04 12 planet spinning 3 (a) (i) nearby 0.8 cm planets ................................................................................................................... B1 / stars (any sensible comments, 1 mark each, maximum 2) B2 (ii) (max.) kinetic energy = 2.56 mJ ........................................................................... C1 v(MAX) = ωa ............................................................................................................ C1 2 2 2 2 2 (a) (Thermal) heat required unit2 (a mass to liquid M1 C1 aconvert or ½mω – xof ) solid ............................................ (max.)energy kinetic/ energy = ½mωto -3 2 -2 2 at its2.56 normal melting / without temperature A1 M1 × 10 = ½ ×point 0.130 × ω × any (0.8 change × 10 ) in .......................................................... -1 to ice → water scores max 1 mark) (reference to 1 kg or ω = 24.8 rad s ..................................................................................................... C1 f = ω/2π ................................................................................................................ M1 = 4.0 Hz (3.95 Hz) .............................................................................................. A0 B1 (b) (i) To make allowance for heat gains from the atmosphere

of production droplets from funnel (b) (ii) (i) e.g. line constant parallel torate x-axis at 2.56 mJof........................................................................... B1 constant mass of water collected per minute in beaker suggestion, 1 mark) B1 B1 (ii) (any 1 4.0sensible Hz ................................................................................................................ C1 B1 (iii) mass by heater 5 minutes = 64.7 – ½ × 16.6 = 56.4 g 2 0.50melted cm (allow ±0.03incm) ................................................................................ –3 56.4 × 10 × L = 18 C1 L = 320 kJ kg–1 A1 m = 64.7, giving L =from 278 sphere kJ kg–1, scores max 1 mark 4 (a) (i) (Use eitherof lines directed away use = 48.1, giving L = 374 kg–1, scores max 2 marks) or of mlines go from positive to kJ negative or line shows direction of force on positive charge ....................................... M1 13 so positively charged ............................................................................................ A1 3 (a) acceleration / force (directly) proportional to displacement M1 and either either all directed towardsto) fixed pointfrom centre (ii) lines (appear radiate & displacement in opposite directions A1 B1 oror acceleration all lines are normal to surface of sphere ...................................................

B1 B1 (b) (i) maximum / minimum height / 8 mm above cloth / 14 mm below cloth (b) tangent to curve ........................................................................................................... in correct position and direction ................................................................................... B1 A1 (ii) 1. a = 11 mm 2. ω = 2πf C1 Page 3 Mark Scheme-12 Syllabus Paper -9 2π × 4.5 (c) (i) V == (0.76 × 10 ) / (4π × 8.85 × 10 × 0.024) ..................................................... C1 GCE A/AS LEVEL – October/November 2008 9702 04 28.3Vrad s–1 (do not allow 1 s.f.) A1 A1 == 285 ...........................................................................................................

(i) v = ωa C1 (ii) negative charge is induced on (inside of) box ...................................................... M1 –3 = 28.3 × 11 × 10 formula applies–1 to isolated (point) charge = 0.31 m s (do not allow©1UCLES s.f.) 2008 A1 OR less work done moving test charge from infinity .......................................... A1 so potential is lower .............................................................................................. A1 (ii) v = ω √(a2 – y2) y = 3 mm C1 –3 2 2 √(11 – 3 ) C1 = 28.3 × 10 (d) either gravitational–1field is always attractive = 0.30 m must s (allow 1 s.f.) towards both box and sphere .............................. A1B1 or field lines be directed

[1]

[1] [2]

[2]

[6] [1] [1]

[1] [2]

[3]

[2]

[2] [1]

[1] [2] [1]

[2] [2]

(c)

4

(a) ∆U = q + w

(b) either

or

5

(allow correct word equation)

kinetic energy constant because temperature constant potential energy constant because no intermolecular forces so no change in internal energy kinetic energy and potential energy both constant (M1) so no change in internal energy (A1) reason for either constant k.e. or constant p.e. given (A1)

(a) change/loss in kinetic energy = change/gain in electric potential energy 2 ×COLLEGE ½mv2 = q2 / 4πε0 r CEDAR © UCLES 2007 2 × ½ × 2 × 1.67 × 10–27 × v2 = (1.6 × 10–19 )2 / (4π × 8.85 × 10–12 × 1.1 × 10–14 ) v = 2.5 × 106 m s–1

[2] [3]

[3] [1]

B1

[1]

M1 M1 A1

[3]

B1 C1

129

M1 A0

[3]

Page Page 33 14

43

Mark Mark Scheme: Scheme: Teachers’ Teachers’ version version GCE A/AS LEVEL – May/June 2009 2009 GCE A/AS LEVEL – October/November

Syllabus Syllabus 9702 9702

Paper Paper 04 41

(a) ω t (allow any subject if allat terms B1 (a) (i) e.g. (θ two=)objects of different masses samegiven) temperature (M1) same material would have different amount of heat (A1) (ii) =) r sinωt (allow any subject if all terms given) B1 e.g. (SQ temperature shows direction of heat transfer (M1) from high to low regardless of objects (A1) e.g. when substance melts/boils (M1) (b) this is the input solution equation achange = –ω2 x M1 heat but of nothe temperature (A1) 2 aany = two, –ω xM1 is the (defining) equation of s.h.m. A1 + A1 each, max 4 ………………………………..…………...........................

[1] [1]

[2] [4]

(c) = ω / 2πlosses (to the surroundings) .................................................................M1 C1 (b) (i) (i) fenergy either as the temperature rises = 4.7 / increase 2π or= 0.75 rise Hz is zero when heat loss = heat input ............................................... A1

[2]

(ii) videa of(rinput = maximum rate of heat loss ............................................. C1 (ii) = rω mustpower be identified) power m × c × ∆θ / ∆t = 4.7 = × 12 54 × c × 3.7 / 60 ..................................................................................... C1 cm s–1 A1 = = 560.96 c = 910 J kg-1 K-1 ............................................................................................... A1

[2] [3]

[Total: 9] (a) (i) ratio of charge (on body) and its potential B1 [1] (do not allow reference to plates of a capacitor) 15 4 (a) (ii) (i) (potential amplitudeat=surface 0.2 mmof sphere .......................................................................................... A1 [1] M1 =) V = Q / 4πε0 r C = Q / V = 4πε0 r A0 [1] (ii) period = 1.2 ms ................................................................................................. C1 frequency = 830 Hz .......................................................................................... A1 [2] (b) (i) C = 4 × π × 8.85 × 10–12 × 0.36 = 4.0 × 10–11 F (allow 1 s.f.) A1 [1] (b) (i) any two of zero, 0.6 ms and 1.2 ms .................................................................... A1 [1] (ii) Q = CV (ii) any= two ms, 0.9 ×ms, × 7.0 1051.5 ms ...................................................................... A1 [1] 4.0of × 0.3 10–11 –5 A1 [1] = 2.8 × 10 C

5

(c) either v = ωx0 = 2πfx0 m no s-1 free electrons = insulator 2π × 830/ not × 0.2 × 10-3 = 1.05 (c) plastic is an a conductor / has B1 or slope graph(on= an 1.0insulator) m s-1 ……(allow ± 0.1 m s-1) ....................................... C1 charges do notofmove B1 2 EK = ½mv either so no single value for the potential 2 × 2.5 ×cannot 10-3 × 1.05 ..................................................................................... C1 or = ½charge be considered to be at centre B1 -3 = 1.4 × 10 J ...................................................................................................... A1

[3] [3]

C1 (d) either energy = ½CV2 or energy = ½QV and C = Q/V –11 5 2 (d) energy (i) large= / ½ maximum vibration B1 × {(7.0 of × 10 ) – (2.5.............................................................. × 105 )2 )} C1 × 4 × 10amplitude when frequency equals natural frequency of vibration ...................... B1 A1 = impressed 8.6 J

[2] [3]

(ii) e.g. metal panels on machinery vibrate / oscillate ........................................... (M1) motor in machine impresses frequency on panel ......................................(A1) e.g. car suspension system vibrates / oscillates................................................. (M1) going over bumps would give large amplitude vibrations .............................(A1) any feasible example, M1 + A1 ...............................................................................




[2]

[Total: 12]

130

CEDAR COLLEGE

© © UCLES UCLES 2009 2009

M1 (a) work done moving unit mass from infinity to the point [2] (iii) realisation that total internal energy is the total kinetic energy C1A1 –21 23 C1 energy = 6.0 × 10 × 68 × 6.02 × 10 = 2.46 × 105 J A1 [3] B1 (b) (i) at R, φ = 6.3 × 107 J kg–1 (allow ± 0.1 × 107) 16 φ = GM / R –11 6 6.3 × 107 /=backward (6.67 × 10 × M) / (6.4 × 10(between ) 3 (a) (i) to-and-fro and forward motion two limits) B1C1 [1] 24 M = 6.0 × 10 kg (allow 5.95 → 6.14) A1 [3] force / constant energy / constant amplitude (ii) no energy loss or for gain / no external Maximum of 2/3 any value chosen foracting φ not at R B1 [1] 7 –1 7 (ii) change in potential = 2.1 × 10 J kg (allow ± 0.1 × 10 ) C1 (iii) acceleration directed towards a fixed point energy B1B1 loss in potential energy = gain in kinetic B1C1 [2] acceleration m or ½ mv 2to=distance GM / 3Rfrom the fixed point / displacement ½ mv 2 = φ proportional 2 7 ½ v = 2.1 × 10 v = 6.5 × 103 m s–1 ………..……(allow 6.3 → 6.6) A1 [4] (b) acceleration is constant M1 3 (magnitude) –1 (answer 7.9 × 10 m s , based on x = 2R, allow max 3 marks) so cannot be s.h.m. A1 [2] 1

(iii) e.g. speed / velocity / acceleration would be greater deviates / bends from straight path (any sensible ideas, 1 each, max 2) © UCLES 2010 17 2 (a) (i) reduction in energy (of the oscillations) reduction in amplitude / energy of oscillations due to force (always) opposing motion / resistive forces any two of the above, max 2

(ii) amplitude is decreasing (very) gradually / oscillations would continue (for a long time) /many oscillations light damping

(b) (i) frequency = 1 / 0.3 = 3.3 Hz allow points taken from time axis giving f = 3.45 Hz (ii) energy

= ½ mv 2 and v = ωa = ½ × 0.065 × (2π/0.3)2 × (1.5 × 10–2)2 = 3.2 mJ

(c) amplitude reduces exponentially / does not decrease linearly so will be not be 0.7 cm

B1 B1

[2]

(B1) (B1) (B1)

[2]

M1 A1

[2]

A1

[1]

C1 M1 A0

[2]

M1 A1

[2]

© UCLES 2010

CEDAR COLLEGE

131

Page 3

Mark Scheme: Teachers’ version GCE AS/A LEVEL – October/November 2010

18 3 (a) (i) resonance

Syllabus 9702

B1

[1]

A1

[1]

(b) (i) a = (–)ω2x and ω = 2πf a = 4π2 × 4.62 × 16 × 10–3 = 13.4 m s–2

C1 C1 A1

[3]

(ii) F = ma = 150 × 10–3 × 13.4 = 2.0 N

C1

(ii) amplitude 16 mm and frequency 4.6 Hz

(c) line always ‘below’ given line and never zero peak is at 4.6 Hz (or slightly less) and flatter

4

(a) charge / potential (difference) (ratio must be clear) Page 3 Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 19 (b) (i) (i) 8.0 V = cm Q / 4πε0r 3 (a)

Syllabus 9702

(ii) 2πf C=Q / V = 4πε0r and 4πε0 is constant (ii) = 220 so C ∝ f = 35 r (condone unit)

4

Paper 41

(iii) line drawn mid-way between AB and CD (allow ±2 mm) (c) (i) r = C / 4πε0r –12 ) / (4π × 8.85 × 10–12) ωa × 10 (iv) vr ==(6.8 –2 = 6.1 × 10 m = 220 × 4.0 = 880 cm s–1 (ii) Q = CV = 6.8 × 10–12 × 220 = 1.5 × 10–9 C (b) (i) 1. line drawn 3 cm above AB (allow ±2 mm) 2. arrow pointing upwards (d) (i) V = Q/C = (1.5 × 10–9) / (18 × 10–12) = 83line V drawn 3 cm above AB (ii) 1. (allow ±2 mm) 2. arrow pointing downwards (ii) either energy = ½CV 2 –12 2 –12 2 2 –=x2½ ) × 6.8 × 10 × 220 – ½ × 18 × 10 × 83 (iii) v = ω√(a∆E –7 2 –8 2 = 220 =× 1.65 √(4.0× 10 – 2.0–)6.2 × 10 –7 1.03 = 760 =cm s–1 × 10 J or energy = ½QV (incorrect value for x, 0/2 marks) ∆E = ½ × 1.5 × 10–9 × 220 – ½ × 1.5 × 10–9 × 83 = 1.03 × 10–7 J (a) (i) work done moving unit positive charge from infinity to the point (ii) charge / potential (difference)

(ratio must be clear)

(b) (i) capacitance = (2.7 × 10–6 ) / (150 × 103 ) (allow any appropriate values) (allow 1.8 ±0.05) capacitance = 1.8 × 10–11 (ii) either energy = ½CV 2 or energy = ½QV and Q = CV × 103 )2 2010 or ½ × 2.7 × 10–6 × 150 × 103 energy = ½ × 1.8 × 10–11 × (150 © UCLES = 0.20 J

CEDAR COLLEGE

A1

[2]

M1 A1

[2]

B1 [1] Paper 43 B1 A1

[1] [1]

M1 C1 A0 A1

[1] [2]

B1 C1 C1 C1 A1 A1

[1]

[3] [2]

A1 B1 B1

[1] [1] [1]

A1 B1 B1 C1 C1 C1 A1 (C1) (C1) (A1) M1 A1

[1] [1]

[3] [2]

B1

[1]

[2]

C1 A1

[2]

C1

132

A1

[2]

substitutes temperature as 298 K either 1.1 × 105 × 6.5 × 10–2 = N × 1.38 × 10–23 × 298 or 1.1 × 105 × 6.5 × 10–2 = n × 8.31 × 298 and n = N / 6.02 × 1023 N = 1.7 × 1024

C1 C1 A1

[4]

20

3

(a) acceleration / force proportional to displacement from a fixed point acceleration / force (always) directed towards that fixed point / in opposite direction to displacement

(b) (i) Aρg / m is a constant and so acceleration proportional to x negative sign shows acceleration towards a fixed point / in opposite direction to displacement (ii) ω 2 = (Aρg / m) ω = 2πf (2 × π × 1.5)2 = ({4.5 × 10–4 × 1.0 × 103 × 9.81} / m) m = 50 g Page 3

4 3

3

4

4

5

5

Mark Scheme: Teachers’ version GCE AS/A LEVEL – October/November 2011 (a) work done in bringing unit positive charge 21 from infinity (to that point) (a) (i) amplitude remains constant

Syllabus 9702

M1 A1

[2]

B1 B1

[2]

C1 C1 C1 A1

[4]

Paper 41 M1 A1 [2] B1 [1]

(ii) amplitude decreases gradually (b) (i) field strength is potential gradient Page 3 light damping Mark Scheme: Teachers’ version Syllabus GCE AS/A LEVEL – May/June 2012 9702 (ii) field strength proportional to force (on particle Q) (iii) period = 0.80 s potential gradient proportional to gradient of (potential energy) graph frequencyequal = 1.25 Hz (period not 0.8 s,energy then 0/2) (a) (numerically to) quantity of (thermal) required to change so force is proportional to the gradient of the graph the state of unit mass of a substance without any change of temperature (b) (i) (induced) is proportional to latent heat of fusion/vaporisation) (Allow 1 mark e.m.f. for definition of specific rate of change/cutting of (magnetic) flux (linkage)

M1 B1 A1 Paper B1 41 C1 B1 A1 A0 M1 A1 M1 A1

a current is induced coil × 2 × 60 = 288000 J (b) (ii) either energy suppliedin=the 2400 as magnet moves in coil energy required for evaporation = 106 × 2260 = 240000 J current in resistor to a heating effect Examinations 2011 © University ofrise International JCambridge difference = gives 48000 thermal energy is derived from energy of rate of loss = 48000 / 120 = 400 Woscillation of the magnet or energy required for evaporation = 106 × 2260 = 240000 J power required for evaporation = 240000 / (2 × 60) = 2000 W (a) (i) zero rate field of (strength) inside spheres loss = 2400 – 2000 = 400 W

C1 C1

22 (ii) (a) a T a

either field strength is zero in opposite directions andfields ω = are 2π/T =or (–)ω2x the = 0.60 s at a point between the spheres = (4π2 × 2.0 × 10–2 ) / (0.6)2 = 2.2 m s–2 (b) (i) field strength is (–) potential gradient (not V/x) 1. field strength has maximum value (b) (ii) sinusoidal wave with all values positive atpositive, x = 11.4allcm all values peaks at EK and energy = 0 at t = 0 period = 0.30 s 2. field strength is zero either at x = 7.9 cm (allow ±0.3 cm) or at 0 to charge 1.4 cm acting or 11.4 to 12 cm (a) force per unit positive on a cm stationary charge

2 (a) or Bqv(cosθ) (b) (i) (i) Bqv(sinθ) E = Q / 4πε 0r 4 Q = 1.8 × 10 × 102 × 4π × 8.85 × 10–12 × (25 × 10–2 )2 (ii) qE Q = 1.25 × 10–5 C = 12.5 µC

M1 A1 M1 A1

[2] [2] [2]

[2]

[4]

A1 (C1) (C1) (A1) B1

[1] [3]

M1 A1

[2]

B1

[3] [1]

C1 C1 A1

B1 B1 B1

B1 B1

[2] [3]

B1 B1

B1

[2] [1]

C1 M1 A0

B1

[1]

B1

[1] [2]

133

CEDAR (ii) COLLEGE V = Q / 4πε0 r

(b) FB must be opposite–5in direction to FE –12 = (1.25 × 10 ) / (4π × 8.85 × 10 × 25 × 10–2 ) plane of paper so magnetic field into = 4.5 × 105 V (Do not allow use of V = Er unless explained)

[1] [2]

C1 A1

B1 B1

[2] [2]

∆EP = GMm/r1 – GMm/r2 Correct substitution 8.0 × 1018 J (∆EP = GMm/r1 + GMm/r2 is incorrect physics so 0/3)

or

2

C1 B1 A1

23 (a) energy = ½mω2a2 and ω = 2πf = ½ × 37 × 10–3 × (2π × 3.5)2 × (2.8 × 10–2)2 = 7.0 × 10–3 J (allow 2π × 3.5 shown as 7π)

C1 M1 A0

Energy = ½ mv2 and v = rω Correct substitution Energy = 7.0 × 10–3 J

(b) EK = EP ½mω2 (a2 – x2) = ½mω2x2 or EK or EP = 3.5 mJ x = a/√2 = 2.8 /√2 or EK = ½mω2 (a2 – x2) = 2.0 cm (EK or EP = 7.0 mJ scores 0/3) Allow: Page 3

k = 17.9 E = ½ kx2Mark Scheme: Teachers’ version x = 2.0 cm GCE AS/A LEVEL – May/June 2012

(c) (i) graph:

(ii) graph:

(C1) (M1) (A0)

or EP = ½mω2x2

reasonable curve with maximum at (0,7.0) end-points of line at (–2.8, 0) and (+2.8, 0)

© University of Cambridge International Examinations 2012 inverted version of (ii) with intersections at (–2.0, 3.5) and (+2.0, 3.5) (Allow marks in (iii), but not in (ii), if graphs K & P are not labelled)

(iii) graph:

(d) gravitational potential energy Page 3 Mark Scheme Syllabus GCE AS/A LEVEL – May/June 2013 9702 3 (a) sum of potential energy and kinetic energy of atoms/molecules/particles 24 reference to random (distribution) 3 (a) (i) ω = 2π / T = 2π / 0.69 9.1 rad s–1 (b) (i) as = lattice structure is ‘broken’/bonds broken/forces between (allow use reduced of f = 1.5(not Hz to give ω = separate) 9.4 rad s–1) molecules molecules no change in kinetic energy, potential energy increases (ii) internal 1. x = energy 2.1 cosincreases 9.1t 2.1 and 9.1 numerical values of cos (ii) eitheruse molecules/atoms/particles move faster/ is increasing or kinetic energy increases with temperature (increases) 10–2 × 9.1 (allowkinetic ecf of energy value ofincreases x0 from (ii)1.) 2. change v0 = 2.1 no in ×potential energy, –1 v0 =energy 0.19 mincreases s internal v = v0 sin 9.1t (allow cos 9.1t if sin used in (ii)1.)

(a) (i) as r decreases, energy decreases/work got out (due to) 2 2 so½point negatively charged attraction or ½ is mω x0 (b) energy = either mv02mass 2 = either ½ × 0.078 × 0.19 or ½ × 0.078 × 9.12 × (2.1 × 10–2)2 (ii) electric = 1.4potential × 10–3 J energy = charge × electric potential electric field strength is potential gradient field strength = gradient of potential energy graph/charge CEDAR COLLEGE 4 (a) (i) V = q / 4πε0R (b) tangent drawn at (4.0, 14.5) (ii) (capacitance is) ratio of charge and potential or q /V –24 gradient 3.6= ×4πε 100R C == q /V

C1 C1 A1

[3]

(C1) Syllabus (C1) Paper (A1) 9702 42

horizontal line, y-intercept = 7.0 mJ with end-points of line at +2.8 cm and –2.8 cm

4

[2]

B1

[1]

B1 B1

[2]

M1 A1

[2]

B1

[1]

M1 A1

Paper 42

[2] C1 A1

B1 M1 A1

B1 M1 A1

[3] B1 B1

[2]

B1 B1

[3] [2]

M1 A1 B1 B1 A0

[2]

[2] C1 A1

[2]

[2] B1 134 [1]

B1 A2

M1 A0

[1]

Page 3

Mark Scheme Cambridge International AS/A Level – October/November 2014

25 4 (a) acceleration / force proportional to displacement (from a fixed point) either acceleration and displacement in opposite directions or acceleration always directed towards a fixed point

Syllabus 9702

M1

(b) (i) g and r are constant so a is proportional to x negative sign shows a and x are in opposite directions (ii) ω 2 = g / r and ω = 2π / T ω 2 = 9.8 / 0.28 = 35 Page 3 Mark Scheme T = 2π / √35 = 1.06 s Cambridge AS / A Level – May / June 2015 time interval τ =International 0.53 s

Syllabus 9702

B1 B1

[2]

C1 Paper 41 [3] A1

[3]

A1

[3]

C1 A1 M1 A1

[2]

(a) (i) frequency at which object is made to vibrate/oscillate

B1

[1]

(b) (i) inside the sphere, the potential would be constant (ii) frequency at which object vibrates when free to do so

B1 B1

[1] [1]

B1 B1 M1 B1 A1

[2] [3]

(ii) for point charge, Vx is constant (iii) maximum amplitude of determines vibration of oscillating co-ordinates clear and two valuesbody of Vx at least 4 cm apart when forced frequency equals natural frequency (of vibration) conclusion made clear

(b) (c) e.g. q =vibration 4πε0Vx of quartz/piezoelectric crystal (what is vibrating) either for accurate timing q = 4π × 8.85 × 10–12 × 180 × 1.0 × 10–2 or maximise amplitude of ultrasound waves (why it is useful) = 2.0 × 10–10 C

5

[2]

C1 C1 M1 A1 A1

successive decrease in peak height 2. either 70 + h = 0.26 × 330 or 110 + h = 0.38 × 330 h = 17 / 16 / 15 W 5 (a) work done in moving unit positive charge from infinity (to the point) 26

6

A1

C1

(ii) 1. P + h = mL or substitution of one set of values 70) = constant (0.38 – 0.26)L (c) sketch:(110 time–period (or increases very slightly) –1 J g L = 330 drawn line always ‘inside’ given loops

4

Paper 41

(c) e.g. vibrating metal panels (what is vibrating) (a) F =either BILsinθ place strengthening struts across the panel –2 =or 2.6 × 10–3 shape/area × 5.4 × 4.7 ×of10 × sin 34°it is reduced) change panel (how –4 = 3.69 × 10 N (allow 1 mark for use of cos 34°) (a) (magnitude of electric field strength is the potential gradient use of gradient at x = 4.0 cm (b) peakgradient current == 4.5 1.7 ×× 10 √24 N C–1 (allow ± 0.3 × 10 4 ) = 2.4 A or max. force = 2.6 × 10–3 × 2.4 × 4.7 × 10–2 × sin 34° –4 V Q= 1.64 × 10 NQ and E = leading to E = V = 2 x 4πε0 x 4πε x variation = 2 × 1.64 × 10–4 0 = 3.3 × 10–4 N E = 1.8 × 103 / 0.04 = 4.5 × 104 N C–1

(b) (i) 3.6 × 103 V

M1

M1 A1 A1

–9

[2] [2]

M1 C1 A1 A1

[2] [2]

B1 M1 C1 A1

C1 (B1) A1 [3] (M1) (A1) [3]

A1

CEDAR (ii) COLLEGE capacitance = Q /©VCambridge International Examinations 2014

[2]

[1]

C1 135

3

= (8.0 × 10 ) / (3.6 × 10 ) = 2.2 × 10–12 F

A1

[2]

Page 3

Mark Scheme Cambridge International AS/A Level – May/June 2016

27 3 (a) acceleration/force proportional to displacement (from fixed point) Page 3 Mark Scheme Cambridge International AS/A Level – May/June 2016 acceleration/force and displacement in opposite directions

Syllabus 9702

Paper 41

Syllabus 9702

M1 Paper 42 A1 [2]

(ii) loss is 5 / 100 × 150 mol = 7.5 mol (b) maximum displacements/accelerations are different or ∆Nis =curved/not 4.52 × 1024a straight line graph

C1 B1

t = (7.5 × 6.02 × 1023 ) / 1.5 × 1019 or = 2π / T and T = 0.8 s (c) (i) ω t = 4.52 × 1024 / 1.5 × 1019

C1 C1

ω = 7.9 rad5s–1 = 3.0 × 10 s

B1

3

A1 A1

(ii) a = (–)ω2 x = 7.852 × 1.5 × 10–2 (a) no net energy transfer between the bodies or = 0.93 m s–2 or 0.94 m s–2 bodies are at the same temperature

[2] [3]

C1 A1 B1

(iii) ∆E = ½ mω2 (x02 – x2)

[2] [1]

C1

(b) (i) thermocouple, platinum/metal resistance thermometer, pyrometer = ½ × 120 × 10–3 × 7.852 × {(1.5 × 10–2)2 – (0.9 × 10–2)2}

4

[2]

B1 C1

[1]

(ii) thermistor, thermocouple = 5.3 × 10–4 J

B1 A1

[1] [3]

(c) (i) change = 11.5 K (a) (i) product of speed and density

B1 M1

[1]

B1 A1

[1] [2]

(ii) final temperature = 311.2 K reference to speed in medium (and density of medium) 28

4

intensity and/to incident intensity and ω = 2π /T (a) (ii) (i) α: T =ratio 0.60ofs reflected

B1 C1

Z 2: (specific) –1 (10.47) rad s acoustic impedances of media (on each side of boundary) B1 A1 ω1 =and 10 Z

(ii) energy = ½mω2x0 2 or ½mv2 and v = ωx0 (b) in muscle: IM = I0 e–µx × 10–22 )× (2.0 × 10–2)2 × (10.5) ==½I0×exp(–23 120 × 10×–33.4

[2] [2]

C1

IM / I0 = 0.457= 2.6 × 10–3 J

C1

at boundary: α = (6.3 – 1.7)2 / (6.3 + 1.7)2 = 0.33in correct directions (b) sketch: smooth curve

C1 A1

[2]

C1 B1

= [(1 Ipeak T /IM at f – α) =] 0.67

C1 M1

× 0.67 Iamplitude T / I0 = 0.457 never zero and line extends from 0.7fto 1.3f = 0.31

A1 A1

(c) sketch: peaked line always below a peaked line A

[3] [5]

peak not as sharp and at (or slightly less than) frequency of peak in line A

M1 A1

[2]




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Mark Scheme Cambridge International AS/A Level – May/June 2016

29 3 (a) acceleration/force proportional to displacement (from fixed point)

Syllabus 9702

M1

acceleration/force and displacement in opposite directions

A1

(b) maximum displacements/accelerations are different

B1

(c) (i) ω = 2π / T and T = 0.8 s

A1

(ii) a = (–)ω2 x = 7.852 × 1.5 × 10–2

A1

(iii) ∆E = ½ mω2 (x02 – x2)

(ii) α: ratio of reflected intensity and/to incident intensity




C1 A1

[3]

M1 A1

[2]

B1

Z1 and Z2: (specific) acoustic impedances of media (on each side of boundary) B1

(b) in muscle: IM = I0 e–µx = I0 exp(–23 × 3.4 × 10–2)

[2]

C1

= ½ × 120 × 10–3 × 7.852 × {(1.5 × 10–2)2 – (0.9 × 10–2)2}

reference to speed in medium (and density of medium)

[2]

C1

= 0.93 m s–2 or 0.94 m s–2

(a) (i) product of speed and density

[2]

C1

ω = 7.9 rad s–1

= 5.3 × 10–4 J

[2]

B1

graph is curved/not a straight line

4

Paper 43

[2]

C1

IM / I0 = 0.457

C1

at boundary: α = (6.3 – 1.7)2 / (6.3 + 1.7)2 = 0.33

C1

IT /IM = [(1 – α) =] 0.67

C1

IT / I0 = 0.457 × 0.67 = 0.31

A1

[5]

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A2 ELECTRIC FIELDS


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1 May/June 2002 , Question #5 , qp_4




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2 May/June 2003 , Question #4 , qp_4

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3 May/June 2005 , Question #5, qp_4 5

10

An isolated conducting sphere of radius r is given a charge +Q. This charge may be assumed to act as a point charge situated at the centre of the sphere, as shown in Fig. 5.1.

+Q

r Fig. 5.1 Fig. 5.2. shows the variation with distance x from the centre of the sphere of the potential V due to the charge +Q.

V

0

0

r

2r

3r

4r

x

Fig. 5.2 (a) State the relation between electric field and potential. ......................................................................................................................................[1]

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For Examiner’s Use

11 (b) Using the relation in (a), on Fig. 5.3 sketch a graph to show the variation with distance x of the electric field E due to the charge +Q.

For Examiner’s Use

E

0 0

r

2r

3r

4r

x [3]

Fig. 5.3

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4 May/June 2007 , Question #3 , qp_4 3

For Examiner’s Use

8

Two charged points A and B are separated by a distance of 6.0 cm, as shown in Fig. 3.1. 6.0 cm A

B d Fig. 3.1

The variation with distance d from A of the electric field strength E along the line AB is shown in Fig. 3.2. 20 E / kV m–1 15

10

5

0 0

2

4

6

d /cm position of A

position of B Fig. 3.2

An electron is emitted with negligible speed from A and travels along AB. (a) State the relation between electric field strength E and potential V. .......................................................................................................................................... ..................................................................................................................................... [2]

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9 (b) The area below the line of the graph of Fig. 3.2 represents the potential difference between A and B.

For Examiner’s Use

Use Fig. 3.2 to determine the potential difference between A and B.

potential difference = …………………………. V [4] (c) Use your answer to (b) to calculate the speed of the electron as it reaches point B.

speed = …………………………. m s–1 [2] (d) (i)

Use Fig. 3.2 to determine the value of d at which the electron has maximum acceleration. d = …………………… cm [1]

(ii)

Without any further calculation, describe the variation with distance d of the acceleration of the electron. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [2]

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5 Oct/Nov 2007 , Question #4 , qp_4 4

10

A small charged metal sphere is situated in an earthed metal box. Fig. 4.1 illustrates the electric field between the sphere and the metal box.

For Examiner’s Use

A

Fig. 4.1 (a) By reference to Fig. 4.1, state and explain (i)

whether the sphere is positively or negatively charged, .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2]

(ii)

why it appears as if the charge on the sphere is concentrated at the centre of the sphere. .................................................................................................................................. ..............................................................................................................................[1]

(b) On Fig. 4.1, draw an arrow to show the direction of the force on a stationary electron situated at point A. [2]

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11 (c) The radius r of the sphere is 2.4 cm. The magnitude of the charge q on the sphere is 0.76 nC. (i)

For Examiner’s Use

Use the expression V=

Q 4

0r

to calculate a value for the magnitude of the potential V at the surface of the sphere.

V = ...............................................V [2] (ii)

State the sign of the charge induced on the inside of the metal box. Hence explain whether the actual magnitude of the potential will be greater or smaller than the value calculated in (i). .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[3]

(d) A lead sphere is placed in a lead box in free space, in a similar arrangement to that shown in Fig. 4.1. Explain why it is not possible for the gravitational field to have a similar shape to that of the electric field. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[1]

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6 May/June 2008 , Question #4 , qp_4 (i)

11

State the value of x at which the potential is zero. x = ........................................... cm [1]

(ii)

For Examiner’s Use

Use your answer in (i) to determine the charge at B.

charge = ........................................... C [3] (c) A small test charge is now moved along the line AB in (b) from x = 5.0 cm to x = 27 cm. State and explain the value of x at which the force on the test charge will be maximum. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ...................................................................................................................................... [3]

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7 Oct/Nov 2009 , Question #5 , qp_41 5

10

(a) Define electric potential at a point. ..........................................................................................................................................

For Examiner’s Use

.......................................................................................................................................... .................................................................................................................................... [2] (b)

An α-particle is emitted from a radioactive source with kinetic energy of 4.8 MeV. The α-particle travels in a vacuum directly towards a gold ( 197 79Au) nucleus, as illustrated in Fig. 5.1. gold nucleus

path of α - particle Fig. 5.1

The α-particle and the gold nucleus may be considered to be point charges in an isolated system. (i)

Explain why, as the α-particle approaches the gold nucleus, it comes to rest. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................ [2]

(ii)

For the closest approach of the α-particle to the gold nucleus determine 1. their separation,

separation = ........................................... m [3]

150

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11 2. the magnitude of the force on the α-particle.

For Examiner’s Use

force = .......................................... N [2]

151

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8 May/June 2010 , Question #4 , qp_41 4

8

(a) Explain what is meant by the potential energy of a body. ..........................................................................................................................................

For Examiner’s Use

.......................................................................................................................................... ...................................................................................................................................... [2] 2

(b) Two deuterium ( 1 H) nuclei each have initial kinetic energy EK and are initially separated by a large distance. The nuclei may be considered to be spheres of diameter 3.8 × 10–15 m with their masses and charges concentrated at their centres. The nuclei move from their initial positions to their final position of just touching, as illustrated in Fig. 4.1. 2 1H

2 1H

kinetic energy EK

kinetic energy EK

initially

3.8 × 10–15 m 2 1H

finally

2 1H

at rest Fig. 4.1 (i)

For the two nuclei approaching each other, calculate the total change in 1. gravitational potential energy,

energy = ............................................ J [3] 2. electric potential energy.

energy = ............................................ J [3] © UCLES 2010

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9 (ii)

Use your answers in (i) to show that the initial kinetic energy EK of each nucleus is 0.19 MeV.

For Examiner’s Use

[2] (iii)

The two nuclei may rebound from each other. Suggest one other effect that could happen to the two nuclei if the initial kinetic energy of each nucleus is greater than that calculated in (ii). .................................................................................................................................. .............................................................................................................................. [1]

153

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9 May/June 2010 , Question #7 , qp_41 7

13

Negatively-charged particles are moving through a vacuum in a parallel beam. The particles have speed v. The particles enter a region of uniform magnetic field of flux density 930 μT. Initially, the particles are travelling at right-angles to the magnetic field. The path of a single particle is shown in Fig. 7.1.

negatively-charged

For Examiner’s Use

arc of radius 7.9 cm

particles, speed v

uniform magnetic field, flux density 930 μT Fig. 7.1 The negatively-charged particles follow a curved path of radius 7.9 cm in the magnetic field. A uniform electric field is then applied in the same region as the magnetic field. For an electric field strength of 12 kV m–1, the particles are undeviated as they pass through the region of the fields. (a) On Fig. 7.1, mark with an arrow the direction of the electric field.

[1]

(b) Calculate, for the negatively-charged particles, (i)

the speed v,

v = ....................................... m s–1 [3] (ii)

the ratio

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charge . mass

ratio = .................................... C kg–1 [3] 9702/41/M/J/10 [Turn over

154

as illustrated in Fig. 4.1. particle P

particle Q

x

10 May/June 2011 , Question #4 , qp_41

Fig. 4.1

The variation with separation x of the electric potential energy EP of particle Q is shown in Fig. 4.2. 0

0

2

4

6

8

10

12

14

16

x / 10–10 m –1

–2 E P / eV –3

–4 Fig. 4.2 (i)

State how the magnitude of the electric field strength is related to potential gradient. .................................................................................................................................. .............................................................................................................................. [1]

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9 (ii)

Use your answer in (i) to show that the force on particle Q is proportional to the gradient of the curve of Fig. 4.2.

For Examiner’s Use

.................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2] (c) The magnitude of the charge on each of the particles P and Q is 1.6 × 10–19 C. Calculate the separation of the particles at the point where particle Q has electric potential energy equal to –5.1 eV.

separation = ............................................ m [4] (d) By reference to Fig. 4.2, state and explain (i)

whether the two charges have the same, or opposite, sign, .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

(ii)

the effect, if any, on the shape of the graph of doubling the charge on particle P. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

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11 Oct/Nov 2011 , Question #4 , qp_41 4

10

Two small charged metal spheres A and B are situated in a vacuum. The distance between the centres of the spheres is 12.0 cm, as shown in Fig. 4.1. 12.0 cm sphere A

sphere B

P

x Fig. 4.1 (not to scale) The charge on each sphere may be assumed to be a point charge at the centre of the sphere. Point P is a movable point that lies on the line joining the centres of the spheres and is distance x from the centre of sphere A. The variation with distance x of the electric field strength E at point P is shown in Fig. 4.2. 150 E / 106 N C–1 100

50

0

0

2

4

6

8

x / cm

10

12

–50

–100

–150

–200 Fig. 4.2

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For Examiner’s Use

11 (a) State the evidence provided by Fig. 4.2 for the statements that (i)

For Examiner’s Use

the spheres are conductors, .................................................................................................................................. .............................................................................................................................. [1]

(ii)

the charges on the spheres are either both positive or both negative. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

(b) (i)

State the relation between electric field strength E and potential gradient at a point. .................................................................................................................................. .............................................................................................................................. [1]

(ii)

Use Fig. 4.2 to state and explain the distance x at which the rate of change of potential with distance is 1. maximum, .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2] 2. minimum. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

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12 Oct/Nov 2011 , Question #5 , qp_41 5

12

Positively charged particles are travelling in a vacuum through three narrow slits S1, S2 and S3, as shown in Fig. 5.1. S1

S2

For Examiner’s Use

S3

beam of charged particles direction of electric field Fig. 5.1 Each particle has speed v and charge q. There is a uniform magnetic field of flux density B and a uniform electric field of field strength E in the region between the slits S2 and S3. (a) State the expression for the force F acting on a charged particle due to (i)

the magnetic field, .............................................................................................................................. [1]

(ii)

the electric field. .............................................................................................................................. [1]

(b) The electric field acts downwards in the plane of the paper, as shown in Fig. 5.1. State and explain the direction of the magnetic field so that the positively charged particles may pass undeviated through the region between slits S2 and S3. .......................................................................................................................................... .......................................................................................................................................... ...................................................................................................................................... [2]

159

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13 May/June 2012 , Question #5 , qp_41 5

11

(a) Define electric field strength. ..........................................................................................................................................

For Examiner’s Use

..................................................................................................................................... [1] (b) An isolated metal sphere is to be used to store charge at high potential. The charge stored may be assumed to be a point charge at the centre of the sphere. The sphere has a radius of 25 cm. Electrical breakdown (a spark) occurs in the air surrounding the sphere when the electric field strength at the surface of the sphere exceeds 1.8 × 104 V cm–1. (i)

Show that the maximum charge that can be stored on the sphere is 12.5 μC.

[2] (ii)

Calculate the potential of the sphere for this maximum charge.

potential = ............................................. V [2]

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14 May/June 2012 , Question #4 , qp_42 4

10

A charged point mass is situated in a vacuum. A proton travels directly towards the mass, as illustrated in Fig. 4.1. charged point mass

For Examiner’s Use

proton r Fig. 4.1

When the separation of the mass and the proton is r, the electric potential energy of the system is UP . The variation with r of the potential energy UP is shown in Fig. 4.2.

0

0

2

4

6

8

r / cm

10

–10 UP / 10–26 J –20

–30

–40

–50 Fig. 4.2

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11 (a) (i)

Use Fig. 4.2 to state and explain whether the mass is charged positively or negatively.

For Examiner’s Use

.................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2] (ii) The gradient at a point on the graph of Fig. 4.2 is G. Show that the electric field strength E at this point due to the charged point mass is given by the expression Eq = G where q is the charge at this point. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2] (b) Use the expression in (a)(ii) and Fig. 4.2 to determine the electric field strength at a distance of 4.0 cm from the charged point mass.

field strength = ........................................ V m–1 [4]

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15 May/June 2012 , Question #6 , qp_42 6

14

(a) Describe the main principles of the determination of the charge on an oil drop by Millikan’s experiment. You may draw a diagram if you wish.

For Examiner’s Use

.......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ...................................................................................................................................... [7] (b) In an experiment to determine the fundamental charge, values of charge on oil drops were found by a student to be as shown below. 3.2 × 10–19 C;

6.4 × 10–19 C;

12.8 × 10–19 C;

16 × 10–19 C;

3.1 × 10–19 C;

9.7 × 10–19 C;

6.3 × 10–19 C.

State the value, to two significant figures, of the fundamental charge that is suggested by these values of charge on oil drops.

fundamental charge = .............................................. C [1] © UCLES 2012

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16 Oct/Nov 2012 , Question #3 , qp_43 3

8

(a) State what is meant by a line of force in (i)

For Examiner’s Use

a gravitational field, .................................................................................................................................. .............................................................................................................................. [1]

(ii)

an electric field. .................................................................................................................................. .............................................................................................................................. [2]

(b) A charged metal sphere is isolated in space. State one similarity and one difference between the gravitational force field and the electric force field around the sphere. similarity: .......................................................................................................................... .......................................................................................................................................... difference: ........................................................................................................................ .......................................................................................................................................... .......................................................................................................................................... [3] (c) Two horizontal metal plates are separated by a distance of 1.8 cm in a vacuum. A potential difference of 270 V is maintained between the plates, as shown in Fig. 3.1. 0V

proton 1.8 cm

+270 V Fig. 3.1 A proton is in the space between the plates. Explain quantitatively why, when predicting the motion of the proton between the plates, the gravitational field is not taken into consideration.

[4] © UCLES 2012

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17 Oct/Nov 2013 , Question #4 , qp_41 4

8

An α-particle and a proton are at rest a distance 20 μm apart in a vacuum, as illustrated in Fig. 4.1. 20 m -particle

proton

P x Fig. 4.1

(a) (i)

State Coulomb’s law. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

(ii)

The α-particle and the proton may be considered to be point charges. Calculate the electric force between the α-particle and the proton.

force = ............................................. N [2] (b) (i)

Define electric field strength. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

165

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9 (ii)

A point P is distance x from the α-particle along the line joining the α-particle to the proton (see Fig. 4.1). The variation with distance x of the electric field strength Eα due to the α-particle alone is shown in Fig. 4.2.

For Examiner’s Use

300 E 200 electric field strength / V m–1 100

0

0

2

4

6

8

10

12

14

x/ m

16

EP –100

–200

–300 Fig. 4.2 The variation with distance x of the electric field strength EP due to the proton alone is also shown in Fig. 4.2. 1.

Explain why the two separate electric fields have opposite signs.

.................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2] 2.

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On Fig. 4.2, sketch the variation with x of the combined electric field due to the α-particle and the proton for values of x from 4 μm to 16 μm. [3]

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18 Oct/Nov 2013 , Question #3 , qp_43 3

8

(a) Define electric potential at a point.

8

3 .......................................................................................................................................... (a) Define electric potential at a point. For

For Examiner’s Use

Examiner’s Use .......................................................................................................................................... ..........................................................................................................................................

..........................................................................................................................................

...................................................................................................................................... [2] ...................................................................................................................................... [2]

(b) Two point charges A and B are separated by a distance of 20 nm in a vacuum, as (b) Two point charges illustrated in Fig. 3.1. A and B are separated by a distance of 20 nm in a vacuum, as illustrated in Fig. 3.1.

20 nm 20 nm

P P

AA

BB

xx Fig. 3.1

Fig. 3.1 A point P is a distance x from A along the line AB.

A point P variation is a distance x fromxAofalong the line AB. VA due to charge A alone is shown The with distance the electric potential The variation with distance x of the electric potential VA due to charge A alone is shown in Fig. 3.2. in Fig. 3.2. 0.8 potential 0.8 V/V

potential V/V

VA

0.6

VB

VA

0.6

VB

0.4

0.4 0.2

0.2

0

2

4

6

8

10

12

14

16

18

x / nm

0

Fig. 3.2

2

The variation with distance x of the VB due alone is also16 4 6 8 electric potential 10 12to charge B14 shown in Fig. 3.2.

18

x / nm

Fig. 3.2 The variation with distance x of the electric potential VB due to charge B alone is also shown in Fig. 3.2. © UCLES 2013

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9 (i)

State and explain whether the charges A and B are of the same, or opposite, sign. ..................................................................................................................................

For Examiner’s Use

.................................................................................................................................. .............................................................................................................................. [2] (ii)

By reference to Fig. 3.2, state how the combined electric potential due to both charges may be determined. .................................................................................................................................. .............................................................................................................................. [1]

(iii)

Without any calculation, use Fig. 3.2 to estimate the distance x at which the combined electric potential of the two charges is a minimum. x = .......................................... nm [1]

(iv)

The point P is a distance x = 10 nm from A. An α-particle has kinetic energy EK when at infinity. Use Fig. 3.2 to determine the minimum value of EK such that the α-particle may travel from infinity to point P.

EK = ............................................. J [3]

168

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19 Oct/Nov 2013 , Question #6 , qp_43 6

14

A particle has mass m and charge +q and is travelling with speed v through a vacuum. The initial direction of travel is parallel to the plane of two charged horizontal metal plates, as shown in Fig. 6.1.

For Examiner’s Use

+V metal plate

path of particle

metal plate Fig. 6.1 The uniform electric field between the plates has magnitude 2.8 × 104 V m–1 and is zero outside the plates. The particle passes between the plates and emerges beyond them, as illustrated in Fig. 6.1. (a) Explain why the path of the particle in the electric field is not an arc of a circle. .......................................................................................................................................... .......................................................................................................................................... ...................................................................................................................................... [1] (b) A uniform magnetic field is now formed in the region between the metal plates. The magnetic field strength is adjusted so that the positively charged particle passes undeviated between the plates, as shown in Fig. 6.2. +V

path of particle

region of uniform electric and magnetic fields path of particle

Fig. 6.2

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15 (i)

State and explain the direction of the magnetic field. ..................................................................................................................................

For Examiner’s Use

.................................................................................................................................. .............................................................................................................................. [2] (ii)

The particle has speed 4.7 × 105 m s–1. Calculate the magnitude of the magnetic flux density. Explain your working.

magnetic flux density = ............................................. T [3] (c) The particle in (b) has mass m, charge +q and speed v. Without any further calculation, state the effect, if any, on the path of a particle that has (i)

mass m, charge –q and speed v, .............................................................................................................................. [1]

(ii)

mass m, charge +q and speed 2v, .............................................................................................................................. [1]

(iii)

mass 2m, charge +q and speed v. .............................................................................................................................. [1]

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20 May/June 2014 , Question #8 , qp_41 8

16

(a) State what is meant by quantisation of charge. ................................................................................................................................................... .............................................................................................................................................. [1] (b) A student carries out an experiment to determine the elementary charge. A charged oil drop is positioned between two horizontal metal plates, as shown in Fig. 8.1. + 680 V

oil drop 7.0 mm

Fig. 8.1 The plates are separated by a distance of 7.0 mm. The lower plate is earthed. The potential of the upper plate is gradually increased until the drop is held stationary. The potential for the drop to be stationary is 680 V. The weight of the oil drop, allowing for the upthrust of the air, is 4.8 × 10−14 N. Calculate the value for the charge on the oil drop.

charge = ..................................................... C [2]

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17 (c) The student repeats the experiment and determines the following values for the charge on oil drops. 3.3 × 10−19 C

4.9 × 10−19 C

9.7 × 10−19 C

3.4 × 10−19 C

Use these values to suggest a value for the elementary charge. Explain your working.

elementary charge = ..................................................... C [2]

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21 May/June 2014 , Question #4 , qp_42 4

9

A helium nucleus contains two protons. In a model of the helium nucleus, each proton is considered to be a charged point mass. The separation of these point masses is assumed to be 2.0 × 10−15 m. (a) For the two protons in this model, calculate (i)

the electrostatic force,

electrostatic force = ..................................................... N [2] (ii)

the gravitational force.

gravitational force = ..................................................... N [2] (b) Using your answers in (a), suggest why (i)

there must be some other force between the protons in the nucleus, ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3]

(ii)

this additional force must have a short range. ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2]

© UCLES 2014

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173

22 May/June 2014 , Question #6 , qp_42 6

12

(a) Explain the use of a uniform electric field and a uniform magnetic field for the selection of the velocity of a charged particle. You may draw a diagram if you wish.

................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3] (b) Ions, all of the same isotope, are travelling in a vacuum with a speed of 9.6 × 104 m s−1. The ions are incident normally on a uniform magnetic field of flux density 640 mT. The ions follow semicircular paths A and B before reaching a detector, as shown in Fig. 6.1.

detector

A

vacuum

B

uniform magnetic field, flux density 640 mT Fig. 6.1 Data for the diameters of the paths are shown in Fig. 6.2. path

diameter / cm

A B

6.2 12.4

Fig. 6.2 The ions in path B each have charge +1.6 × 10−19 C. © UCLES 2014

CEDAR COLLEGE

9702/42/M/J/14

174

13 (i)

Determine the mass, in u, of the ions in path B.

mass = ..................................................... u [4] (ii)

Suggest and explain quantitatively a reason for the difference in radii of the paths A and B of the ions. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3]

© UCLES 2014

CEDAR COLLEGE

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175

23 May/June 2014 , Question #8 , qp_43 8

16

(a) State what is meant by quantisation of charge. ................................................................................................................................................... .............................................................................................................................................. [1] (b) A student carries out an experiment to determine the elementary charge. A charged oil drop is positioned between two horizontal metal plates, as shown in Fig. 8.1. + 680 V

oil drop 7.0 mm

Fig. 8.1 The plates are separated by a distance of 7.0 mm. The lower plate is earthed. The potential of the upper plate is gradually increased until the drop is held stationary. The potential for the drop to be stationary is 680 V. The weight of the oil drop, allowing for the upthrust of the air, is 4.8 × 10−14 N. Calculate the value for the charge on the oil drop.

charge = ..................................................... C [2]

© UCLES 2014

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176

17 (c) The student repeats the experiment and determines the following values for the charge on oil drops. 3.3 × 10−19 C

4.9 × 10−19 C

9.7 × 10−19 C

3.4 × 10−19 C

Use these values to suggest a value for the elementary charge. Explain your working.

elementary charge = ..................................................... C [2]

© UCLES 2014

CEDAR COLLEGE

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177

24 May/June 2016 , Question #6 , qp_41 6

15

A solid metal sphere of radius R is isolated in space. The sphere is positively charged so that the electric potential at its surface is VS. The electric field strength at the surface is ES. (a) On the axes of Fig. 6.1, show the variation of the electric potential with distance x from the centre of the sphere for values of x from x = 0 to x = 3R. 1.0 Vs 0.8 Vs potential 0.6 Vs 0.4 Vs 0.2 Vs 0 0

R

2R

3R distance x

Fig. 6.1 [3] (b) On the axes of Fig. 6.2, show the variation of the electric field strength with distance x from the centre of the sphere for values of x from x = 0 to x = 3R. 1.0 Es 0.8 Es field strength 0.6 Es 0.4 Es 0.2 Es 0 0

R

2R

3R distance x

Fig. 6.2 [3] [Total: 6]

© UCLES 2016

CEDAR COLLEGE

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178

25 May/June 2016 , Question #6 , qp_42 6

13

(a) By reference to electric field lines, explain why, for points outside an isolated spherical conductor, the charge on the sphere may be considered to act as a point charge at its centre. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) Two isolated protons are separated in a vacuum by a distance x. (i)

Calculate the ratio electric force between the two protons . gravitational force between the two protons

ratio = ......................................................... [3] (ii)

By reference to your answer in (i), suggest why gravitational forces are not considered when calculating the force between charged particles. ........................................................................................................................................... ...................................................................................................................................... [1] [Total: 6]

© UCLES 2016

CEDAR COLLEGE

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179

26 May/June 2010 , Question #4 , qp_41 4

8

(a) Explain what is meant by the potential energy of a body. ..........................................................................................................................................

For Examiner’s Use

.......................................................................................................................................... ...................................................................................................................................... [2] 2

(b) Two deuterium ( 1 H) nuclei each have initial kinetic energy EK and are initially separated by a large distance. The nuclei may be considered to be spheres of diameter 3.8 × 10–15 m with their masses and charges concentrated at their centres. The nuclei move from their initial positions to their final position of just touching, as illustrated in Fig. 4.1. 2 1H

2 1H

kinetic energy EK

kinetic energy EK

initially

3.8 × 10–15 m 2 1H

finally

2 1H

at rest Fig. 4.1 (i)

For the two nuclei approaching each other, calculate the total change in 1. gravitational potential energy,

energy = ............................................ J [3] 2. electric potential energy.

energy = ............................................ J [3] © UCLES 2010

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9702/41/M/J/10

180

9 (ii)

Use your answers in (i) to show that the initial kinetic energy EK of each nucleus is 0.19 MeV.

For Examiner’s Use

[2] (iii)

The two nuclei may rebound from each other. Suggest one other effect that could happen to the two nuclei if the initial kinetic energy of each nucleus is greater than that calculated in (ii). .................................................................................................................................. .............................................................................................................................. [1]

181

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9702/41/M/J/10

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Oct/Nov 2012 , Question #3 , qp_43 27 Oct/Nov 2012 , Question #3 , qp_43

Q17-

3

8

(a) State what is meant by a line of force in (i)

For Examiner’s Use

a gravitational field, .................................................................................................................................. .............................................................................................................................. [1]

(ii)

an electric field. .................................................................................................................................. .............................................................................................................................. [2]

(b) A charged metal sphere is isolated in space. State one similarity and one difference between the gravitational force field and the electric force field around the sphere. similarity: .......................................................................................................................... .......................................................................................................................................... difference: ........................................................................................................................ .......................................................................................................................................... .......................................................................................................................................... [3] (c) Two horizontal metal plates are separated by a distance of 1.8 cm in a vacuum. A potential difference of 270 V is maintained between the plates, as shown in Fig. 3.1. 0V

proton 1.8 cm

+270 V Fig. 3.1 A proton is in the space between the plates. Explain quantitatively why, when predicting the motion of the proton between the plates, the gravitational field is not taken into consideration.

[4] © UCLES 2012

CEDAR COLLEGE

9702/43/O/N/12

182

28 Oct/Nov 2005 , Question #5 , qp_4 5

10

(a) An electron is accelerated from rest in a vacuum through a potential difference of 1.2 104 V. Show that the final speed of the electron is 6.5 107 m s–1.

[2] (b) The accelerated electron now enters a region of uniform magnetic field acting into the plane of the paper, as illustrated in Fig. 5.1.

+ +

+

+ +

+

+

path of electron

+

+

magnetic field into plane of paper

Fig. 5.1 (i)

Describe the path of the electron as it passes through, and beyond, the region of the magnetic field. You may draw on Fig. 5.1 if you wish. path within field: ........................................................................................................ ................................................................................................................................... path beyond field: .................................................................................................... .............................................................................................................................. [3]

183

CEDAR COLLEGE © UCLES 2005

9702/04/O/N/05

For Examiner’s Use

11 (ii)

State and explain the effect on the magnitude of the deflection of the electron in the magnetic field if, separately, 1.

For Examiner’s Use

the potential difference accelerating the electron is reduced, ........................................................................................................................... ........................................................................................................................... ...................................................................................................................... [2]

2.

the magnetic field strength is increased. ........................................................................................................................... ........................................................................................................................... ...................................................................................................................... [2]

184

CEDAR COLLEGE © UCLES 2005

9702/04/O/N/05

[Turn over

2 (a)

(-1 for each error or omission) ........................................ B2

[2]

ANSWERS

heat lost by liquid gold = 0.95m x 129 x ∆T.................................. C1 heat gained (silver) = 0.05m x 235 x (1340 – 300) + 0.05m x 105 000..C1, C1 122.5m∆T = 17 470m

(b) 1

∆T = 143 K.......................................................................................C1 temperature = 143 + 1340 = 1483 K................................................A1

[5]

(c)

e.g. thermocouple/resistance thermometer .................................. B1

[1]

3 (a)

f0 is at natural frequency of spring (system) ................................. B1 this is at the driver frequency ....................................................... B1 (allow 1 mark for recognition that this is resonance)

[2]

(b)

line: amplitude less at all frequencies ......................................... B1 peak flatter .......................................................................... B1 peak at f0 or slightly below f0 ................................................ B1

(c)

(aluminium) sheet cuts the magnetic flux/field.............................. B1 (so) currents/e.m.f. induced in the (metal) sheet .......................... B1 these currents dissipate energy ...................................................M1 less energy available for the oscillations ...................................... A1 so amplitude smaller .................................................................... A0 (‘current opposes motion of sheet’ scores one of the last two marks)

24 (a)

(b)

field causes forces on the electrons .............................................M1 and the nucleus in opposite directions ......................................... A1 (field causes) electrons (to be) stripped off the atom.................... B1 (i)

Page 3

E = Q/4!ε0r2 .....................................................................................C1 20 x 103 x 102 = Q/(4! x 8.85 x 10-12 x 0.212 ....................................C1 Scheme Syllabus charge = 9.8 x 10-6 CMark .......................................................................A1 A/AS LEVEL EXAMINATIONS - JUNE 2003

(ii)

9702

[3]

[4]

[3]

Paper [3] 04

V =© Q/4!ε 0r University of Cambridge Local Examinations Syndicate 2003 = (9.8 x 10-6 )/(4! x 8.85 x 10-12 x 0.21) ........................................C1 = 4.2 x 105 V................................................................................A1

[2]

(c)

e.g. sphere not smooth, humid air, etc ............................................B1

[1]

5 (a)

centripetal force = mv2 /r...................................................................B1 magnetic force F = Bqv....................................................................B1 (hence) mv2 /r = Bqv .........................................................................B1 r = mv/Bq .........................................................................................A0

[3]

r! /r" = (m! /m") x (q" /q!) ...................................................................C1 = (4 x 1.66 x 10-27)/(9.11 x 10-31 x 2) = 3.64 x 103 ...............................................................................A2

[3]

(b)

(c)

(d)

(i)

r! = (4 x 1.66 x 10-27 x 1.5 x 106 )/(1.2 x 10-3 x 2 x 1.6 x 10-19 ) = 25.9 m ......................................................................................A2

(ii)

r" = 25.9 x 3.64 x 103 = 7.13 x 10-3 m ..............................................A1

(i)

deflected upwards............................................................................B1 but close to original direction ...........................................................B1

CEDAR COLLEGE (ii)

6 (a)

[3]

opposite direction to "-particle and ‘through side’ ...........................B1

185[3]

greater binding energy gives rise to release of energy ................... M1 so must be yttrium ...........................................................................A1

[2]

not influenced by another planet etc. M = 0.0384 kg 2

(a) (ii) (on melting,) bonds molecules are broken/weakened peakbetween height/amplitude decreasing or molecules further apart/are able to slide over one another kinetic energy unchanged so no temperature change 35 (a) potential energy increased/changed field strength = potential gradientso [- energy sign notrequired required] [allow E = ∆V/∆x but not E = V/d] (b) thermal energy/heat required to convert unit mass of solid to liquid (b) No field for x < r with no change in temperature/ at its normal boiling point for x > r, curve in correct direction, not going to zero discontinuity at x = r (vertical line required) (c) (i) thermal energy lost by water = 0.16 × 4.2 x 100 = 67.2 kJ 6 (a) (i) flux/field in core 67.2 = 0.205 × Lmust be changing so that an e.m.f./current is induced in the secondary L = 328 kJ kg–1 (ii) power more energy = VI (than calculated) melts ice so, (calculated) L is lowersothan acceptedISvalue power is constant if VSthe increases, decreases output

34

(b) (i) same shape and phase as IP graph (a) field strength = potential gradient correct sign OR directions discussed (ii) same frequency (b)

7 (a) (c)

(b)

correct phase w.r.t. Fig. 6.3 area is 21.2 cm2 ± 0.4 cm2 (iii) ½π rad±or (if outside 0.490° cm2 but within ± 0.8 cm2, allow 1 mark) 2 1.0 cm represents (1.0 × 10–2 × 2.5 × 103 =) 25 V potential difference = 530 V curve levelling out (at 1.4 µg) correct shape judged by masses at nT½ 2 ½mv[for = second qV mark, values must be marked on y-axis) ½ × 9.1 × 10–31 × v2 = 1.6 × 10–19 × 530 –1 v = 1.37 107×ms 10-6 × 6.02 × 1023)/56 (i) N0 = ×(1.4 = 1.5 × 1016

(d) (i) (ii) (ii)

(c)

d=0 A = λN -5 -1 λacceleration = ln2/(2.6 ×decreases 3600) (= 7.4 × 10 s ) then increases 12 A = 1.11 × 10 Bq analysis (e.g. minimum at 4.0 cm) some quantitative (any suggestion that acceleration becomes zero or that there is a 1/10 of original mass0/2) of Manganese remains deceleration scores 0.10 = exp(-ln2 × t/2.6) t = 8.63 hours © UCLES 2007 [use of 1/9, giving answer 8.24 hrs scores 1 mark]

A1 B1

B1

B1 [3]

[1] B1 B1 [1]B1

[3]

M1 A1

[2]

B1 B1 B1

[3]

M1 A1

C1 C1 [2]A1

[3]

M1 A1

M1 [2]A1

[2]

B1

[1] M1 A1

[2]

M1 A1 B1

[2] C2 [1] C1 A1

[4]

M1 A1

[2]

C1 A1

C1 A1 [2]

[2]

B1

[1]

B1 [3]B1

[2]

C1 C1 A1 C1 A1

[2]

© University of Cambridge International Examinations 2005

CEDAR COLLEGE

[1]

186

(ii) 1 4.0 Hz ................................................................................................................ B1 2 0.50 cm (allow ±0.03 cm) ................................................................................ B1 45

3

[2]

Page 3 Mark Scheme Syllabus Paper (a) (i) either lines directed away from sphere GCE A/AS LEVEL – May/June 2008 9702 04 or lines go from positive to negative line shows direction of force on positive charge ....................................... M1A1 (a) (i) or amplitude = 0.5 cm [1] so positively charged ............................................................................................ A1 [2] (ii) period = 0.8 s A1 [1] (ii) either all lines (appear to) radiate from centre or all lines are normal to surface of sphere ................................................... B1 [1] (b) (i) ω = 2π / T C1 –1 = 7.85 rad s (b) tangent to curve ........................................................................................................... B1B1 correct use of v = ω √(x02 – x2) in correct position and direction B1 [2] –2 2 ................................................................................... –2 2 = 7.85 × √({0.5 × 10 } – {0.2 × 10 } ) –1 = 3.6 cm s A1 [3] (if tangent drawn -9 or clearly implied-12 (B1) (c) (i) V = (0.76 × 10–1 ) / (4π × 8.85 × 10 × 0.024) ..................................................... C1 (A2) 3.6=± 285 0.3 cm s V ........................................................................................................... A1 [2] but allow 1 mark for > ±0.3 but Ğ ±0.6 cm s–1) (ii) negative charge is induced on (inside of) box ...................................................... M1 (ii) formula d = 15.8 cm to isolated (point) charge A1 [1] applies OR less work done moving test charge from infinity .......................................... A1 so potential is lower .............................................................................................. A1 [3] (c) (i) (continuous) loss of energy / reduction in amplitude (from the oscillating system) B1 by forcefield acting in opposite direction to the motion / friction / (d) eithercaused gravitational is always attractive forces or viscous field lines must be directed towards both box and sphere .............................. B1B1 [1][2]

(ii) same period / small increase in period line displacement always less than that on Fig.3.2 (ignore first T/4) peak progressively smaller

4

6

B1 M1 A1

[3]

(a) work done moving unit positive charge from infinity to the point

M1 A1

[2]

(b) (i) x = 18 cm

A1

[1]

C1 C1 A1

[3]

(c) field strength = (–) gradient of graph force = charge × gradient / field strength or force ∝ gradient force largest at x = 27 cm

B1 B1 B1

[3]

(a) at t = 1.0 s, V = 2.5 V energy = ½CV 2 0.13 = ½ × C × (8.02 – 2.52) C = 4500 µF

C1 C1 M1 A0

[3]

(b) use of two capacitors in series in all branches of combination connected into correct parallel arrangement

M1 A1

[2]

(ii) VA + VB = 0 (3.6 × 10–9) / (4πε0 × 18 × 10–2) + q / (4πε0 × 12 × 10–2) = 0 q = –2.4 × 10–9 C © UCLES 2007 (use of VA = VB giving 2.4 × 10 –9 C scores one mark)

5

187

CEDAR COLLEGE

© UCLES 2008

Page 4

7 5

Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2009

Syllabus 9702

Paper 41

(a) work done per / on unit positive charge .....................................................................M1 moving charge from infinity to the point ..................................................................... A1

[2]

(b) (i) α-particle and gold nucleus repel each other ..................................................... B1 all kinetic energy of α-particle converted into electric potential energy .............. B1

[2]

(ii) 1 potential energy = (79 × 2 × {1.6 × 10-19}2) / (4π × 8.85 × 10-12 × d) .............. C1 kinetic energy = 4.8 × 1.6 × 10-13 = 7.68 × 10-13 J ........................................... C1 equating to give d = 4.7 × 10-14 m ..................................................................... A1

[3]

(ii) 2 F = Qq / 4πε0d × 1 / d = 7.68 × 10-13 × 1 / (4.7 × 10-14) .............................. C1 = 16 N ....................................................................................................... A1

[2]

[Total: 9]

6

48

7

5

(a) concentric circles …(at least three lines) ................................................................M1 with increasing separation ......................................................................................... A1 correct direction clear ................................................................................................ B1

[3]

Page Mark Scheme: version Syllabus Paper[1] (b) (i)3 correct position to left of wire Teachers’ .............................................................................. B1 GCE AS/A LEVEL – May/June 2010 9702 41 (ii) B = (4π × 10-7 × 1.7) / (2π × 1.9 × 10-2) ............................................................. C1 (a) ability to=do work 1.8 × 10-5 T ................................................................................................. A1B1 [2] as a result of the position/shape, etc. of an object B1 [2] (c) distance ∝ current ................................................................................................... C1 = GMm C1 (b) (i) 1 ∆E current = (2.8 × 1.7/ r gpe / 1.9) = 2.5 A =........................................................................................................ A1C1 [2] (6.67 × 10–11 × {2 × 1.66 × 10–27}2) / (3.8 × 10–15) = 1.93 × 10–49 J A1 [3] [Total: 8] 2 ∆Eepe = Qq / 4πε0r C1 = (1.6 × 10–19)2 / (4π × 8.85 × 10–12 × 3.8 × 10–15) C1 (a) e.g. more (output)= power available –14 6.06 × 10 J A1 [3] e.g. less ripple for same smoothing capacitor any sensible suggestion ............................................................................................ B1 [1] (ii) idea that 2EK = ∆Eepe – ∆Egpe B1 –14 EK = 3.03 × 10 J –13 = (3.03 × 10–14half-wave ) / 1.6 × 10rectification (b) (i) curve showing ................................................................. B1M1 [1] = 0.19 MeV A0 [2] (ii) similar to (i) but phase shift of 180° .................................................................... B1 [1] (iii) fusion may occur / may break into sub-nuclear particles B1 [1] (c) (i) correct symbol, connected in parallel with R ...................................................... B1 [1] B1 (a) (i) VH depends on angle between (plane of) probe and B-field max when plane capacitor and B-field normal other VHcapacitor (ii) either 1 larger / second in are parallel withtoReach ..................................... B1 [1] or VH zero (not increase R)when plane and B-field are parallel depends sine of angle between plane and B-field [2] or V 2 same Hpeak valueson........................................................................................... B1B1 correct shape giving less ripple .......................................................................... B1 [2] M1 (ii) 1 calculates VHr at least three times [Total: 7] to 1 s.f. constant so valid or approx constant so valid or to 2 s.f., not constant so invalid A1 [2]

2

straight line passes through origin

B1

[1]

© UCLES 2009

CEDAR(b) COLLEGE (i) e.m.f. induced is proportional / equal to

rate of change of (magnetic) flux (linkage) constant field in coil / flux (linkage) of coil does not change

188 M1 A1 B1

[3]

in 0.012 kg of carbon-12

79

3

Page 4 Mark Scheme: Teachers’ version Syllabus (b) pV = NkT or pVGCE = nRT AS/A LEVEL – May/June 2010 9702 substitutes temperature as 298 K 5 ×the 6.5page × 10–2 = N × 1.38 × 10–23 × 298 either pointing 1.1 × 10up (a) arrow or 1.1 × 105 × 6.5 × 10–2 = n × 8.31 × 298 and n = N / 6.02 × 1023 N = 1.7 × 1024 (b) (i) Eq = Bqv v = (12 × 103) / (930 × 10–6) (a) acceleration / force to displacement from a fixed point = 1.3 × 107proportional m s–1 acceleration / force (always) directed towards that fixed point / in opposite direction (ii) Bqv to = displacement mv 2 / r

10 4

[2]

Paper C1 41 C1 B1 [1] C1 A1 [4] C1 C1 M1A1 [3]

A1 C1

[2]

C1 A1

[3]

B1 M1

[2]

A1 C1 C1 C1 C1 A1

[2]

A1

[2]

(a) (ii) workEdone=inhc bringing unit positive charge /λ from infinity (to that point) λ = (6.63 × 10–34 × 3.0 × 108) / (1.08 × 10–11) = 1.84 × 10–14 m

M1 A1 C1 A1

[2] [2]

(b) (i) field strength is potential gradient (iii) λ = h/p = (6.63 proportional × 10–34) / (1.84 × 10–14 ) particle Q) (ii) pfield strength to force (on –20 = 3.6 × 10 N s potential gradient proportional to gradient of (potential energy) graph so force is proportional to the gradient of the graph Page 3 Mark Scheme: Teachers’ version Syllabus B 2011 GCE AS/A LEVELSection – May/June 9702

B1

[1]

7

8

A1

–2

–6

q/m = (1.3 × 10 ) / (7.9 × 10 × 930 × 10 ) = 1.8 × 1011 C kg–1 (b) (i) Aρg / m is a constant and so acceleration proportional to x negative sign shows acceleration towards a fixed point / in opposite direction to displacement (a) momentum conservation hence momenta of photons are equal (but opposite) same momentum so same energy (ii) ω 2 = (Aρg / m) ω = 2πf (2 × π=× (∆)mc 1.5)2 =2 ({4.5 × 10–4 × 1.0 × 103 × 9.81} / m) (b) (i) (∆)E m = 50 = g1.2 × 10–28 × (3.0 × 108)2 = 1.08 × 10–11 J

(a) point=X5.1 shown (c) (i) energy × 1.6correctly × 10–19 (J) potential energy = Q1Q2 / 4πε0r –19very large / infinite gain (ii) has 5.1 op-amp × 1.6 × 10 = (1.6 × 10–19)2 / 4π × 8.85 × 10–12 × r –10 non-inverting r = 2.8 × 10 m input is at earth (potential) / earthed / at 0 V © not University of Cambridge International if amplifier is to saturate, inverting input mustExaminations be (almost) 2011 at earth potential / 0 (V) same potential as inverting input (d) (i) work is got out as x decreases so opposite sign (b) (i) total input resistance = 1.2 kΩ gain be (= doubled –4.2 / 1.2) = –3.5 (ii) (amplifier) energy would (voltmeter) reading = –3.5 × –1.5 gradient would be increased = 5.25 V (total disregard of signs or incorrect sign in answer, max 2 marks) 5 (a) region (of space) where there is a force (ii) (lesson bright so) resistance of LDR increases either / produced by magnetic pole gain decreases or (amplifier) on / produced by current carrying conductor / moving charge (voltmeter) reading decreases

9

(b) (i) force on particle is (always) normal to velocity / direction of travel speed of particle is constant (ii) magnetic force provides the centripetal force mv2 / r = Bqv r = mv / Bq CEDAR COLLEGE

© UCLES 2010

(c) (i) direction from ‘bottom to top’ of diagram (ii) radius proportional to momentum

B1

[4]

B1 C1 B1 A1 [2] A0 [2] Paper 41

C1B1 C1 C1M1 A1M1

[1]

A1 M1 A1 C1 B1C1 B1 A1

[3]

[2] [3]

M1 M1 A1M1 A1

[2] [3]

[4]

[2]

B1 B1

[2]

B1 M1 A0

[2]

189 B1 C1

[1]

11 4

4

as magnet moves in coil light damping current in resistor gives rise to a heating effect (iii) thermal period =energy 0.80 sis derived from energy of oscillation of the magnet frequency = 1.25 Hz (period not 0.8 s, then 0/2)

A1 A1 M1 A1 C1 A1

(a) (i) zero field (strength) inside spheres (b) (i) (induced) e.m.f. is proportional to (ii) either field strength isof zero rate of change/cutting (magnetic) flux (linkage) or the fields are in opposite directions at is a point between (ii) a current induced in the the coil spheres as magnet moves in coil current in resistor gives rise to a heating effect (b) (i) field strength is is (–)derived potential gradient V/x) thermal energy from energy(not of oscillation of the magnet

B1 M1 A1 M1 A1 M1 A1 M1 B1 A1

[1] [4]

(ii) 1. field strength has maximum value x =(strength) 11.4 cm inside spheres (a) (i) zeroat field

B1 B1 B1

[2] [1]

B1 M1 B1 A1

[2] [2]

B1 B1

[1] [1]

B1 B1 B1

[1] [2]

opposite (b) FB must 2. be field strengthinisdirection zero to FE plane of paper so magnetic field into either at x = 7.9 cm (allow ±0.3 cm) or at 0 to 1.4 cm or 11.4 cm to 12 cm

B1 B1 B1 B1

[2] [2]

(a) (i) Bqv(sinθ) or Bqv(cosθ)

B1

[1]

B1

[1]

B1 B1

[2]

field strength is zero (ii) 2. either field strength is zero either at x = 7.9 (allowdirections ±0.3 cm) or the fields are in cm opposite or at a at 0 to 1.4 cm or 11.4 point between the spherescm to 12 cm

5

(a) or Bqv(cosθ) (b) (i) (i) Bqv(sinθ) field strength is (–) potential gradient (not V/x) (ii) (ii) qE 1. field strength has maximum value at x = 11.4 cm

12 5

(ii) qE

(b) FB must be opposite in direction to FE so magnetic field into plane of paper

© University of Cambridge International Examinations 2011

© University of Cambridge International Examinations 2011

CEDAR COLLEGE

190

[2] [4] [2] [1]

[2] [2]

13 5

3

(c) (i) graph: horizontal line, y-intercept = 7.0 mJ with end-points of line at (b) sinusoidal wave with values positive cmall and –2.8 cm +2.8 all values positive, all peaks at EK and energy = 0 at t = 0 s period = 0.30reasonable (ii) graph: curve with maximum at (0,7.0) end-points of line at (–2.8, 0) and (+2.8, 0) (a) force per unit positive charge acting on a stationary charge (iii) graph: inverted version of (ii) with intersections at (–2.0, 3.5) and (+2.0, 3.5) 2 (b) (i) (Allow E = marks Q / 4πεin 0 r (iii), but not in (ii), if graphs K & P are not labelled) Q = 1.8 × 104 × 102 × 4π × 8.85 × 10–12 × (25 × 10–2 )2 Q = 1.25 × 10–5 C = 12.5 µC (d) gravitational potential energy (ii) V = Q / 4πε0 r = (1.25 × 10–5 ) / (4π × 8.85 × 10–12 × 25 × 10–2 ) 5 (a) sum of=potential and kinetic energy of atoms/molecules/particles 4.5 × 10energy V reference toallow random (Do not use(distribution) of V = Er unless explained)

B1 B1 M1 A1 C1 M1 A0 B1

[2] [1]

C1 M1 A1 A1

[2] [2]

(b) (i) as lattice structure is ‘broken’/bonds broken/forces between molecules reduced (not molecules separate) no change in kinetic energy, potential energy increases internal energy increases

B1 M1 A1

[3]

B1 M1 A1

[3]

M1 A1

[2]

B1 B1 A0

[2]

(ii) either molecules/atoms/particles move faster/ is increasing or kinetic energy increases with temperature (increases) no change in potential energy, kinetic energy increases internal energy increases 14 4

(a) (i) as r decreases, energy decreases/work got out (due to) attraction so point mass is negatively charged (ii) electric potential energy = charge × electric potential electric field strength is potential gradient field strength = gradient of potential energy graph/charge

B1 B1 B1 B1 B1

[1]

[3] [2] [1] [2]

© University of Cambridge International Examinations 2012

(b) tangent drawn at (4.0, 14.5) B1 A2 gradient = 3.6 × 10–24 (for < ±0.3 allow 2 marks, for < ±0.6 allow 1 mark) field strength = (3.6 × 10–24) / (1.6 × 10–19) A1 = 2.3 × 10–5 V m–1 (allow ecf from gradient value) (one point solution for gradient leading to 2.3 × 10–5 Vm–1 scores 1 mark only)

[4]

© University of Cambridge International Examinations 2012

CEDAR COLLEGE

191

15 6

7

(c) new reading is 2.4√2 g either changes between +3.4 g and –3.4 g or total change is 6.8 g

C1

(a) oil drop charged by friction/beta source between parallel metal plates plates are horizontal adjustable potential difference/field between plates until oil drop is stationary mg = q × V/d symbols explained oil drop viewed through microscope m determined from terminal speed of drop (when p.d. is zero) (any two extras, 1 each)

B1 B1

A1

(1) B1 B1 B1 (1) (1) (1) B2

[7]

(b) 3.2 × 10–19 C

A1

[1]

(a) minimum energy to remove an electron from the metal/surface

B1

[1]

C1 (b) gradient = 4.17 × 10–15 (allow 4.1 → 4.3) or h = 4.1 to 4.3 × 10–15 eV s A1 h = 4.15 × 10–15 × 1.6 × 10–19 A0 = 6.6 × 10–34 J s Page 3 Mark Scheme Syllabus GCE AS/A LEVEL – October/November 2012 9702 3 16

(c) graph: straight line parallel to given line (a) (i) (tangent to line gives) direction of force on a (small test) mass with intercept at any higher frequency intercept at between 6.9 × 1014 Hz and 7.1 × 1014 Hz (ii) (tangent to line gives) direction of force on a (small test) charge charge is positive

(b) similarity: e.g. radial fields lines normal to surface greater separation of lines with increased distance from sphere field strength ∝ 1 / (distance to centre of sphere)2 (allow any sensible answer) difference: © University of Cambridge International Examinations 2012 e.g. gravitational force (always) towards sphere electric force direction depends on sign of charge on sphere / towards or away from sphere e.g. gravitational field/force is attractive electric field/force is attractive or repulsive (allow any sensible comparison)

4

[2]

[2] Paper 43 B1

B1 B1

M1 A1

[1]

[3] [2]

B1

B1 B1 (B1) (B1) [3]

(c) gravitational force = 1.67 × 10–27 × 9.81 = 1.6 × 10–26 N electric force = 1.6 × 10–19 × 270 / (1.8 × 10–2) = 2.4 × 10–15 N electric force very much greater than gravitational force

A1 C1 A1 B1

[4]

(a) force on proton is normal to velocity and field provides centripetal force (for circular motion)

M1 A1

[2]

(b) magnetic force = Bqv centripetal force = mrω2 or mv2/r CEDAR COLLEGE v = rω Bqv = Bqrω = mrω2 ω = Bq/m

B1 B1 192 B1 A1

[4]

1

17 4

2

y = 12.0→13.0) A1 (a) force proportional to product of the two masses and inversely proportional to the square of their separation (ii) at max. amplitude potential energy is total energy B1M1 either reference point total energy to = 4.0 mJmasses or separation >> ‘size’ of masses B1 A1

(b) force provides centripetal (a) gravitational (i) force2 proportional totheproduct of force (two) charges and inversely / R = mRω2to square of separation GMmproportional where m is theto mass the planet reference pointofcharges 3 2 GM = R ω (ii) F = 2 × (1.6 × 10–19)2 / {4π × 8.85 × 10–12 × (20 × 10–6)2} F = 1.15 × 10–18 N (c) ω = 2π / T either Mstar / MSun = (Rstar / RSun)3 × (TSun / Tstar)2 Mstar 43 ×charge (½)2 × 2.0 × 1030 (b) (i) force per=unit × 1031 kg on either=a3.2 stationary charge 2 or or aMpositive Rstar3 / GT2 star = (2π) charge = {(2π)2 × (6.0 × 1011)3} / {6.67 × 10–11 × (2 × 365 × 24 × 3600)2} 31 = 3.2 × 10 (ii) 1. electric field is a kg vector quantity electric fields are in opposite directions (a) (i) sum charges of kineticrepel and potential energies of the molecules reference to random distribution Any two of the above, 1 each (ii) for gas, noalways intermolecular 2. ideal graph: line betweenforces given lines so no potential energy (only kinetic) crosses x-axis between 11.0 µm and 12.3 µm reasonable shape for curve

5

18 3

B1 M1 M1 A1 A1 A0 C1 A1 C1

M1C1 A1 (C1) A1 (C1) (A1)

[2]

[2]

[2] [3] [2]

[3] [2]

M1 B2 A1

[2]

M1M1 A1 A1 A1

[2] [3]

(b) (i) either change in kinetic energy = 3/2 × 1.38 × 10–23 × 1.0 × 6.02 × 1023 × 180 C1 = 2240 J (a) (i) field shown as right to left B1 A1 or R = kNA energy 3/2 × 1.0 × 180 (C1) (ii) lines are more =spaced out×at8.31 ends B1 = 2240 J (A1)

[2] [1]

[1]

in internal (b) (ii) Hall increase voltage depends onenergy angle = heat supplied + work done on system 2240 = energy supplied – 1500 either between field and plane of probe J to plane of probe energy supplied = 3740 or maximum when field normal or zero when field parallel to plane of probe

M1B1 C1 A1 A1

[3] [2]

(a) work done bringing unit positive charge infinity (toe.m.f. the point) (c) from (i) (induced) proportional to rate of change of (magnetic) flux (linkage) (allow rate of cutting of flux) (b) (i) either both potentials are positive / same sign so same sign (ii) e.g. move coil towards/away from solenoid or gradients e.g. rotate coil are positive & negative (so fields in opposite directions) socurrent same sign e.g. vary in solenoid

M1 M1A1 A1

[2] [2]

e.g. insert iron core into solenoid (ii) the individual potentials are summed (any three sensible suggestions, 1 each)

M1 A1 (M1) (A1)

[2]

B1

[1] [3]

(iii) allow value of x between 10 nm and 13 nm

A1

[1]

(allow 0.42 V → 0.44 V) (iv) V = 0.43 V energy = 2 × 1.6 × 10–19 × 0.43 = 1.4 × 10–19 J

M1 A1 A1

[3]

B3

© Cambridge International Examinations 2013 © Cambridge International Examinations 2013

CEDAR COLLEGE

193

Page 4

19 6

Mark Scheme GCE A LEVEL – October/November 2013

Syllabus 9702

(a) either constant speed parallel to plate or accelerated motion / force normal to plate / in direction field so not circular

B1 A0

[1]

(b) (i) direction of force due to magnetic field opposite to that due to electric field magnetic field into plane of page

B1 B1

[2]

(ii) force due to magnetic field = force due to electric field Bqv = qE B =E/v = (2.8 × 104) / (4.7 × 105) Page 4 Mark Scheme = 6.0 × 10–2 T GCE AS/A LEVEL – May/June 2014

B1 C1 Syllabus 9702

(b) energy lost (as thermal energy) in resistance/wires/battery/resistor / not deviated (c) (i) no change (award only if answer in (a)(i) > answer in (a)(ii)2)

(ii) deviated upwards 7

(a) graph: VH increases from zero when current switched on not deviated (iii) no change constant V then /non-zero H

VH returns to zero when current switched off

7

(a) (i) minimum photon energy to remove an (from the surface) (b) (i) minimum (induced) energy e.m.f. proportional to electron rate of change of (magnetic) flux (linkage) (ii) either maximum KE is photon energy – work function energy max KE when electron ejected (ii) or pulse as current is being switched on from the surface energies lower than max because energy required to bring electron to zero e.m.f. when current in coil the surface pulse in opposite direction when switching off

20 8

(allow ±0.05 × 1015) (b) threshold frequency = 1.0 1015 Hz (a) (i) discrete and equal amounts (of× charge) work function energyof 1.6 = hf×0 10–19C/elementary charge/e allow: discrete amounts –34 15 × 1.0 × 10charge/e = integral multiples of 1.66.63 × 10×–1910 C/elementary –19 = 6.63 × 10 J (allow alternative approaches based on use of co-ordinates of points on the=line) (b) weight qV / d 4.8 × 10–14 = (q × 680)/(7.0 × 10–3) (ii) sketch: –19 straight line with same gradient q = 4.9 × 10 C displaced to right

8 9

Paper 43

Paper A1 [3] 41 B1 B1

[1] [1]

B1

[1]

B1 B1 B1 B1

[1]

B1 B1 M1

[2]

A1

[2]

B1 B1 B1 B1 B1

[2] [3]

[3]

C1 B1 C1

[1]

A1

[3]

C1 M1 A1 A1

[2] [2]

B1 M0 B1

[2]

(iii) intensity determines number of photons arriving per unit time (c) elementary charge = 1.6 × 10–19 C (allow 1.6 × 10–19 C to 1.7 × 10–19 C ) intensity determines number of electrons per unit time (not energy) either the values are (approximately) multiples of this or it is a common factor it is the highest common factor (a) probability of decay (of a nucleus) / fraction of number of nuclei in sample that decay per unit time (a) e.g. no time delay between illumination and emission (allow λ =(dN / dt) / N with symbols explained – (M1), (A1) ) max. (kinetic) energy of electron dependent on frequency max. (kinetic) energy of electron independent of intensity rate of emission of electrons dependent on/proportional to intensity / 235 (b) (i) = (1.2 ×statements, 6.02 × 1023)one (anynumber three separate mark each, maximum 3) = 3.1 × 1021

C1 A1

[2]

M1 A1

[2]

C1 B3 A1

[3] [2]

(b) (i) (photon) interaction with electron may be below surface energy required to bring electron to surface

B1 B1

[2]

CEDAR COLLEGE

© Cambridge International Examinations 2013

194

to hold nucleus together (c) e.g. some microwave leakage from the cooker e.g.(Do container for ifthe is (a) alsoorheated > FE in one of the forces not calculated in (a)) not allow FGwater (any sensible suggestion) 21 4 =

(a) =

5

(a)

= =

= = (b)

(b)

5

22 6

(ii) outside nucleus there is repulsion between protons either attractive2 force must act only in nucleus Q1Qshort (i) or FE R 2 / 4πεrange, 0r if not all nuclei would stick together = = R 8.99 × 109 × (1.6 × 10–19)2 / (2.0 × 10–15)2 R 58 N only curve with decreasing gradient / r2 xR 0 and does not reach zero (ii) FG R Gm 1m2near acceptable value = R 6.67 × 10–11 × (1.67 × 10–27)2 / (2.0 × 10–15)2 –35 = Rline 4.7less × 10than N4.0 cm do not allow A1 mark) (if graph (no credit if graph line has positive and negative values of VH) (i) force of repulsion (much) greater than force of attraction must be some other force of attraction tofrom hold 0nucleus together graph: to 2T, two cycles of a sinusoidal wave all peaks above 3.5 mV > FE in ortoone the forces not calculated in (a)) not allow if FG(allow / 5.0 mV 4.8(a) mV 5.2 of mV) peaks(Do at 4.95

(ii) outside nucleus there is repulsion between protons force must actfield only in nucleus / flux is changing / cutting (c) e.m.f.either inducedattractive in coil when magnetic or if not short range, all nuclei would stick together either at each position, magnetic field does not vary so no e.m.f. is induced in the coil / no reading on the millivoltmeter (a) or only at curve decreasing eachwith position, switchgradient off current and take millivoltmeter reading acceptable value near xR 0 and doescoil notfrom reach zero or at each position, rapidly remove field and take meter reading (if graph line less than 4.0 cm do not allow A1 mark) (no credit graph linefields has positive and negative (a) electric andif magnetic normal to each other values of VH)

A1 M1 C1 A1

[1]

[2] [2]

[2]

A1

[2]

B1 M1 A1 M1 C1 A1

[3]

[3]

B1 B1 B1

[2]

M1 A1 B1

[2] [2]

B1

Paper B1 42

either each (i) m RatBqr / v position, magnetic field does not vary reading on) /the so =no e.m.f. is induced the× coil 10–19/ no × 6.2 × 10–2 (9.6millivoltmeter × 104 ) R=(640 × 10–3 ×in1.6 –26 or = at each position, off current and take millivoltmeter reading R=6.61 × 10 switch kg –26rapidly –27 coil from © Cambridge International Examinations 2014 or = at each position, field and take meter reading R=(6.61 × 10 ) / (1.66remove × 10 )u = R=40 u

C1 C1 C1 B1 A1

(a) electric and magnetic fields normal to each other

[3] [3]

[2] [4]

B1

or m constant and q ∝ 1 / r (ii) q / m ∝ 1 / r eitherq / mcharged enters for A is particle twice that for Bregion normal to both fields or correct B direction w.r.t. E for zero deflection ions in path A have (same mass but) twice the charge (of ions in path B) for no deflection, v R E / B

B1 B1 B1 B1 B1

[3] [3]

credit if magnetic field region (a) (no angle subtended at the centre of clearly a circlenot overlapping with electric field region) by an arc equal in length to the radius

B1 B1

[2]

(b) (i) arc R distance × angle diameter R 3.8 × 105 × 9.7 × 10–6 © Cambridge International Examinations 2014 = = R 3.7 km

C1

(ii) Mars is (much) further from Earth / away (answer must be comparative) angle (at telescope is much) smaller CEDAR COLLEGE

8

C1 B1

(no credit if magnetic field region clearly not overlapping with electric field region) Page 4 Mark Scheme Syllabus / flux is2014 changing / cutting 9702 (c) e.m.f. induced in coilGCE when A magnetic LEVEL – field May/June

6

[3]

B1

M1 B1 C1 B1 A1

= = = =

=

B1

either charged particle enters region normal to both fields (b) or graph: from B0 direction to 2T, twow.r.t. cycles of zero a sinusoidal wave correct E for deflection mV all peaks above 3.5 for no deflection, v R E / B peaks at 4.95 / 5.0 mV (allow 4.8 mV to 5.2 mV)

(b) = = = =

7

A1

(a) photon energy R hc / λ R (6.63 × 10–34 × 3.0 × 108) / (590 × 10–9 ) R 3.37 × 10–19 J

A1

[2]

B1 B1

[2]

195 C1 C1

(ii) pulse as current is being switched on zero e.m.f. when current in coil pulse in opposite direction when switching off

23 8

B1 B1 B1

(a) discrete and equal amounts (of charge) Page 4allow: discrete amounts of 1.6Mark Scheme C/elementary charge/e × 10–19 –19 Cambridge International AS/A Level – May/June integral multiples of 1.6 × 10 C/elementary charge/e 2016

5

(a) (i) 1011

1.25

1.50

1011 0110 1000 1110 0101 0011 0001 (c) elementary charge = 1.6 × 10–19 C (allow 1.6 × 10–19 C to 1.7 × 10–19 C ) eitherAllthe values are (approximately) 6 correct, 2 marks. 5 correct, 1multiples mark. of this or it is a common factor it is the highest common factor (b) sketch: 6 horizontal steps of width 0.25 ms shown

(a) e.g. noat time delayheights between and emission steps correct andillumination all steps shown max. (kinetic) energy of electron dependent on frequency max. (kinetic) energy of electron steps shown in correct time intervalsindependent of intensity rate of emission of electrons dependent on/proportional to intensity (any three separate statements, one mark each, maximum 3) (c) increase sampling frequency/rate

(b) (i) (photon) interaction with electron may be below surface so that step width/depth is reduced energy required to bring electron to surface

24 6

C1 A1

A2 C1 A1

A1

B3

[2]

[3]

[3]

M1

B1 A1 B1

A1

[2]

[4]

B1

smooth curve through (R, VS) and (2R, 0.5VS)

B1

smooth curve continues to (3R, 0.33VS)

B1

[3]

B1

smooth curve through (R, E) and International (2R, 0.25E)Examinations 2014 © Cambridge

B1

smooth curve continues to (3R, 0.11E)

B1

(a) line has non-zero intercept/line does not pass through origin

B1

charge is/should be proportional to potential (difference) or charge is/should be zero when p.d. is zero (therefore there is a systematic error)

B1

196

© Cambridge International Examinations 2016

[2]

A1

so that step height is reduced

CEDAR COLLEGE

[2]

M1

M1

(a) sketch: from x = 0 to x = R, potential is constant at VS

[1]

M0

increase number of bits (in each number)

(b) sketch: from x = 0 to x = R, field strength is zero

7

B1 [1] Paper 41 A1

(b) weight = qV / d (ii) 4.8 × 10–14 = (q × 680)/(7.0 × 10–3) 0.50 0.75 1.00 q = 4.9 ×010–190.25 C

9

Syllabus 9702

[3]

[3]

[2]

or maximum attenuation per unit length calculated to be 2.2 dB km–1 2.2 dB km–1 > 2.0 dB km–1 so yes

256

(a) lines perpendicular to surface or lines are radial

[2]

M1

lines appear to come from centre

A1

(b) (i) FE = (1.6 × 10–19 )2 / 4πε0 x2

[2]

C1

FG = G × (1.67 × 10–27 )2 / x2

C1

FE / FG = (1.6 × 10–19 )2 × (8.99 × 109 ) / [(1.67×10–27 )2 × (6.67×10–11)] = 1.2 (1.24) × 1036

A1

[3]

B1

[1]

(ii) FE !" FG Page 3

264

(M1) (A1)

Mark Scheme: Teachers’ version GCE AS/A LEVEL – May/June 2010

Syllabus 9702

(a) ability to do work as a result of the position/shape, etc. of an object

Paper 41 B1 B1

[2]

= GMm / r = (6.67 × 10–11 × {2 × 1.66 × 10–27}2) / (3.8 × 10–15) = 1.93 × 10–49 J

C1 C1 A1

[3]

= Qq / 4πε0r = (1.6 × 10–19)2 / (4π × 8.85 × 10–12 × 3.8 × 10–15) = 6.06 × 10–14 J

C1 C1 A1

[3]

© Cambridge International Examinations 2016

(b) (i) 1

2

5

∆Egpe

∆Eepe

(ii) idea that 2EK = ∆Eepe – ∆Egpe EK = 3.03 × 10–14 J = (3.03 × 10–14) / 1.6 × 10–13 = 0.19 MeV

M1 A0

[2]

(iii) fusion may occur / may break into sub-nuclear particles

B1

[1]

(a) (i) VH depends on angle between (plane of) probe and B-field either VH max when plane and B-field are normal to each other or VH zero when plane and B-field are parallel or VH depends on sine of angle between plane and B-field (ii) 1

2

B1

B1

[2]

calculates VHr at least three times to 1 s.f. constant so valid or approx constant so valid or to 2 s.f., not constant so invalid

M1 A1

[2]

straight line passes through origin

B1

[1]

M1 A1 B1

[3]

197 B3

[3]

(b) (i) e.m.f. induced is proportional / equal to rate of change of (magnetic) flux (linkage) constant field in coil / flux (linkage) of coil does not change (ii) e.g. vary current (in wire) / switch current on or off / use a.c. current rotate coil CEDAR COLLEGE move coil towards / away from wire (1 mark each, max 3)

6

B1

(a) all four diodes correct to give output, regardless of polarity

M1

Page 3

27

3

Mark Scheme GCE AS/A LEVEL – October/November 2012

Syllabus 9702

Paper 43

(a) (i) (tangent to line gives) direction of force on a (small test) mass (ii) (tangent to line gives) direction of force on a (small test) charge charge is positive

(b) similarity: e.g. radial fields lines normal to surface greater separation of lines with increased distance from sphere field strength ∝ 1 / (distance to centre of sphere)2 (allow any sensible answer)

A LEVEL – NOVEMBER 2005

28

(b) flux through coil = BA sinθ ……………………………………………….. flux linkage = BAN sinθ ………………..………………………………… and non zero between the poles and zero outside (b) graph: VH constant sharp increase/decrease at ends of magnet (c) (i) (induced) e.m.f. proportional to ………………………………..…..…..… rate of change of flux (linkage) …………………….…………………….

7

(a) (i)

CEDAR COLLEGE

(ii)

graph:

two square sections in correct positions, zero elsewhere ….. pulses in opposite directions …………………………………… amplitude of second about twice amplitude of first …………..

energy required to separate the nucleons in a nucleus ..………..…….. Cambridge International Examinations 2012 nucleons separated© to infinity / completely …………..……………….. S shown at peak

M1 A1

[2

B1 B1 (B1) (B1)

[3

A1 C1 A1 B1 Paper

Syllabus 9702

5 4 (a)(a) force ½mv = qV is……(or verbal explanation) …..……………..…… on2 proton normal some to velocity and field provides centripetal v2 = (for 1.6 circular × 10-19 ×motion) 1.2 × 104 ………………………… ½ × 9.11 × 10-31 ×force v = 6.49 × 107 m s–1 ………….………………………………………… (b) magnetic force = Bqv 2 /r force =circular mrω2 or (b)(i) centripetal within field: arcmv ………………..………………..…………….. in ‘downward’ direction ……………..………………….. v = rω 2 beyond straight, with no ‘kink’ on leaving field ………………… Bqv = Bqrωfield: = mrω ω = Bq/m (ii) 1. v is smaller …………………………………………………………………. deflection is larger ………………………………………………………… (magnetic) force is larger ………………………………………………… 5 (a) 2. either φ = BA sinθ deflection is larger(through ……………………………………………………….. where A is the area which flux passes) θ is the angle between B and (plane of) A 6 (a) or (numerically equal to) force per unit length …………………….…….… straight conductor carrying unit current ……………………………. φ =on BA normal the field where A istoarea normal………………………………………………………… to B

(ii)

[1

B1

difference: e.g. gravitational force (always) towards sphere electric force direction depends on sign of charge on sphere / towards or away from sphere e.g. gravitational field/force is attractive electric field/force is attractive or repulsive (allow any sensible comparison)

(c) gravitational force = 1.67 × 10–27 × 9.81 = 1.6 × 10–26 N electric force = 1.6 × 10–19 × 270 / (1.8 × 10–2) = 2.4 × 10–15 N electric force Page 2 force very much greater than Markgravitational Scheme

B1

…..………………………………………………………

[4

4

B1 B1 A0 B1 B1 B1

M1 A1 B1 B1 B1 A1

M1 A1 M1 A1

M1 A1 B1 B1 B1 M1

198 A1 B1

[4

M1

A1 M1 A1 (M1) A1 (A1) B1 B1

[2

M1 A1

[2

[2

A2 IDEAL GASES


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1 May/June 2002 , Question #3 , qp_4

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200

CEDAR COLLEGE

201

2 Oct/Nov 2003 , Question #3 , qp_4

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202

3 May/June 2004 , Question #2 , qp_4 2

For Examiner’s Use

5

The pressure p of an ideal gas is given by the expression p =

1 Nm 2 . 3 V

(a) Explain the meaning of the symbol . .......................................................................................................................................... .................................................................................................................................... [2] (b) The ideal gas has a density of 2.4 kg m–3 at a pressure of 2.0 temperature of 300 K. (i)

105 Pa and a

Determine the root-mean-square (r.m.s.) speed of the gas atoms at 300 K.

r.m.s. speed = .................................. m s–1 [3] (ii)

Calculate the temperature of the gas for the atoms to have an r.m.s. speed that is twice that calculated in (i).

temperature = ......................................... K [3]

203

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4 Oct/Nov , 2002 , Question #4 , qp_4

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204

5

5 Oct/Nov 2005 , Question #2 , qp_4 2

The air in a car tyre has a constant volume of 3.1 10–2 m3. The pressure of this air is 2.9 105 Pa at a temperature of 17 °C. The air may be considered to be an ideal gas.

For Examiner’s Use

(a) State what is meant by an ideal gas. .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [2] (b) Calculate the amount of air, in mol, in the tyre.

amount = ……………………………. mol [2] (c) The pressure in the tyre is to be increased using a pump. On each stroke of the pump, 0.012 mol of air is forced into the tyre. Calculate the number of strokes of the pump required to increase the pressure to 3.4 105 Pa at a temperature of 27 °C.

number = ……………………………. [3]

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205 [Turn over

6 May/June 2006 , Question #2 , qp_4 2

For Examiner’s Use

6

(a) The equation pV = constant

T

relates the pressure p and volume V of a gas to its kelvin (thermodynamic) temperature T. State two conditions for the equation to be valid. 1. ..................................................................................................................................... .......................................................................................................................................... 2. ..................................................................................................................................... ..................................................................................................................................... [2] (b) A gas cylinder contains 4.00 104 cm3 of hydrogen at a pressure of 2.50 107 Pa and a temperature of 290 K. The cylinder is to be used to fill balloons. Each balloon, when filled, contains 7.24 103 cm3 of hydrogen at a pressure of 1.85 105 Pa and a temperature of 290 K. Calculate, assuming that the hydrogen obeys the equation in (a), (i)

the total amount of hydrogen in the cylinder,

amount = ……………………….. mol [3] (ii)

the number of balloons that can be filled from the cylinder.

number = ……………………….. [3] © UCLES 2006

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206

7 Oct/Nov 2007 , Question #2 , qp_4 2

6

(a) An amount of 1.00 mol of Helium-4 gas is contained in a cylinder at a pressure of 1.02 × 105 Pa and a temperature of 27 °C. (i)

Calculate the volume of gas in the cylinder.

volume = ............................................ m3 [2] (ii)

Hence show that the average separation of gas atoms in the cylinder is approximately 3.4 × 10–9 m.

[2] (b) Calculate (i)

the gravitational force between two Helium-4 atoms that are separated by a distance of 3.4 × 10–9 m,

force = .............................................. N [3]

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For Examiner’s Use

7 (ii)

the ratio weight of a Helium-4 atom . gravitational force between two Helium-4 atoms with separation 3.4 × 10–9 m

ratio = ...................................................[2] (c) Comment on your answer to (b)(ii) with reference to one of the assumptions of the kinetic theory of gases. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2]

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208 [Turn over

8 May/June 2008 , Question #2 , qp_4 2

6

(a) Explain qualitatively how molecular movement causes the pressure exerted by a gas. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ...................................................................................................................................... [3] (b) The density of neon gas at a temperature of 273 K and a pressure of 1.02 × 105 Pa is 0.900 kg m–3. Neon may be assumed to be an ideal gas. Calculate the root-mean-square (r.m.s.) speed of neon atoms at (i)

273 K,

speed = ........................................... m s–1 [3] (ii)

546 K.

speed = ........................................... m s–1 [2]

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For Examiner’s Use

7 (c) The calculations in (b) are based on the density for neon being 0.900 kg m–3. Suggest the effect, if any, on the root-mean-square speed of changing the density at constant temperature.

For Examiner’s Use

.......................................................................................................................................... .......................................................................................................................................... ...................................................................................................................................... [2]

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210

9 Oct/Nov 2009 , Question #2 , qp_41 2

5

An ideal gas occupies a container of volume 4.5 × 103 cm3 at a pressure of 2.5 × 105 Pa and a temperature of 290 K.

For Examiner’s Use

(a) Show that the number of atoms of gas in the container is 2.8 × 1023.

[2] (b) Atoms of a real gas each have a diameter of 1.2 × 10–10 m. (i)

Estimate the volume occupied by 2.8 × 1023 atoms of this gas.

volume = ......................................... m3 [2] (ii)

By reference to your answer in (i), suggest whether the real gas does approximate to an ideal gas. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................ [2]

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10 Oct/Nov 2009 , Question #2 , qp_42 2

6

(a) State what is meant by the internal energy of a gas. ..........................................................................................................................................

For Examiner’s Use

.......................................................................................................................................... .................................................................................................................................... [2] (b) The first law of thermodynamics may be represented by the equation

!U = q + w. State what is meant by each of the following symbols. +!U ................................................................................................................................. +q ..................................................................................................................................... +w .................................................................................................................................... [3] (c) An amount of 0.18 mol of an ideal gas is held in an insulated cylinder fitted with a piston, as shown in Fig. 2.1. piston

gas insulated cylinder Fig. 2.1 Atmospheric pressure is 1.0 × 105 Pa. The volume of the gas is suddenly increased from 1.8 × 103 cm3 to 2.1 × 103 cm3. For the expansion of the gas, (i)

calculate the work done by the gas and hence show that the internal energy changes by 30 J,

[3] © UCLES 2009

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212

11 May/June 2010 , Question #2 , qp_41 2

5

(a) Some gas, initially at a temperature of 27.2 °C, is heated so that its temperature rises to 38.8 °C. Calculate, in kelvin, to an appropriate number of decimal places, (i)

For Examiner’s Use

the initial temperature of the gas,

initial temperature = ............................................. K [2] (ii)

the rise in temperature.

rise in temperature = ............................................ K [1] (b) The pressure p of an ideal gas is given by the expression p = 13 ρ!c 2" where ρ is the density of the gas. (i)

State the meaning of the symbol !c 2". .................................................................................................................................. .............................................................................................................................. [1]

(ii)

Use the expression to show that the mean kinetic energy <EK> of the atoms of an ideal gas is given by the expression <EK> = 32 kT. Explain any symbols that you use. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [4]

213

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6 (c) Helium-4 may be assumed to behave as an ideal gas. A cylinder has a constant volume of 7.8 × 103 cm3 and contains helium-4 gas at a pressure of 2.1 × 107 Pa and at a temperature of 290 K. Calculate, for the helium gas, (i)

the amount of gas,

amount = ......................................... mol [2] (ii)

the mean kinetic energy of the atoms,

mean kinetic energy = .............................................. J [2] (iii)

the total internal energy.

internal energy = .............................................. J [3]

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214

For Examiner’s Use

12 Oct/Nov 2010 , Question #2 , qp_42 2

(a) (i)

6

State the basic assumption of the kinetic theory of gases that leads to the conclusion that the potential energy between the atoms of an ideal gas is zero.

For Examiner’s Use

.................................................................................................................................. .............................................................................................................................. [1] (ii)

State what is meant by the internal energy of a substance. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

(iii)

Explain why an increase in internal energy of an ideal gas is directly related to a rise in temperature of the gas. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

(b) A fixed mass of an ideal gas undergoes a cycle PQRP of changes as shown in Fig. 2.1. 10

P

8 volume / 10–4 m3 6

4

2

0

R

Q

0

5

10

15

20

25 30 5 pressure / 10 Pa

Fig. 2.1

215

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13 Oct/Nov 2010 , Question #2 , qp_43 2

6

(a) State the basic assumptions of the kinetic theory of gases. ..........................................................................................................................................

For Examiner’s Use

.......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ...................................................................................................................................... [4] (b) Use equations for the pressure of an ideal gas to deduce that the average translational kinetic energy <EK> of a molecule of an ideal gas is given by the expression <EK> =

3 RT 2 NA

where R is the molar gas constant, NA is the Avogadro constant and T is the thermodynamic temperature of the gas.

[3] 2

(c) A deuterium nucleus 1H and a proton collide. A nuclear reaction occurs, represented by the equation 2 1H

(i)

+

1 1p

3 2 He

+ c.

State and explain whether the reaction represents nuclear fission or nuclear fusion. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [2]

216

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7 (ii)

For the reaction to occur, the minimum total kinetic energy of the deuterium nucleus and the proton is 2.4 × 10–14 J. Assuming that a sample of a mixture of deuterium nuclei and protons behaves as an ideal gas, calculate the temperature of the sample for this reaction to occur.

For Examiner’s Use

temperature = ............................................. K [3] (iii)

Suggest why the assumption made in (ii) may not be valid. .................................................................................................................................. .............................................................................................................................. [1]

217

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14 May/June 2011 , Question #2 , qp_41 2

5

(a) State what is meant by the Avogadro constant NA. ..........................................................................................................................................

For Examiner’s Use

.......................................................................................................................................... ...................................................................................................................................... [2] (b) A balloon is filled with helium gas at a pressure of 1.1 × 105 Pa and a temperature of 25 °C. The balloon has a volume of 6.5 × 104 cm3. Helium may be assumed to be an ideal gas. Determine the number of gas atoms in the balloon.

number = ................................................ [4]

218

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15 May/June 2011 , Question #2 , qp_42 2

5

(a) State what is meant by a mole. ..........................................................................................................................................

For Examiner’s Use

.......................................................................................................................................... .................................................................................................................................... [2] (b) Two containers A and B are joined by a tube of negligible volume, as illustrated in Fig. 2.1.

container A 3.1 × 103 cm3 17 °C

container B 4.6 × 103 cm3 30 °C

Fig. 2.1 The containers are filled with an ideal gas at a pressure of 2.3 × 105 Pa. The gas in container A has volume 3.1 × 103 cm3 and is at a temperature of 17 °C. The gas in container B has volume 4.6 × 103 cm3 and is at a temperature of 30 °C. Calculate the total amount of gas, in mol, in the containers.

amount = ........................................ mol [4]

219

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16 Oct/Nov 2011 , Question #2 , qp_42 2

6

(a) One assumption of the kinetic theory of gases is that gas molecules behave as if they are hard, elastic identical spheres. State two other assumptions of the kinetic theory of gases. 1. ...................................................................................................................................... .......................................................................................................................................... 2. ...................................................................................................................................... .......................................................................................................................................... [2] (b) Using the kinetic theory of gases, it can be shown that the product of the pressure p and the volume V of an ideal gas is given by the expression pV = 13 Nm where m is the mass of a gas molecule. (i)

State the meaning of the symbol 1.

N,

.............................................................................................................................. [1] 2.

.

.............................................................................................................................. [1] (ii)

Use the expression to deduce that the mean kinetic energy <EK > of a gas molecule at temperature T is given by the equation <EK> = 32 kT where k is a constant.

[2]

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For Examiner’s Use

7 (c) (i)

State what is meant by the internal energy of a substance. ..................................................................................................................................

For Examiner’s Use

.................................................................................................................................. .............................................................................................................................. [2] (ii)

Use the equation in (b)(ii) to explain that, for an ideal gas, a change in internal energy ΔU is given by

ΔU ∝ ΔT where ΔT is the change in temperature of the gas. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]




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17 May/June 2012 , Question #2 , qp_41 2

6

(a) The kinetic theory of gases is based on some simplifying assumptions. The molecules of the gas are assumed to behave as hard elastic identical spheres. State the assumption about ideal gas molecules based on (i)

For Examiner’s Use

the nature of their movement, .................................................................................................................................. ............................................................................................................................. [1]

(ii)

their volume. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [2]

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7 (b) A cube of volume V contains N molecules of an ideal gas. Each molecule has a component cX of velocity normal to one side S of the cube, as shown in Fig. 2.1.

For Examiner’s Use

side S

cx

Fig. 2.1 The pressure p of the gas due to the component cX of velocity is given by the expression pV = NmcX2 where m is the mass of a molecule. Explain how the expression leads to the relation pV = 13 Nm where is the mean square speed of the molecules.

[3] (c) The molecules of an ideal gas have a root-mean-square (r.m.s.) speed of 520 m s–1 at a temperature of 27 °C. Calculate the r.m.s. speed of the molecules at a temperature of 100 °C.

© UCLES 2012

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r.m.s. speed = ....................................... m s–1 [3] 9702/41/M/J/12 [Turn over

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18 Oct/Nov 2012 , Question #2 , qp_41 2

6

A student suggests that, when an ideal gas is heated from 100 °C to 200 °C, the internal energy of the gas is doubled. (a) (i)

State what is meant by internal energy. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

(ii)

By reference to one of the assumptions of the kinetic theory of gases and your answer in (i), deduce what is meant by the internal energy of an ideal gas. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [3]

(b) State and explain whether the student’s suggestion is correct. .......................................................................................................................................... .......................................................................................................................................... ...................................................................................................................................... [2]

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4 Section A

For Examiner’s Use

Answer all the questions in the spaces provided. 19 Oct/Nov 2012 , Question #1 , qp_43 1

An ideal gas has volume V and pressure p. For this gas, the product pV is given by the expression pV = 13 Nm where m is the mass of a molecule of the gas. (a) State the meaning of the symbol (i)

N, .............................................................................................................................. [1]

(ii)

. .............................................................................................................................. [1]

(b) A gas cylinder of volume 2.1 × 104 cm3 contains helium-4 gas at pressure 6.1 × 105 Pa and temperature 12 °C. Helium-4 may be assumed to be an ideal gas. (i)

Determine, for the helium gas, 1.

the amount, in mol,

amount = ......................................... mol [3] 2.

the number of atoms.

number = .................................................. [2]

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5 (ii)

Calculate the root-mean-square (r.m.s.) speed of the helium atoms.

For Examiner’s Use

r.m.s. speed = ....................................... m s–1 [3]

226

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20 May/June 2013 , Question #2 , qp_41 2

6

(a) State what is meant by an ideal gas. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ...................................................................................................................................... [3] (b) Two cylinders A and B are connected by a tube of negligible volume, as shown in Fig. 2.1.

cylinder A tap T

cylinder B

2.5 × 103 cm3 3.4 × 105 Pa

1.6 × 103 cm3

300 K

4.9 × 105 Pa

tube Fig. 2.1 Initially, tap T is closed. The cylinders contain an ideal gas at different pressures. (i)

Cylinder A has a constant volume of 2.5 × 103 cm3 and contains gas at pressure 3.4 × 105 Pa and temperature 300 K. Show that cylinder A contains 0.34 mol of gas.

[1]

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7 (ii) Cylinder B has a constant volume of 1.6 × 103 cm3 and contains 0.20 mol of gas. When tap T is opened, the pressure of the gas in both cylinders is 3.9 × 105 Pa. No thermal energy enters or leaves the gas.

For Examiner’s Use

Determine the final temperature of the gas.

temperature = .............................................. K [2] (c) By reference to work done and change in internal energy, suggest why the temperature of the gas in cylinder A has changed. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ...................................................................................................................................... [3]

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228

21 May/June 2013 , Question #2 , qp_42 2

5

(a) The volume of an ideal gas in a cylinder is 1.80 × 10–3 m3 at a pressure of 2.60 × 105 Pa and a temperature of 297 K, as illustrated in Fig. 2.1.

For Examiner’s Use

ideal gas 1.80 × 10–3 m3 2.60 × 105 Pa 297 K

Fig. 2.1 The thermal energy required to raise the temperature by 1.00 K of 1.00 mol of the gas at constant volume is 12.5 J. The gas is heated at constant volume such that the internal energy of the gas increases by 95.0 J. (i)

Calculate 1. the amount of gas, in mol, in the cylinder,

amount = ........................................... mol [2] 2. the rise in temperature of the gas.

temperature rise = .............................................. K [2]

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6 (ii)

Use your answer in (i) part 2 to show that the final pressure of the gas in the cylinder is 2.95 × 105 Pa.

For Examiner’s Use

[1] (b) The gas is now allowed to expand. No thermal energy enters or leaves the gas. The gas does 120 J of work when expanding against the external pressure. State and explain whether the final temperature of the gas is above or below 297 K. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [3]

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22 Oct/Nov 2013 , Question #2 , qp_43 2

(a) (i)

6

State what is meant by the internal energy of a system. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

(ii)

Explain why, for an ideal gas, the internal energy is equal to the total kinetic energy of the molecules of the gas. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

(b) The mean kinetic energy <EK> of a molecule of an ideal gas is given by the expression <EK> = 32 kT where k is the Boltzmann constant and T is the thermodynamic temperature of the gas. A cylinder contains 1.0 mol of an ideal gas. The gas is heated so that its temperature changes from 280 K to 460 K. (i)

Calculate the change in total kinetic energy of the gas molecules.

change in energy = ............................................. J [2]

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23 May/June 2014 , Question #2 , qp_41 2

5

(a) Explain what is meant by the Avogadro constant. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) Argon-40 ( 40 18Ar) may be assumed to be an ideal gas. A mass of 3.2 g of argon-40 has a volume of 210 cm3 at a temperature of 37 °C. Determine, for this mass of argon-40 gas, (i)

the amount, in mol,

amount = ................................................. mol [1] (ii)

the pressure,

pressure = ................................................... Pa [2] (iii)

the root-mean-square (r.m.s.) speed of an argon atom.

r.m.s. speed = ............................................... m s−1 [3]

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232

24 May/June 2014 , Question #2 , qp_42 2

6

A constant mass of an ideal gas has a volume of 3.49 × 103 cm3 at a temperature of 21.0 °C. When the gas is heated, 565 J of thermal energy causes it to expand to a volume of 3.87 × 103 cm3 at 53.0 °C. This is illustrated in Fig. 2.1.

3.49 × 103 cm3 21.0 °C

565 J

3.87 × 103 cm3 53.0 °C

Fig. 2.1 (a) Show that the initial and final pressures of the gas are equal.

[2] (b) The pressure of the gas is 4.20 × 105 Pa. For this heating of the gas, (i)

calculate the work done by the gas,

work done = ..................................................... J [2]

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7 (ii)

use the first law of thermodynamics and your answer in (i) to determine the change in internal energy of the gas.

change in internal energy = ..................................................... J [2] (c) Explain why the change in kinetic energy of the molecules of this ideal gas is equal to the change in internal energy. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3]

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234

25 Oct/Nov 2014 , Question #3 , qp_43 3

10

A fixed mass of gas has an initial volume of 5.00 × 10−4 m3 at a pressure of 2.40 × 105 Pa and a temperature of 288 K. It is heated at constant pressure so that, in its final state, the volume is 14.5 × 10−4 m3 at a temperature of 835 K, as illustrated in Fig. 3.1. initial state

final state

5.00 × 10–4 m3 2.40 × 105 Pa 288 K

14.5 × 10–4 m3 2.40 × 105 Pa 835 K

Fig. 3.1 (a) Show that these two states provide evidence that the gas behaves as an ideal gas.

[3] (b) The total thermal energy supplied to the gas for this change is 569 J. Determine (i)

the external work done,

work done = ..................................................... J [2]

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11 (ii)

the change in internal energy of the gas. State whether the change is an increase or a decrease in internal energy.

change in internal energy = ........................................................... J ........................................................................................................ [2]

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236

26 May/June 2015 , Question #2 , qp_41 2

6

(a) State what is meant by internal energy. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) The variation with volume V of the pressure p of an ideal gas as it undergoes a cycle ABCA of changes is shown in Fig. 2.1. 4.0 p / 105 Pa B 3.5

3.0

2.5

2.0

1.5 A 1.0 3.0

4.0

C 5.0

6.0

7.0 V / 10

8.0 m3

Fig. 2.1 The temperature of the gas at A is 290 K. The temperature at B is 870 K.

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237

7 Determine (i)

the amount, in mol, of gas,

amount = .................................................. mol [2] (ii)

the temperature of the gas at C.

temperature = ..................................................... K [2] (c) Explain why the change from C to A involves external work and a change in internal energy. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2]

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238

27 May/June 2015 , Question #2 , qp_42 2

7

In a sample of gas at room temperature, five atoms have the following speeds: 1.32 × 103 m s–1 1.50 × 103 m s–1 1.46 × 103 m s–1 1.28 × 103 m s–1 1.64 × 103 m s–1. For these five atoms, calculate, to three significant figures, (a) the mean speed,

mean speed = ................................................. m s–1 [1] (b) the mean-square speed,

mean-square speed = ................................................ m2 s–2 [2] (c) the root-mean-square speed.

root-mean-square speed = ................................................. m s–1 [1]

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239

mass ....................................................................................... B1 acceleration = 9.77 m s-2 .............................................................. B1 2 (a)

(b)

(i)

a,ω and x identified ………(-1 each error or omission) ................. B2

(ii)

(-)ve because a and x in opposite directions OR a directed towards mean position/centre................................ B1

[3]

forces in springs are k(e + x) and k(e – x) .................................... C1 resultant = k(e + x) – k(e – x) ......................................................M1 = 2kx ............................................................................ A0

[2]

F = ma ....................................................................................... B1 a = -2kx/m .................................................................................... A0 (-)ve sign explained...................................................................... B1

[2]

ω2 = 2k/m ..................................................................................... C1 (2πf)2 = (2 x 120)/0.90 .................................................................. C1 f = 2.6 Hz ..................................................................................... A1

[3]

atom held in position by attractive forces atom oscillates, not just two forces OR 3D not 1D force not proportional to x any two relevant points, 1 each, max 2 ........................................ B2

[2]

(i)

1

(ii)

(iii)

(c)

23 (a)

(i)

(ii)2 Page

(a)

(a)

(b)

[3]

∆U = q + w symbols identified correctly ..........................................................M1 directions correct.......................................................................... A1 q is zero ....................................................................................... B1 Mark Scheme Syllabus w is positive ∆U =EXAMINATIONS w and U increases A/ASOR LEVEL - JUNE.................................... 2004 9702B1 ∆U is rise in kinetic energy of atoms ............................................M1 and mean kinetic energy ∝ T ....................................................... A1 charge is quantised/enabled electron charge to be measured (allow one of the last two marks if states ‘U increases so T rises’)

all are (approximately) n x (1.6 x 10-19 C) so e = 1.6 x 10-19 C (allow 2 sig. fig. only © University of Cambridge Local Examinations Syndicate 2003 summing charges and dividing ten, without explanation scores 1/2 Total mean (value of the) square of the speeds (velocities) of the atoms/particles/molecules

(b)

3 2

ANSWERS

pV/T = constant............................................................................ C1 T = (6.5 x 106 x 30 x 300)/(1.1 x 105 x 540)................................. C1 = 985 K .................................................................................... A1 (if uses °C, allow 1/3 marks for clear formula)

3 (b)

1

[2]

(i)

[2] Paper 04

B1

[4] [1]

M1 A1

[2] [3]

M1 A1

[2]

C1

1

ρ < c2 >

p= 3

(ii)

= 3 x 2 x 105/2.4 = 2.5 x 105 r.m.s speed = 500 ms-1 new = 1.0 x 106 or increases by factor of 4 ∝ T or 3/2 kT = 1/2 m T = {(1.0 x 106) / (2.5 x 105)} x 300 = 1200 K

C1 A1 C1 C1 A1 Total

3

(a)

(i) (ii)

(b)

B1 B1

[2]

ω = 2π/(1.26 x 108) or 2π/T

C1 A1

[2]

-1

= 4.99 x 10 rad s allow 2 s.f.: 1.59π x 10-8 scores 1/2

240

CEDAR COLLEGE

(i)

[3] [8]

(force) = GM1M2/(R1 + R2)2 (force) = M1R1 ω 2 or M2R2 ω 2

-8

(c)

[3]

reference to either taking moments (about C) or same (centripetal) force M1R1 = M2R2 or M1R1 ω 2 = M2R2 ω 2

B1 B1

1

(a)

4

GM / R2 = Rω2 …………….…………………...…..………………….. ω = 2π / (24 × 3600) ………………………………..……..…………… 6.67 × 10–11 × 6.0 × 1024 = R3 × ω2 R3 = 7.57 × 1022 ………………………………………………………… R = 4.23 × 107 m ………………………………………………………..

(b)(i) ∆Φ = GM/Re – GM/Ro …...………………………………………….….. = (6.67 × 10–11 × 6.0 ×Mark 1024)Scheme ( 1 / 6.4 × 106 – 1 / 4.2 × 107) Page 1 Syllabus 7 –1 = 5.31 × 10GCE J kgA/AS …………………………………………………. Level – May/June 2006 9702 ∆EP = 5.31 × 107 × 650 …………………………………………………. 10 = 3.45 × 10 J …………………………………………………….. 1 (a) centripetal force is provided by gravitational force 2 2 mv / r = GMm / r (c) e.g. satellite will already have some speed in the correct direction … hence v = √(GM / r) 25 (a) obeys the law2 pV = constant × T ………………………..……………….. (b) (i) at E ½mvof) = allK (= values p,GMm V and/ T2r ………………………………………………. - GMm (b) (ii)n E=P =(2.9 × 105/ ×r 3.1 × 10–2) / (8.31 × 290) …..………………..………... = 3.73 mol ………………………………………………………………. (iii) ET = - GMm / r + GMm / 2r 3 .4 290 = - GMm / 2r. (c) at new pressure, n n = 3.73 × × 2 .9

62

(a)

statement, words or symbols …..…………………………...….. (ii)correct EK = GMm / 2r and r decreases so (EK and) v increases (b)(i) w = p∆V ………………………………………………………………….. 5 = 1.03 × 10 × (2.96 × 10–2 – 1.87 × 10–5) (a) e.g. fixed mass/ amount of gas = (–) 3050 J …………………..……………….…………..…………… ideal gas (any two, 1 each) 4 (ii) q = 4.05 × 10 J …………………………………………………………. (b) (i) n = pV / RT 4 (iii) ∆U = =4.05 J -6 )…no e.c.f. from (a)………………… (2.5××10 107 –×3050 4.00 =× 37500 104 x 10 / (8.31 × 290) penalise 2 sig.fig. once only = 415 mol (c)

4

3

4

M1 A0

[3]

C1 Paper C1 04 C1 A1 B1

B1

B1 A0

M1 A1 B1 C1 B1 A1 C1 A1

[4] [1] [2] [1] [2] [1]

[2] [2]

300

(c) (i) if ET decreases then=-4.23 GMmmol / 2r ….………………………………………. becomes more negative or GMm / 2r becomes larger change = 0.50 mol ……………………………………………………….… so r decreases number of strokes = 0.50 / 0.012 = 42 (must round up for mark) …….

3

C1 C1

number of molecules = NA ………………………………………………. 5 7 (ii)energy volume=of gas at 1.85 ×× 10 × 104 ) / (1.85 × 105 ) ) = (2.5 × 10 6 × 4.00 37500 / (6.02 1023Pa 3 = 5.41 × 10 cm = 6.2 ×6 10–20 J (accept 1 sig.fig.) …………………………..…. 4 3 so, 5.41 × 10 = 4.00 × 10 + 7.24 × 10 N = 2πf 741 ………………………………………………………....………….. (a) (i) ω N = (answer = 2π × 740 1400or fails to allow for gas in cylinder, max 2/3 ) = 8800 rad s–1 ………………………………………………………….. (a) gradient of graph is (a measure of) the sensitivity 2 with temperature (ii)thea0gradient = (–)ωvaries x0 ……………………………..………………………………… 2 = (8800) × 0.080 × 10–3 (b) 2040 ±=20 Ω corresponds to 15.0 ± 0.2 °C 6200 m s–2 …………………….……………………………………. T / K = T / °C + 273.15 (allow 273.2) (b) temperature straight lineisthrough 288.2 Korigin with negative gradient …….…………….... end points of line correctly labelled …………………………………….. (a) (i) 1.0 (c) (i) zero displacement ………………………………………………………… (ii) 40 Hz (ii) v = ωx0 ……………………………………………………………………. = 8800 ×fa 0.080 × 10–3 (b) (i) speed = 2π = 0.70 m s–1 ……………………………………………………………. = 2π × 40 × 42 × 10-3 = 10.6 m s-1

C1 C1 M1 A1 A1 B1

[2] [3]

[1]

M1 A1

[2]

B2

[2]

C1 A1

[2]

B1 A1

[1] C1 C1 A1

[1] [3]

C1 C1 C1 C1 A1

A1

[2] [3]

A1

[2]

M1 A1 C1 C1 C1 M1 A1 A1 B1 B1 B1 C1

[2]

A1

[2] [3]

[2] [1]

[1] [1]

C1

A1

[2] A1

[2]

© University of Cambridge International Examinations 2005

(ii) acceleration = 4π2 f2 a = (80π)2 × 42 × 10-3 = 2650 m s-2 CEDAR COLLEGE (c) (i) S marked correctly (on ‘horizontal line through centre of wheel)

C1 A1 241 B1

[2]

(L – 13) × 5/1.8 = 5.0 + 5/9.8 × L × 10-2 × 8.72 .................................................. C1 L = 17.2 cm ........................................................................................................... A1 (constant centripetal force of 5.0 N gives L = 16.6 cm allow 2/4)

72

1

[4]

(a) (i)2 pV = nRT Page Mark Scheme Syllabus Paper V = (8.31 × 300)/(1.02 × LEVEL 105) ............................................................................... C1 04 GCE A/AS – May/June 2008 9702 = 0.0244 m3 (if uses Celsius, then 0/2) .......................................................... A1 [2]

Section A (ii) volume occupied by one atom = 0.0244 / (6.02 × 1023) = 4.06 × 10-26 m3 ............ M1 separation ≈ 3√(4.06 10-26) of ................................................................................ A1 B1 (a) (i) angle (subtended) at×centre circle -9 = 3.44 × 10 m ................................................................................... A0 by an arc equal in length to the radius (of the circle) B1[2] [2] (ii) angle swept out per unit time / rate of change of angle M1 (b) (i) F = GMm / r2 ....................................................................................................... C1 by the string -11 A1 [2] = (6.67 × 10 × {4 × 1.66 × 10-27}2) / (3.44 × 10-9)2 .......................................... C1 [3] = 2.49 × 10-46 N ................................................................................................ A1 (b) friction provides / equals the centripetal force B1 (ii) ratio = (4 ×2 1.66 × 10-27 × 9.8) / 2.49 × 10-46 ........................................................ C1 C1 0.72 W = mdω20 [2] = 2.6 × 10 .......................................................................................................... A1 0.72 mg = m × 0.35ω2 –1 ω = 4.49 (rad s ) C1 n = (ω /2π) × 60 (c) assumption that forces between atoms are negligible ................................................. B1 B1 –1 = 43 min (allow 42) gravitational force to be very small A1 [5] comment e.g. ratio shows e.g. force is very much less than weight e.g. if there are forces, they are not gravitational ....................................... B1 [2] (c) either centripetal force increases as r increases M1 or centripetal force larger at edge so flies off at edge first A1 [2] (F = mrω2 so edge first – treat as special case and allow one mark)

8 2

(a) molecule(s) rebound from wall of vessel / hits walls change in momentum gives rise to impulse / force either (many impulses) averaged to give constant force / pressure or the molecules are in random motion

"

(b) (i) p =

ρ

B1 B1 B1

[3]

C1

!

© UCLES 2007

1.02 × 105 =

"

× 0.900 ×

!

= 3.4 × 105 cRMS = 580 m s–1 (ii) either ∝ T or = 2 × 3.4 ×105 cRMS = 830 m s–1 (allow 820)

(c) cRMS depends on temperature (alone) so no effect

C1 A1

[3]

C1 A1

[2]

B1 B1

[2]

242

CEDAR COLLEGE

© UCLES 2008

1

x = 7.6 × 10 x = 4.2 107 m ................................................................................................. A1 (a) (i) force per×(unit) mass ……(ratio idea essential) ................................................. B1 (use of g = 10 m s-2, loses 1 mark but once only in the Paper)

[3] [1]

(ii) g = GM / R2 ....................................................................................................... C1 [Total: 11] 9.81 = (6.67 × 10-11 × M) / (6.38 × 106 )2 ……(all 3 s.f) ......................................M1 24 M = 5.99 × 10 kg ........................................................................................... A0 [2]

9 2

(a) either pV = NkT or pV = nRT and n = N / NA ..................................................... C1 clear correct substitution e.g. 2 gR = ω2r3 .................................................................. C1 (b) (i) either GM = ω3 2r3 or 5 -6 -23 2.5 × 10 × 4.5 × 10 -11 × 10 = N × 1.38 × 10 × 290 ...............................................M1 24 2 7 3 either 6.67 × 10 x 5.99 × 10 = ω × (2.86 × 10 ) A0 N = 2.8 × 1023 .......................................................................................................... or 9.81 × (6.38 × 106 )2 = ω2 × (2.86 × 107 )3 ............................................... C1 (allow 1 mark for calculation of n = 0.467 mol) -4 -1 ω = 1.3 × 10 rad s ......................................................................................... A1 (use of r = 2.22 × 107m scores max 2 marks) (b) (ii) (i) period volumeof =orbit (1.2=× 2π 10-10 )3 ×....................................................................................... 2.8 × 1023 or 4 πr3 × 2.8 × 1023 ..............................C1 C1 /ω 3 4 3 s (= 13.4 hours) 4.8-7×m10 2.53 ....................................................... × 10-7 m3 4 ..................................... A1 A1 = 4.8=× 10 period for geostationary satellite is 24 hours (= 8.6 × 10 s) ............................. A1 no 4.5 ................................................................................................................... (ii) so either × 103 cm3 >> 0.48 cm3 or ratio of volumes is about 10-4 ................A0 B1 justified because volume of molecules is negligible

........................................... B1

[2] [3]

[2] [3]

[2]

(c) satellite can then provide cover at Poles ...................................................................[Total: B1 [1] 6] [Total: 10]

10 2

(a) sum of kinetic and potential energies of molecules / particles / atoms ......................M1 random (distribution) ................................................................................................. A1

[2]

(b) +∆U: increase in internal energy ............................................................................... B1 +q: heating of / heat supplied to system ................................................................. B1 +w: work done on system ....................................................................................... B1

[3]

(c) (i) work done = p∆V .............................................................................................. C1 = 1.0 × 105 × (2.1 – 1.8) × 10-3 © UCLES 2009 = 30 J ..............................................................................................M1 w = 30 J, q = 0 so ∆U = 30 J .............................................................................. A1

[3]

(ii) these three marks were removed, as insufficient data was given in the question. [Total: 8]

© UCLES 2009

CEDAR COLLEGE

243

centripetal force = mrω2 15 = 3.0/9.8 × 0.85 × ω2 ω = 7.6 rad s–1 11 2

1

3 12 2

C1 C1 A1

(a) (i) 27.2 + 273.15 or 27.2 + 273.2 300.4 K Page 2 Mark Scheme: Teachers’ version (ii) 11.6 K GCE AS/A LEVEL – October/November 2010

Syllabus 9702

Section A (b) (i) ( is the) mean / average square speed (a) force per unit mass (ratio idea essential) (ii) ρ = Nm/V with N explained so, pV = 1/3 Nm and pV = NkT with k explained (b) graph: correct curvature 2 = ½m one gS) & at/ least other > correct fromkinetic (R,1.0energy

(c) = nRT (c) (i) (i) pV fields of Earth and Moon are in opposite directions 7 –3 × 7.8 × 10 = n × by 8.3subtraction × 290 2.1 × 10resultant either field found of the field strength nor= 68 any mol other sensible comment so there is a point where it is zero (ii) mean energy = 3/2 kT (allowkinetic FE = –F M for 2 marks) = 3/2 × 1.38 × 10–23 × 290 2 (ii) GME / x2 = GMM / (D –=x)6.0 × 10–21 J 24 22 (6.0 × 10 ) / (7.4 × 10 ) = x2 / (60RE – x)2 x = 54 RE that total internal energy is the total kinetic energy (iii) realisation energy = 6.0 × 10–21 × 68 × 6.02 × 1023 (iii) =graph: = 05 Jat least ⅔ distance to Moon 2.46 ×g 10 gE and gM in opposite directions correct curvature (by eye) and gE > gM at surface (a) (i) to-and-fro / backward and forward motion (between two limits)

[4]

C1 A1 [2] Paper A1 41 [1]

B1 B1 B1 B1 B1 M1 B1 A1

[4] [2]

M1 C1 A1 A1 A0

[2] [2]

C1 C1 A1 C1 A1 C1 C1 B1 A1 M1 A1 B1

[1] [1]

[2]

[3] [3]

[3] [1]

(a) (ii) (i) no noenergy forces (of between atoms / molecules / particles B1 [1] acting / constant energy / constant amplitude lossattraction or gain / or norepulsion) external force B1 [1] (ii) sum of kinetic and potential energy of atoms / molecules M1 due to random motion A1 [2] (iii) acceleration directed towards a fixed point B1 B1 [2] acceleration proportional to distance from the fixed point / displacement (iii) (random) kinetic energy increases with temperature M1 no potential energy (so increase in temperature increases internal energy) A1 [2] (b) acceleration is constant (magnitude) M1 so cannot be s.h.m. A1 [2] (b) (i) zero

A1

(ii) work done = p∆V –4 = 4.0 × 105 × 6 × 10 © UCLES 2010 = 240 J (ignore any sign)

[1]

C1 A1

[2]

B3

[3]

(iii) change

work done / J

heating / J

increase in internal energy / J

P→Q Q→R R→P

+240 0 –840

–600 +720 +480

–360 +720 –360

(correct signs essential) (each horizontal line correct, 1 mark – max 3 ) CEDAR COLLEGE

244

T = 165 years (ii) speed = (2π × 1.08 × 108) / (0.615 × 365 × 24 × 3600) = 35 km s–1

13 2

(a) atoms / molecules / particles behave as elastic (identical) spheres volume of atoms / molecules negligible compared to volume of containing vessel time of collision negligible to time between collisions no forces of attraction or repulsion between atoms / molecules atoms / molecules / particles are in (continuous) random motion (any four, 1 each)

(b) pV =

1 3

Nm and pV = nRT or pV = NkT

Nm = nRT or = NkT n = N/NA or k = R/NA 3 <EK> = × R/NA × T 2 1 3

and

protonproportional and deuterium nucleus have equal kinetic energies (a) (ii) (i) force to product ofwill masses –14 23 3 × 8.31 / (6.02 10 ) ×ofTseparation 1.2 × inversely 10 = 2 proportional force to ×square T = 5.8 × 10 K 9 than 1.16 radius / diameter of 1 Sun / planet (ii) separation (use of E =much 2.4 × greater 10–14 giving × 10 K scores mark) (iii) either inter-molecular / atomic / nuclear forces exist (b) (i) e.g. or field ∝ 1nucleus /r2 or force proton andstrength deuterium are positively charged / repel potential ∝ 1 / r

(ii) e.g. gravitational force (always) attractive electric force attractive or repulsive

(a) number of atoms of carbon-12 in 0.012 kg of carbon-12

© UCLES 2010

(b) pV = NkT or pV = nRT substitutes temperature as 298 K either 1.1 × 105 × 6.5 × 10–2 = N × 1.38 × 10–23 × 298 or 1.1 × 105 × 6.5 × 10–2 = n × 8.31 × 298 and n = N / 6.02 × 1023 N = 1.7 × 1024

3

(a) acceleration / force proportional to displacement from a fixed point acceleration / force (always) directed towards that fixed point / in opposite direction to displacement

(b) (i) Aρg / m is a constant and so acceleration proportional to x negative sign shows acceleration towards a fixed point / in opposite direction to displacement (ii) ω 2 = (Aρg / m) ω = 2πf (2 × π × 1.5)2 = ({4.5 × 10–4 × 1.0 × 103 × 9.81} / m) m = 50 g CEDAR COLLEGE

4

C1 A1

[2]

(1) (1) (1) (1) (1) B4

[4]

B1 B1 A0

8

14 2

[2]

B1

<EK> = ½m

Page 2 Mark Scheme: Teachers’ version Syllabus (c) (i) reaction represents either build-up of nucleus from light nuclei GCE AS/A LEVEL – May/June 2011 9702 or build-up of heavy nucleus from nuclei Section A so fusion reaction 1

A1

(a) work done in bringing unit positive charge

[3]

Paper 41 M1 A1 [2] B1 B1 B1 C1 A1 B1

B1

[2] [3] [1]

B1

[1] [1]

B1 B1

[2]

M1 A1

[2]

C1 C1 C1 A1

[4]

M1 A1

[2]

B1 B1

[2]

C1 C1 C1 A1 245

[4]

M1

GMm/r = mrω (must be in terms of ω) or3 2 FE = 2.30 × 10–28 × R –2 (C1) a constant r ω = GM and GM is FG = 1.86 × 10–64 × R –2 (C1) (A1) FE / FG = 1.2 × 1036 for Phobos, ω = 2π/(7.65 × 3600) = 2.28 × 10–4 rad s–1 (a) amount of substance 6 3 (9.39 × 10 ) × (2.28 × 10–4)2 = 6.67 × 10–11 × M containing same number23of particles as in 0.012 kg of carbon-12 M = 6.46 × 10 kg

B1 B1

C1

(b) (i) 1.

15 2

2. (9.39 × 106)3 × (2.28 × 10–4)2 = (1.99 × 107)3 × ω2 (b) pV = nRT –5 –1 = 7.30 3.1 rad × 10s–3) / (8.31 × 290) amount = ω(2.3 × 105××10 –5 ) T5 × = 4.6 2π/ω = –32π/(7.30 × 10 × 10 ) / (8.31 × 303) + (2.3 × 10 = 8.6 × 104 s = 0.296 + 0.420 = 0.716 mol = 23.6 hours (give full credit for starting equation pV = NkT and N = nNA) (ii) either almost ‘geostationary’ or satellite would take a long time to cross the sky

3 162

M1 C1 A1 A1 C1

(ii) either pV = NkT or pV = nRT and links n and k (c) combined capacitance of2 Y & Z = 20 µF or total capacitance = 6.67 µF and <EK> = ½m p.d. across capacitor X = 8 V or p.d. across combination = 12 V 3 –6 × 8 or to 6.67 10=–6 × kT 12 charge = 10 × 10 clear algebra leading <E×K> 2 = 80 µC

C1 C1 A1 A1

[3] [4]

B1

[1]

[3]

[2] [2]

C1 B1

[1]

B1

[2] [1]

C1 M1 C1 A1 A1

[2] [3]

M1 A1

[2]

A1

(c) (i) sum of potential energy and kinetic energy of molecules/atoms/particles reference to random (distribution) (ii) no intermolecular forces so no potential energy (change in) internal energy is (change in) kinetic energy and this is © University of in Cambridge International Examinations 2011 proportional to (change )T

B1 B1

© University of Cambridge International Examinations 2011

CEDAR COLLEGE

[2] [3]

C1 C1

(a) charges on plates are equal and opposite M1 so no resultant charge A1 (a) energy e.g. moving random there (rapid)ismotion molecules/atoms/particles storedinbecause chargeofseparation B1 no intermolecular forces of attraction/repulsion volume of molecules/atoms/particles negligible compared to volume of container = Q / V (b) (i) capacitance C1 time of collision negligible 10time between collisions = (18 × 10–3) / to (1 each, max 2) = 1800 µF A1 B2 (ii) use of area under graph or energy = ½CV2 (b) (i) 1. number of (gas) molecules –3 energy = 2.5 × 15.7 × 10 or energy = ½ × 1800 × 10–6 × (102 – 7.52) = 39 mJ 2. mean square speed/velocity (of gas molecules)

[3]

246

[2]

v = 2 × GM/r C1 C1 = (2 × 4.3 × 1013) / (3.4 × 106) A1 v = 5.0 × 103 m s–1 (Use of diameter instead of radius to give v = 3.6 × 103 m s–1 scores 2 marks) 172

1

random motion Mark constant velocity untilScheme hits wall/other molecule GCE AS/A LEVEL – October/November 2012 (ii) (total) volume of molecules is negligible Section compared to volume of containing vesselA or (a) force radius/diameter is proportional toofthe product ofisthe masses and a molecule negligible inversely proportional to the square of the separation compared to the average intermolecular distance either point masses or separation >> size of masses (a) (i) either Page 2 or

Syllabus 9702

(b) either molecule has component of velocity in three directions 2 2 2 the centripetal force (b) (i)or gravitational + cprovides c2 = cXforce Y + cZ 2 2 2 = GMm/r EK = ½mv mv /rmotion random andand averaging, so = = 2 GMm/2r EKX= > = 3 (ii) 1. ∆EK = ½ × 4.00 × 1014 × 620 × ({7.30 × 106}–1 – {7.34 × 106}–1) = 9.26 × 107 J (ignore any sign in answer) (allow 1.0 × 108 J if evidence that EK evaluated separately for each r) (c) ∝ T or crms ∝ ! temperatures are 300 K14and 373×K({7.30 × 106}–1 – {7.34 × 106}–1) × 10 × 620 2. ∆EP = 4.00 –1 crms = 580 m = s1.85 × 108 J (ignore any sign in answer) 8 (Do not(allow allow any marks J) of temperature in units of ºC instead of K) 1.8 or 1.9 ×for 10use

(iii) either (7.30 × 106)–1 – (7.34 × 106)–1 or ∆EK is positive / EK increased speed has increased 18 2

© University Cambridge International Examinations 2012 / molecules / particles (a) (i) sum of potential energy of and kinetic energy of atoms reference to random

(ii) no intermolecular forces no potential energy internal energy is kinetic energy (of random motion) of molecules (reference to random motion here then allow back credit to (i) if M1 scored)

3

B1Paper 41 M1 A1 (M1) (A1) M1 A1

M1 B1 M1 M1 A1 A0 A0 C1 A1 C1 C1 C1 A1 A1

[1]

[2] [2]

[2] [3] [2]

[3] [2]

M1 A1

[2]

M1 A1

[2]

B1 B1 B1

[3]

(b) kinetic energy ∝ thermodynamic temperature either temperature in Celsius, not kelvin so incorrect or temperature in kelvin is not doubled

B1 B1

[2]

(a) temperature of the spheres is the same no (net) transfer of energy between the spheres

B1 B1

[2]

(b) (i) power = m × c × ∆θ where m is mass per second 3800 = m × 4.2 × (42 – 18) m = 38 g s–1

C1 C1 A1

[3]

M1 A1

[2]

M1 A1 M1 A1

[4]

(ii) some thermal energy is lost to the surroundings so rate is an overestimate

4

[3]

(a) straight line through origin shows acceleration proportional to displacement negative gradient shows acceleration and displacement in opposite directions

CEDAR COLLEGE

247

Page 2

Mark Scheme GCE AS/A LEVEL – October/November 2012

Syllabus 9702

Paper 43

Syllabus 9702

Paper B1 [1] 41

Section A 19 1

Page Mark Scheme (a) (i)2 number of molecules GCE AS/A LEVEL – May/June 2013

(ii) mean square speed

B1

[1]

B1 C1 B1 C1

[2]

A1

[3]

Section A 1

(a) space area / volume (b) region (i) 1. ofpV = nRT where a mass experiences 104 × 10–6 ) / (8.31 × 285) n = (6.1 × 105 × 2.1a×force

n = 5.4 mol

2

20 2

(b) (i) force proportional 2. either N = nNAto product of two masses force=inversely proportional to the square of their separation 5.4 × 6.02 × 1023 either reference to point masses or separation >> ‘size’ of masses 24 = 3.26 × 10 or 2 (ii) field pV strength = NkT = GM / x or field strength ∝!"!#!x! 4 8 2 –6 –23 (6.1 ×× 10 1085)2× /2.1 ratioN= =(7.78 (1.5× ×10 10× ) 10 ) / (1.38 × 10 × 285) 24 N= =273.26 × 10

M1 M1 C1 A1 A1

[3]

C1 (C1) C1 (A1) A1

[2] [3]

(ii) either 6.1 × 105 × 2.1 × 10–2 = 1 /3 × 3.25 × 1024 × 4 × 1.66 × 10–27 × 1.78 × 106 force = mRω2 and ω = 2π / T = centripetal (c) (i) either × 103 m s–1 = mv2 / R and v = 2πR /T cRMS = 1.33 or centripetal force or gravitational force –27 provides the3 centripetal force 1 2 2 > = /2 × 1.38 2 /2 × 4GMm × 1.66/ R × 210 × 4π 2 M R3 / GT 3 –1 c RMS = 1.33 × 10 m s (allow working to be given in terms of acceleration)

C1 C1 A1 B1 B1 (C1) M1 (C1) A0 (A1)

[3] [3]

(ii) M = {4π2 × (1.5 × 1011)3} / {6.67 × 10–11 × (3.16 × 107)2} (a) (i) 1. 0.1 s, 0.3 s,300.5 s, etc (any two) = 2.0 × 10 kg

C1 A1 A1

[1] [2]

2. either 0, 0.4 s, 0.8 s, 1.2 s or (a) obeys the0.2 equation s, 0.6 s,pV 1.0=s constant (any two)× T or pV = nRT p, V and T explained at allperiod values=of0.4 p,sV and T/fixed mass/n is constant (ii)

M1 A1 A1 A1 C1

[3]

A1

[2]

5 3 –6 × 2.5 × 10 × 8.31 × 300 (b) (iii) (i) 3.4 × 10 or ½=πnrad phase difference = 90×°10 n = 0.34 mol

M1 B1 A0

[1] [1]

for total (b) (ii) frequency = 5mass/amount 2.4 – 2.5 Hz of gas 3.9 × 10 × (2.5 + 1.6) × 103 × 10–6 = (0.34 + 0.20) × 8.31 × T T = 360 K

B1 C1 A1

[1]

frequency = (1/0.4 =) 2.5 Hz

(c) e.g. attach sheet of card to trolley increases damping / frictional force (c) when tap opened e.g. reduce oscillator amplitude gas passed (from cylinder B)to tosystem cylinder A reduces power/energy input work done on gas in cylinder A (and no heating) so internal energy and hence temperature increase

M1 A1 (M1) B1 (A1) M1 A1

© Cambridge International Examinations 2012

248

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© Cambridge International Examinations 2013

[1]

[2]

[2] [3]

212

1

= 7.57 × 1022 R = 4.2 × 107 m (missing out 3600 gives 1.8 × 105 m and scores 2/3 marks) Page 2 Mark Scheme Syllabus GCE A LEVEL – October/November 2013 9702 (a) (i) 1. pV = nRT Section = n × A8.31 × 297 1.80 × 10–3 × 2.60 × 105 n = 0.19 mol (a) force proportional to product of the two masses and inversely proportional to the square of their separation 2. ∆q = mc∆T either reference to point masses or separation >> ‘size’ of masses 95.0 = 0.190 × 12.5 × ∆T ∆T = 40 K (allow 2 marks for the correct answerforce with clear logic shown) (b) gravitational force provides centripetal 2 2 GMm / R = mRω (ii) p/T where m =isconstant the5 mass of the planet 3 ×2 10 ) / 297 = p / (297 + 40) (2.6 GM = R ω p = 2.95 × 105 Pa

C1 A1 Paper 43

C1 A1 M1 A1 B1 A1 B1 M1 A1 M1 A0 A0

(c) ω = 2π / T C1 3 J / 25 J 2 (b) either changeMin internal energy is 120 B1 / M = (R / R ) × (T / T ) star Sun star Sun Sun star 3 2 30 internalMenergy decreases / ∆U is negative / kinetic energy of molecules decreases M1 C1 star = 4 × (½) × 2.0 × 10 so temperature lower A1 = 3.2 × 1031 kg A1 2 2 3 or Mstar = (2π) Rstar / GT (C1) 2 11 3 –11 2 = {(2π) × (6.0 × 10 ) } / {6.67 × 10 × (2 × 365 × 24 × 3600) } (C1) = 3.2 × 1031 kg (A1)

22 2

(a) (i) sum of kinetic and potential energies of the molecules reference to random distribution (ii) for ideal gas, no intermolecular forces so no potential energy (only kinetic)

3

(a) work done bringing unit positive charge from infinity (to the point)

(b) (i) either both potentials are positive / same sign so same sign or gradients are positive & negative (so fields in opposite directions) so same sign

[2]

[2]

[2]

[3] [1]

[3] [3]

M1 A1

[2]

M1 A1

[2]

(b) (i) either change in kinetic energy = 3/2 × 1.38 × 10–23 × 1.0 × 6.02 × 1023 × 180 C1 = 2240 J A1 or R = kNA energy = 3/2 × 1.0 × 8.31 × 180 (C1) = 2240 J (A1) (ii) increase in internal energy = heat supplied + work done on system 2240 = energy supplied – 1500 Cambridge International Examinations 2013 J energy supplied =©3740

[3]

[2]

B1 C1 A1

[3]

M1 A1

[2]

M1 A1 (M1) (A1)

[2]

(ii) the individual potentials are summed

B1

[1]

(iii) allow value of x between 10 nm and 13 nm

A1

[1]

(allow 0.42 V → 0.44 V) (iv) V = 0.43 V energy = 2 × 1.6 × 10–19 × 0.43 = 1.4 × 10–19 J

M1 A1 A1

[3]

249

CEDAR COLLEGE

© Cambridge International Examinations 2013

change in potential = (1/5 – 1/6) × (6.3 × 107) change in energy = (1/5 – 1/6) × (6.3 × 107) × 1.3 = 2.7 × 106 J

or

23 2 (a) the number of atoms in 12 g of carbon-12

(C1) (C1) (A1) [4]

M1 A1

[2]

A1

[1]

(ii) pV = nRT p × 210 × 10–6 = 0.080 × 8.31 × 310 p = 9.8 × 105 Pa (do not credit if T in °C not K)

C1 A1

[2]

(iii) either pV = 1/3 × Nm N = 0.080 × 6.02 × 1023 (= 4.82 × 1022) and m = 40 × 1.66 × 10–27 (= 6.64 × 10–26) 9.8 × 105 × 210 × 10–6 = 1/3 × 4.82 × 1022 × 6.64 × 10–26 × = 1.93 × 105 cRMS = 440 m s–1

C1 C1

(b) (i) amount = 3.2/40 = 0.080 mol

or

or

Nm = 3.2 × 10–3 9.8 × 105 × 210 × 10–6 = 1/3 × 3.2 × 10–3 × = 1.93 × 105 cRMS = 440 m s–1 1/2 m = 3/2 kT 1/2 × 40 × 1.66 × 10–27 = 3/2 × 1.38 × 10–23 × 310 = 1.93 × 105 cRMS = 440 m s–1

A1 (C1) (C1) (A1) (C1) (C1) (A1)

(if T in °C not K award max 1/3, unless already penalised in (b)(ii))

© Cambridge International Examinations 2014

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250

[3]

0 induced e.m.f. gives rise to currents which generate a magnetic field (B1) so stone could not enter into orbit A1 the magnetic field opposes the motion of the magnet so amplitude decreases (B1)

as magnet moves, is cut by cup / aluminium giving rise to induced e.m.f. (expressions in (a) flux 2and 2 (b)(i) must be dimensionally correct) C1 either (in cup)use of ½mω x0 and x0 = 0.75 cm or x0 is halved so ¼ energy B1 to give new energy = 1.6 mJ gives rise to currents and heating of the cup B1 (a) use induced of kelvin e.m.f. temperatures B1 mJ A1 either loss in energy = 6.4 – 1.6 or loss = ¾ × 6.4 giving loss = 4.8 energy oscillations of magnet amplitude decreases B1 correctfrom (11.87), V / T is constant soso pressure is constant M1 boththermal values of (V / T)derived or induced rise to currents whichofgenerate a magnetic field (B1) (allow use ofe.m.f. n R1.gives Do not allow other values n.) (c) q = mc∆θ the magnetic field opposes the motion of the magnet so amplitude decreases (B1) –3 10–3done = 6.2 R × 10 910 × ∆θ C1 (b) 4.8 (i) ×work p∆V × –4 2 2 5 3 –6 ∆θ = 8.5 × 10 K A1 x0 =–0.75 cm×or halved so ¼ energy C1 (ii) either use ofR½mω 0 and × (3.87 3.49) 10x0×is10 C1 4.2 × x10 energy = = = to = give new R 160 J= 1.6 mJ A1

[2] [3]

(b) (i) (ii)

24 2

=

2

[2] [2]

[3]

[2] [2]

(a) smooth curve with decreasing not =starting atgiving x = 0 loss = 4.8 mJ either in energy 6.4 gradient, – 1.6ofor loss ¾ × 6.4 (do notloss allow use of V= instead ∆V) end of line not at g = 0 or horizontal

M1 A1 A1

[2] [2]

=

(ii) increase / change in internal energy (c) q = mc∆θ (b) 4.8 straight line with×positive gradient × 10–3 = 6.2 10–3 × 910 × ∆θ –4 line starts at origin ∆θ = = == 8.5 × 10 K

C1 M1 C1 A1 A1 A1

[2] [2] [2]

2

(c) (a) (c)

B1 M1 B1 B1 A1 M1 B1 A1

[2] [3] [3]

R heating of system N work done on system R 565 – 160 R 405 J

sinusoidal shape smooth curve with decreasing gradient, not at x = 0 internal energy R sum ofpeak kinetic energy andstarting potential / trough height constant energy / EK N EP onlyof positive values and end line not at g = 0 or horizontal no intermolecular forces 4 ‘loops’ no potential energy (so ∆U R ∆EK)

(b) straight line with positive gradient = (2.40 × 105 × 5.00 × 10–4) / 288 = 0.417 (a) line initially, pVat/ Torigin starts (a) resonance finally, pV / T = (2.40 × 105 × 14.5 × 10–4) / 835 = 0.417 ideal gas because pV / T is constant (allow 2 marks for two determinations of V / T and then 1 mark for V / T and p (c) sinusoidal shape (b) Pt R mc ∆θ constant, so values ideal) and peak / trough height constant only positive × 2 × 60 R 0.28 × c × (98 – 25) 4750 ‘loops’ c R 4400 J kg–1 K–1

3 3

25 3

/ 3) × 10–4) / 288 = 0.417 (use of ∆θ N 273× max. / T73 = (2.40 105 × 15.00 (a) initially, pV R 5 s not 120× 10 s max. 2 / 3)× 10–4) / 835 = 0.417 (use ofpV t R/ T2 = finally, (2.40 × 14.5 ideal gas because pV / T is constant Page 3(allow 2 marks for two determinations Mark Scheme Syllabus p of V / T and then 1 mark for V / T and © Cambridge International Examinations 2014 Cambridge International AS/A Level – October/November 2014 9702 constant, so ideal) (b) (i) work done = p∆V = 2.40 105 × (14.5 – 5.00) × Examinations 10–4 © ×Cambridge International 2014 = 228 J (ignore sign, not 2 s.f.) (ii) ∆U = q + w = 569 – 228 = 341 J increase

4

M1 M1 A1 B1 M1 A1 B1 C1 B1 C1 B1 A1

[2] [1]

[3]

[3] [3]

M1 M1 A1 [3] Paper 43

C1 A1

[2]

M1 A1

[2]

toInternational displacement (from a fixed point) (a) acceleration / force proportional © Cambridge Examinations 2014 either acceleration and displacement in opposite directions or acceleration always directed towards a fixed point

M1 A1

[2]

(b) (i) zero & 0.625 s or 0.625 s & 1.25 s or 1.25 s & 1.875 s or 1.875 s & 2.5 s

A1

[1]

(ii) 1.

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2.

ω = 2π / T and v0 = ωx0 ω = 2π / 1.25 = 5.03 rad s–1 v0 = 5.03 × 3.2 = 16.1 cm s–1 (allow 2 s.f.) v = ω ( x 02 − x 2 )

C1 C1

251 A1

[3]

Neptune = 1.019 × 1026 kg Page 2alternative method: mass ofMark Scheme Syllabus 25 mass of Uranus Cambridge International AS / = A 8.621 Level ×– 10 Maykg / June 2015 9702 ratio = 1.18 1 (a) (i) 1. F = Gm1m2 / x2 = (6.67 × 10–11 × 2.50 × 5.98 × 1024 ) / (6.37 × 106 )2 26 2 (a) (sum of) potential and kinetic energy of molecules/atoms/particles = 24.6 Nenergy (accept 2 s.f. or more) mention of random motion/distribution 2. F = mxω2 or F = mv 2 / x and v = ωx (accept x or r for distance) = 2.50 × 6.37 × 106 × (2π / 24 × 3600)2 (b) (i) pV = nRT = 0.0842 N (accept 2 s.f. or more) either at A, 1.2 × 105 × 4.0 × 10−3 = n × 8.31 × 290 5 at B, = 24.575 3.6 × 10 × 4.0 × 10−3 = n × 8.31 × 870 (ii) or reading – 0.0842 n = 0.20 mol = 24.5 N (accept only 3 s.f.) Page 2 Mark Scheme Syllabus × 10–3 = 0.20AS × 8.31 × T or– T = (7.75 / 4.0) × 290 (ii) 1.2Cambridge × 105 × 7.75 International / A Level May / June 2015 9702 T = 560 K (b) gravitational force provides the centripetal force –3 2 (Allow from graph: × 10force m3) 1 (a) (i) 1. F = tolerance Gm / x‘equal’ 1m2 is gravitational force to the7.7–7.8 centripetal –11 24 6 2 2 = (6.67 (accept Gm = mxω×2 2.50 or FC× =5.98 FG) × 10 ) / (6.37 × 10 ) 1m2 / x× 10 = 24.6 N (accept 2 s.f. or more) ‘weight’/sensation of weight/contact force/reaction force is difference between FG (c) temperature so internal energy changes/decreases and FC whichchanges/decreases is zero volume constant so work is done or F = mv 2 / pressure) x and v = ωx (accept x or r for distance) 2. Fchanges = mxω2 (at 6 2 = 2.50 × 6.37 × 10 × (2π / 24 × 3600) 2 (a) mean speed = 1.44N×(accept 103 m s–1 = 0.0842 2 s.f. or more) 3 (a) (numerically equal to) quantity of (thermal) energy/heat to change state/phase of unit reading mass = 24.575 – 0.0842 (ii) at constant (b) evidence oftemperature summing of individual speeds only 3squared s.f.) = 24.5 N (accept (allow 1/2 for definition restricted or vaporisation) mean square speed = 2.09 × 106 to m2fusion s–2

3

27 2

(C1) Paper (C1) 42 (A1) [3] M1 M1 A1 A1

[2] [2]

C1 A1

[2]

C1 B1 A1 A1 [2] [2] Paper C1 42 A1 M1 [2] M1 M1 A1 [2] B1 A1 [3] B1 [2] C1 A1 A1

[1] [2]

M1 B1 A1 C1 A1 A1

[2] [2] [2]

(b) (i) at 70 W, mass s–1 = 0.26 g s–1 (b) gravitational force provides the centripetal force at 110 W, mass s–1 = 0.38 g s–1 3 (c) gravitational root-mean-square 1.45 10 m s–1 force force isspeed ‘equal’= to the×centripetal 2 2 (allow ECF from but only error) = mxω or ifFCarithmetic = F G) (accept Gm1m 2 / x (b) ‘weight’/sensation of weight/contact force/reaction force is difference between FG and FC which is zero (a) (numerically equal to) quantity of heat/(thermal) energy to change state/phase of unit mass at constant (a) mean speedtemperature = 1.44 × 103 m s–1 (allow 1/2 for definition restricted to fusion or vaporisation)

A1 M1 A1 A1 M1

[2] [1]

A1

[3]

(b) evidence of summing of individual squared speeds 6 (b) mean (i) constant gradient/straight m2(allow s–2 linear/constant slope) square speed = 2.09 × 10line

C1 B1 A1

[1] [2]

(ii) Pt = mL or power = gradient × L (c) root-mean-square speed = 1.45 × 103 m s–1 of from gradient of graph (allowuse ECF (b) but only if arithmetic error) (or two points separated by at least 3.5 minutes)

C1 A1

[1]

M1 A1 [1] [2] A1

© Cambridge International Examinations 2015

110 × 60equal = L ×to) (372 – 325)of×heat/(thermal) 10–3 / 7.0 (a) (numerically quantity energy to change state/phase of L = 9.80 × 105 J kg–1 (accept 2 s.f.) (allow 9.8 to 9.9 rounded to 2 s.f.) unit mass at constant temperature (iii) energy/heat is lost to the surroundings or vapour condenses on sides (allowsome 1/2 for definition restricted to fusion or vaporisation) so value is an overestimate

3

4

(b) (i) constant gradient/straight line (allow linear/constant slope) (a) displacement (directly) proportional to acceleration/force either acceleration in opposite directions ×L (ii) Pt =displacement mL or power and = gradient or acceleration (always) towards a (fixed) point use of gradient of graph (or two points separated by at least 3.5 minutes) 110 × 60 = L × (372 – 325) × 10–3 / 7.0 International Examinations 2015 (accept 2 s.f.) (allow 9.8 to 9.9 rounded to 2 s.f.) L = 9.80 × 105 J kg©–1 Cambridge

CEDAR COLLEGE

(iii) some energy/heat is lost to the surroundings or vapour condenses on sides so value is an overestimate

M1

A1 [3] M1 A1 [2] M1 A1 [2]

B1 [1] M1 C1 A1 [2] M1

A1

[3]

252 M1 A1

[2]

A2 TEMPERATURE


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253

1 May/June 2002 , Question #2 , qp_4 2

5

For Examiner’s Use

Some water in a saucepan is boiling. (a) Explain why (i)

external work is done by the boiling water, ................................................................................................................................... ................................................................................................................................... ...................................................................................................................................

(ii)

there is a change in the internal energy as water changes to steam. ................................................................................................................................... ................................................................................................................................... ................................................................................................................................... [5]

(b) By reference to the first law of thermodynamics and your answers in (a), show that thermal energy must be supplied to the water during the boiling process. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2]

254

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[Turn over

4

For Examiner’s Use

Answer all the questions in the spaces provided. 2 Oct/Nov 2002 , Question #1 , qp_4 1

A kettle is rated as 2.3 kW. A mass of 750 g of water at 20 °C is poured into the kettle. When the kettle is switched on, it takes 2.0 minutes for the water to start boiling. In a further 7.0 minutes, one half of the mass of water is boiled away. (a) Estimate, for this water, (i)

the specific heat capacity,

specific heat capacity = ........................................ J kg–1 K–1 (ii)

the specific latent heat of vaporisation.

specific latent heat = ........................................ J kg–1 [5] (b) State one assumption made in your calculations, and explain whether this will lead to an overestimation or an underestimation of the value for the specific latent heat. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2]

9702/4 O/N/02

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3 May/June 2003 , Question #2 , qp_4

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256

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257

4 May/June 2016 , Question #2 , qp_42 2

6

(a) State what is meant by (i)

the Avogadro constant NA, ........................................................................................................................................... ...................................................................................................................................... [1]

(ii)

the mole. ........................................................................................................................................... ...................................................................................................................................... [2]

(b) A container has a volume of 1.8 × 104 cm3. The ideal gas in the container has a pressure of 2.0 × 107 Pa at a temperature of 17 °C. Show that the amount of gas in the cylinder is 150 mol.

[1] (c) Gas molecules leak from the container in (b) at a constant rate of 1.5 × 1019 s−1. The temperature remains at 17 °C. In a time t, the amount of gas in the container is found to be reduced by 5.0%. Calculate (i)

the pressure of the gas after the time t,

pressure = ................................................... Pa [2]

© UCLES 2016

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258

7 (ii)

the time t.

t = ....................................................... s [3] [Total: 9]


 3

(a) Explain what is meant by the statement that two bodies are in thermal equilibrium. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [1] (b) Suggest suitable types of thermometer, one in each case, to measure (i)

the temperature of the flame of a Bunsen burner, ...................................................................................................................................... [1]

(ii)

the change in temperature of a small crystal when it is exposed to a pulse of ultrasound energy. ...................................................................................................................................... [1]

(c) Some water is heated so that its temperature changes from 26.5 °C to a final temperature of 38.0 °C. State, to an appropriate number of decimal places, (i)

the change in temperature in kelvin,

change = ..................................................... K [1] (ii)

the final temperature in kelvin.

final temperature = ..................................................... K [1] [Total: 5]

CEDAR © UCLES 2016

COLLEGE

9702/42/M/J/16

259 [Turn over

5 Mar 2016 , Question #3 , qp_42 3

8

(a) Define specific heat capacity. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) A student carries out an experiment to determine the specific heat capacity of a liquid using the apparatus illustrated in Fig. 3.1. liquid out, temperature 25.5 °C

tube

heating coil

liquid in, temperature 19.5 °C

Fig. 3.1 Liquid enters the tube at a constant temperature of 19.5 °C and leaves the tube at a temperature of 25.5 °C. The mass of liquid flowing through the tube per unit time is m. Electrical power P is dissipated in the heating coil. The student changes m and adjusts P until the final temperature of the liquid leaving the tube is 25.5 °C. The data shown in Fig. 3.2 are obtained. m / g s–1

P/W

1.11

33.3

1.58

44.9 Fig. 3.2

(i)

Suggest why the student obtains data for two values of m, rather than for one value. ........................................................................................................................................... .......................................................................................................................................[1]

© UCLES 2016

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260

9 (ii)

Calculate the specific heat capacity of the liquid. Show your working.

specific heat capacity = .......................................... J kg–1 K–1 [3] (c) When the heating coil in (b) dissipates 33.3 W of power, the potential difference V across the coil is given by the expression V = 27.0 sin (395t). The potential difference is measured in volts and the time t is measured in seconds. Determine the resistance of the coil.

resistance = ....................................................... Ω [3] [Total: 9]

© UCLES 2016

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9702/42/F/M/16

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261

THERMOMETRY 6 Oct/Nov 2004 , Question #7 , qp_4 7

14

The e.m.f. generated in a thermocouple thermometer may be used for the measurement of temperature.

For Examiner’s Use

Fig. 7.1 shows the variation with temperature T of the e.m.f. E. 1.5

E / mV 1.0

0.5

0 300

400

500

600

700

T/K Fig. 7.1 (a) By reference to Fig. 7.1, state two disadvantages of using this thermocouple when the e.m.f. is about 1.0 mV. 1. ...................................................................................................................................... 2. ..................................................................................................................................[2] (b) An alternative to the thermocouple thermometer is the resistance thermometer. State two advantages that a thermocouple thermometer has over a resistance thermometer. 1. ...................................................................................................................................... .......................................................................................................................................... 2. ...................................................................................................................................... ......................................................................................................................................[2]

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262

7 May/June 2006 , Question #3 , qp_4 3

7

The electrical resistance of a thermistor is to be used to measure temperatures in the range 12 °C to 24 °C. Fig. 3.1 shows the variation with temperature, measured in degrees Celsius, of the resistance of the thermistor.

For Examiner’s Use

2400

2200 resistance / 2000

1800

1600

1400 12

14

16

18

20

24 22 temperature /

26

Fig. 3.1 (a) State and explain the feature of Fig. 3.1 which shows that the thermometer has a sensitivity that varies with temperature. .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [2] (b) At one particular temperature, the resistance of the thermistor is 2040 ± 20 . Determine this temperature, in kelvin, to an appropriate number of decimal places.

© UCLES 2006

CEDAR COLLEGE

temperature = ……………………… K [3] [Turn over

9702/04/M/J/06

263

.......................................................................................................................................... .......................................................................................................................................... .................................................................................................................................... [4]

8 Oct/Nov 2006 , Question #2 , qp_4 2

A mercury-in-glass thermometer is to be used to measure the temperature of some oil. The oil has mass 32.0 g and specific heat capacity 1.40 J g–1 K–1. The actual temperature of the oil is 54.0 °C. The bulb of the thermometer has mass 12.0 g and an average specific heat capacity of 0.180 J g–1 K–1. Before immersing the bulb in the oil, the thermometer reads 19.0 °C. The thermometer bulb is placed in the oil and the steady reading on the thermometer is taken. (a) Determine (i)

the steady temperature recorded on the thermometer,

temperature = ………………………… °C [3] © UCLES 2006

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9702/04/O/N/06

264

For Examiner’s Use

5 (ii)

the ratio change in temperature of oil . initial temperature of oil

ratio = ………………………… [1] (b) Suggest, with an explanation, a type of thermometer that would be likely to give a smaller value for the ratio calculated in (a)(ii). .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [2] (c) The mercury-in-glass thermometer is used to measure the boiling point of a liquid. Suggest why the measured value of the boiling point will not be affected by the thermal energy absorbed by the thermometer bulb. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [2]

© UCLES 2006

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9702/04/O/N/06

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265

THERMISTORS 9 Oct/Nov 2009 , Question #10 , qp_41

18

10 The circuit of Fig. 10.1 may be used to indicate temperature change.

For Examiner’s Use

+2 V T

P

+5V – +

P

P

–5V

R

G

Fig. 10.1 The resistance of the thermistor T at 16 °C is 2100 Ω and at 18 °C, the resistance is 1900 Ω. Each resistor P has a resistance of 2000 Ω. Determine the change in the states of the light-emitting diodes R and G as the temperature of the thermistor changes from 16 °C to 18 °C.

................................................................................................................................................. ........................................................................................................................................... [4]

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266

10 May/June 2010 , Question #10 , qp_41 3

8

(a) The resistance of a thermistor at 0 °C is 3840 Ω. At 100 °C the resistance is 190 Ω. When the thermistor is placed in water at a particular constant temperature, its resistance is 2300 Ω. (i)

For Examiner’s Use

Assuming that the resistance of the thermistor varies linearly with temperature, calculate the temperature of the water.

temperature = ......................................... °C [2] (ii)

The temperature of the water, as measured on the thermodynamic scale of temperature, is 286 K. By reference to what is meant by the thermodynamic scale of temperature, comment on your answer in (i). .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................ [3]

(b) A polystyrene cup contains a mass of 95 g of water at 28 °C. A cube of ice of mass 12 g is put into the water. Initially, the ice is at 0 °C. The water, of specific heat capacity 4.2 × 103 J kg–1 K–1, is stirred until all the ice melts. Assuming that the cup has negligible mass and that there is no heat exchange with the atmosphere, calculate the final temperature of the water. The specific latent heat of fusion of ice is 3.3 × 105 J kg–1.

temperature = ......................................... °C [4]

267

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ANSWERS 1

Page 2

Mark Scheme A/AS LEVEL EXAMINATIONS - JUNE 2003

1 (a)

work done in bringing/moving unit mass ......................................M1 from infinity to the point.......................................................... ...... A1 (use of 1 kg in the definition – max 1/2)

2

(b)

(c)

Syllabus 9702

potential at infinity defined as being zero........................ ............. B1 forces are always attractive.......................................................... B1 so work got out in moving to point...................... .......................... B1 (max potential is at infinity – allow 1/3) (i)

(ii)

(d)

Paper 04

[2]

[3]

φ = -GM/R change = 6.67 x 10-11 x 6.0 x 1024 x({6.4 x 106}-1- {1.94 x 107}-1) .....C2 change = 4.19 x 107 J kg-1 (ignore sign) .........................................A1 ½mv2 = m φ ................................................................................ C1 v2 = 2 x 4.19 x 107 = 8.38 x 107 v = 9150 m s-1 .............................................................................. A1

[5]

acceleration is not constant.......................................................... B1

[1]

(-1 for each error or omission) ........................................ B2

[2]

32 (a)

(b)

heat lost by liquid gold = 0.95m x 129 x ∆T.................................. C1 heat gained (silver) = 0.05m x 235 x (1340 – 300) + 0.05m x 105 000..C1, C1 122.5m∆T = 17 470m ∆T = 143 K.......................................................................................C1 temperature = 143 + 1340 = 1483 K................................................A1

[5]

(c)

e.g. thermocouple/resistance thermometer .................................. B1

[1]

3 (a)

f0 is at natural frequency of spring (system) ................................. B1 this is at the driver frequency ....................................................... B1 (allow 1 mark for recognition that this is resonance)

[2]

(b)

CEDAR(c) COLLEGE

line: amplitude less at all frequencies ......................................... B1 peak flatter .......................................................................... B1 peak at f0 or slightly below f0 ................................................ B1 (aluminium) sheet cuts the magnetic flux/field.............................. B1 (so) currents/e.m.f. induced in the (metal) sheet .......................... B1 these currents dissipate energy ...................................................M1 less energy available for the oscillations ...................................... A1

[3]

268

B

24

= (2.8 × 1011)2 × (7.0 × 1010 ) × (4.98 × 10–8 )2 / (6.67 × 10–11)

C1

= 2.0 × 1029 kg

A1

[3]

B1

[1]

(a) (i) number of atoms/nuclei in 12 g of carbon-12 (ii) amount of substance

M1

containing NA (or 6.02 × 1023 ) particles/molecules/atoms or which contains the same number of particles/atoms/molecules as there are atoms in 12 g of carbon-12

A1

[2]

A1

[1]

(b) pV = nRT 2.0 × 107 × 1.8 × 104 × 10–6 = n × 8.31 × 290, so n = 149 mol or 150 mol

(c) (i) V and T constant and so pressure reduced by 5.0% pressure = 0.95 × 2.0 × 107

C1

or calculation of new n (= 142.5 mol) and correct substitution into pV = nRT

Page 3

pressure = 1.9 × 107 Pa Mark Scheme Cambridge International AS/A Level – May/June 2016

(ii) loss is 5 / 100 × 150 mol = 7.5 mol or ∆N = 4.52 × 1024

= 3.0 × 105 s

C1 A1

[3]

(a) no net energy transfer between the bodies or bodies are at the same temperature

B1

[1]

(b) (i) thermocouple, platinum/metal resistance thermometer, pyrometer

B1

[1]

B1

[1]

B1

[1]

B1

[1]

(ii) thermistor, thermocouple

(c) (i) change = 11.5 K (ii) final temperature = 311.2 K

4

A1 Paper 42

C1

) / Cambridge 1.5 × 1019 International Examinations 2016 t = (7.5 × 6.02 × 1023 © or t = 4.52 × 1024 / 1.5 × 1019

3

Syllabus 9702

(C1)

(a) (i) T = 0.60 s and ω = 2π / T

C1

ω = 10 (10.47) rad s–1 (ii) energy = ½mω2x0 2

CEDAR COLLEGE

A1 ½mv2 and v = ωx0

or –3

2

C1

269

–2 2

= ½ × 120 × 10 × (10.5) × (2.0 × 10 ) –3

[2]

[2]

(iii) internal energy = 5.9 × 10–21 × 3.2 × 1023 = 1900 (1890) J 53

(a) the (thermal) energy per unit mass to raise the temperature of a substance by one degree

A1

[1]

M1 A1

[2]

(If ratio not clear for M1 mark, allow 1 / 2 marks for an otherwise correct answer)

cancelScheme heat transfer to / from tube / surroundings (b) 1(i) to allow for / determine /Mark Page Syllabus GCE A/AS Level – May/June 2006 9702 (do not allow ‘to stop / prevent’ heat loss) 1

(a) centripetal force is provided by gravitational force (ii)2 / either P =/mc∆θ ±h mv r = GMm r2 –3 = 1.58 henceorv 44.9 = √(GM / r) × 10–3 × c × (25.5 – 19.5) ± h or 33.3 = 1.11 × 10 × c × (25.5 – 19.5) ± h 2 – 33.3) = (1.58 – 1.11) × 10–3 × c × (25.5 – 19.5) (b) (i) E(44.9 K (= ½mv ) = GMm / 2r Page 3 Syllabus c = 4100 (4110) J kg–1 K–1Mark Scheme A LEVEL – NOVEMBER 2004 9702 (ii) EP = - GMm / r –1 g s leading to 5000 J kg –1 K –1) (allow 1 / 3 for use -3of only 33.3 W, 1.11 23 –1 –1 –1 (ii) (allow N =1{(2.24 x 10 )/224} x 6.02 x 10 / 3 for of only (iii) ET = - GMm / r +use GMm / 2r 44.9 W, 1.58 g s leading to 4740 J kg K ) = 6.02 x 1018 = - GMm / 2r. activity = λN -6 = 2.23 6.02 1018 or =x19.1 (c) (i)V0 =if 27 rms (c) ET decreases thenxV-10 GMm / 2r xbecomes more negative 13 Bq 1.3 or x 10 33.3 = 19.12 / R 33.3or=GMm 272 / 2R / 2r =becomes larger R =so 11r Ω decreases (c) A = A0 e-ln2.tlT exp(-In2 . n) (ii)0.1E= K = GMm / 2r and r decreases n = 3.32 and period = 0.30 s 4 (a) amplitude = 1.8vcm so (EK and) increases (n = 3 without working scores 1 mark)

2 76

mV is (a measure of) the sensitivity (b) gradient voltage of = 30 (a) graph the gradient varies with temperature

6 (b) (a) 2040 speed = ZΩ/ ρ ± 20 corresponds to 15.0 ± 0.2 °C 106 / 940(allow (=1490) T / K = T=/ 1.4 °C +× 273.15 273.2) –2 time = (1.1 × 10 × 2) / 1490 temperature is 288.2 K = 1.5 × 10–5 s 4 (a) (i)(time 1.0of 7.4 × 10–6 s is one way only and scores 2 / 3 marks) (use of speed of light is wrong physics and scores 0 / 3 marks) (ii) 40 Hz (b) (i) speed = 2πfa © Cambridge International Examinations 2016 = 2π × 40 × 42 × 10-3 = 10.6 m s-1 CEDAR COLLEGE (ii) acceleration = 4π2 f2 a

= (80π)2 × 42 × 10-3 = 2650 m s-2

[1]

B1 B1 A0 [2] B1 C1 B1 [1] Paper A1 [3] 4 B1 [1]

1 1

C1 A1

[2]

1 1

C1 [4] M1 C1 A1 [2] [3] A1

1 1

M1 [2] A1 A1 [2] [1]

(a) e.g. fixed mass/ amount of gas 2 2 (a) non-linear 1 ½m ωis2 (x or EK = ½mv2 and v = ± ω √(x0 2 – x2) (b) ideal Evariation K = gas 0 – x ) two possible temperatures 1 2 –2 2 –2 2 (any=two, each)× (2π/ 0.30) × [(1.8 × 10 ) – (1.2 × 10 ) ] ½ × 10.080 = 3.2 × 10–3 J (b) e.g. small (b) (i) n =1.pV / RTthermal capacity/measure ∆θ of small object 7 /short = (2.5 × 10response × 4.00 ×time 104 x 10-6 ) / (8.31 × 290) 2 readings taken at a point/physically = 415of) mol / ‘on’ and ‘off’ small / 1’s and 0’s / two values 5 (a) (i) (series ‘highs’ and ‘lows’ 3 can be used to measure temperature with no intermediate values / the values aredifference discrete 4 no power supply required (ii) volume of gas at 1.85 × 105 Pa = (2.5 × 107 × 4.00 × 104 ) / (1.85 × 105 ) (any two, 1 mark each) rate 2 6 (ii) eitheretc.use higher sampling frequency = 5.41 × /10 cm3 6 4 3 each digital number or 5.41 ×use in each sample = 4.00bits × 10 + 7.24 × 10 /N so, 10 more or use more levels in each sample N = 741 (answer 740 or fails to allow for gas in cylinder, max 2/3 )

37

B1 Paper 04

C1 [2] B2 C1 [2] A1 [3] C1 C1 A1 M1 [3] A1 [2]

[2] C1 C1 B1 [3] [1] A1

A1 [1] M1 A1 [2] C1 C1 C1 C1 A1 [3] A1 [3] B1 [1] B1

[1]

C1 A1

[2]

270 C1 A1

[2]

electric forces can be attractive or repulsive for gravitational, work got out as masses come together /mass moves from infinity Page 6 Mark Scheme: Teachers’ version Syllabus for electric, work done on charges if same sign, work got out2009 if opposite sign as charges GCE A/AS LEVEL – October/November 9702 come together

Paper B1 41 B1

[4

idea of heat lost (by oil) = heat gained (by thermometer) 32 x 1.4 x (54 – t) = 12 x 0.18 x (t – 19) (a) resistance of wire = ρL / A ....................................................................................... B1 t = 52.4°C

C1 C1 A1

[3

as crack widens, L increases ....................................................................................M1 (ii) either ratio (= 1.6/54) = 0.030 or (=1.6/327) = 0.0049 and A decreases ............................................................................M1 so resistance increases ............................................................................................. A0

A1

[1

thermistor thermometer (allow ‘resistance thermometer’) because small mass/thermal capacity

B1 B1

82 (a)

9

B1

(b)

Section B (i)

(b) ∆L / L point = ∆Rtemperature / R ........................................................................................................ B1 (c) boiling is constant = (146.2 – 143.0) / 143.0 × 100 .................................................................. C1 further comment e.g./ L heating of bulb ......................................................................................................... would affect only rate of boiling ∆L = 2.24% A1 use of a = – ω 2x clear either ω = √(2k/m) or ω 2 = (2k/m) ω = 2 πf = (1/2 x 300)/0.240) 9 at 16f °C, 10 V+ π)√(2 = 1.00 V and V – = 0.98 V or V+ > V – = 7.96 ≈ 8 Hz 3 (a)

Syllabus 9702

M1 A1 [3]

[2

B1 C1 B1 A0

[4

B1

[1

[4] B1

[1

Paper B1 [Total: 4] 41 B1

10 10 (a) X-ray taken of slice /–1plane / section B1 –1 11 large(i) / 1 T magnetic applied body (allow ‘across’) (1) 4 (a) GMm {(R +field h1) angles – (R + halong 2) } repeated at different B1 r.f. pulse applied (1) {v12 –............................................................................................................... v2 2} images½m / data is processed B1 causes hydrogen nuclei / protons ..................................................................................... (1) combined /xadded to givex (2-D) B1 (b) 2M x 6.67....................................................................................................................... 10–11 {(26.28 106)–1 image – (29.08ofxslice 106)–1} = 53702 – 50902 to resonate (1) –19 repeated successive M xreturn 4.888for xto10 = 2.929 slices x 106 / after relaxation time .................................................B1 (nuclei) equilibrium state (1) 24 toMbuild upx a103-D image B1 = 6.00 kg............................................................................................................ r.f. (pulse) emitted (1) image can be from different anglesthen / rotated (If equation in viewed (a) is dimensionally unsound, 0/3 marks in (b), if dimensionally sound butB1 pulses detected, processed and displayed ...................................................................... (1) incorrect, treat as e.c.f.) max 6 resonant frequency depends on magnetic field strength .................................................. (1) calibrated non-uniform field enables nucleitotorate be of located ................................................ (1) 5 (a) (i) (induced) e.m.f proportional/equal change of flux (linkage)

(b) (i) 16 (allow ‘induced voltage, induced p.d.) A1 flux is custmark as the disc moves any six points, one each .......................................................................................... B6 hence inducing an e.m.f

(b)

[2

[Total:C1 6]

........................................................ B1 at 16 °C, output is positive ................................................................................................M1 diode(i)R is ‘on’ and diode G is ‘off’ .................................................................................... A1 (b) resonance as temperature rises, diode R goes ‘off’ and diode G goes ‘on’ ....................................... B1 Hz 2nd to 3rd marks and also 3rd to 4th marks) (allow(ii)e.c.f.8 from Page(increase 5 Mark Scheme: Teachers’ version (c) amount of) damping LEVEL – May/June without altering (kGCE or) mAS/A …(some indirect reference is2010 acceptable) sensible suggestion

[3]

B1

[3

B1 B1

[2

B1 C1 A1

[3

[6] B1

[1] M1 [6] A0

[2

(ii) evidence of deducting 16 then dividing by 3 C1 [Total: 6] to give [2] (ii) field in disc is not uniform/rate of cutting not same/speed of disc not same (over wholeA1 disc) B1 3 2 so different e.m.f.’s in different parts of disc M1 6 lead5to eddy currents A0

[2

eddy currents dissipate thermal energy in disc

energy derived from oscillation of disc wave varies (in synchrony) with signal 11 (a) frequency of carrier energy of disc depends on amplitude oscillations signal (in synchrony) with displacement of of

(b) advantages e.g.

M1 A1

B1 B1 B1 [2]

B4

[4]

less noise / less interference © UCLES 2006 greater bandwidth / better quality

(1 each, max 2) disadvantages e.g. short range / more transmitters / line of sight more complex circuitry greater expense (1 each, max 2)

12 (a) gain / loss/dB = 10 lg(P1/P2) 190 = 10 lg(18 × 103 / P2) CEDAR COLLEGE or –190 = 10 lg P2 / 18 × 103) power = 1.8 × 10–15 W

C1 © UCLES 2009

271 C1 A1

[3]

[3

A2 THERMODYNAMICS


CEDAR COLLEGE

272

1 May/June 2004 , Question #6 , qp_4 6

For Examiner’s Use

12

The first law of thermodynamics may be expressed in the form U = q + w, where U is the internal energy of the system, U is the increase in internal energy, q is the thermal energy supplied to the system, w is the work done on the system. Complete Fig. 6.1 for each of the processes shown. Write down the symbol ‘+’ for an increase, the symbol ‘–’ to indicate a decrease and the symbol ‘0’ for no change, as appropriate. U

q

w

the compression of an ideal gas at constant temperature

the heating of a solid with no expansion the melting of ice at 0 °C to give water at 0 °C (Note: ice is less dense than water) [6] Fig. 6.1

273

CEDAR COLLEGE

© UCLES 2004

9702/04/M/J/04

2 May/June 2004 , Question #13 , qp_6

For Examiner’s Use

18

13 A certain engine operates on the cycle illustrated in Fig. 13.1. pressure 55 105 Pa

A

B

1020 J

400 J

C

6 1

O 40

150

760

volume / cm3

Fig. 13.1 A mass of gas is firstly compressed adiabatically (O → A) such that 400 J of work is done on the gas. During the stage A → B, fuel is injected into the gas and this causes heating at constant pressure as the fuel burns. The gas and burned fuel then expand adiabatically (B → C) and, during this process, 1020 J of work is done by the gas and burned fuel. Finally, during the stage C → O, energy is wasted. (a) (i)

(ii)

During the stage A → B, 2500 J of energy is supplied to the gas. Show that the work done by the gas as it expands during this stage is 605 J.

Calculate the energy wasted during the stage C → O.

energy = ..................................... J

274

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3 Oct/Nov 2005 , Question #3 , qp_4 3

6

(a) State the first law of thermodynamics in terms of the increase in internal energy U, the heating q of the system and the work w done on the system. .......................................................................................................................................... ..................................................................................................................................... [1] (b) The volume occupied by 1.00 mol of liquid water at 100 °C is 1.87 10–5 m3. When the water is vaporised at an atmospheric pressure of 1.03 105 Pa, the water vapour has a volume of 2.96 10–2 m3. The latent heat required to vaporise 1.00 mol of water at 100 °C and 1.03 105 Pa is 4.05 104 J. Determine, for this change of state, (i)

the work w done on the system,

w = ……………………………. J [2] (ii)

the heating q of the system,

q = ……………………………. J [1] (iii)

the increase in internal energy U of the system.

U = ……………………………. J [1]

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9702/04/O/N/05

275

For Examiner’s Use

7 (c) Using your answer to (b)(iii), estimate the binding energy per molecule in liquid water.

For Examiner’s Use

energy = ………………………………. J [2]

© UCLES 2005 CEDAR COLLEGE

9702/04/O/N/05

[Turn over 276

4 Oct/Nov 2008 , Question #4 , qp_4 10 4

(a) Write down an equation to represent the first law of thermodynamics in terms of the heating q of a system, the work w done on the system and the increase U in the internal energy.

For Examiner’s Use

...................................................................................................................................... [1] (b) The pressure of an ideal gas is decreased at constant temperature. Explain what change, if any, occurs in the internal energy of the gas. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ...................................................................................................................................... [3]

277

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5 May/June 2010 , Question #2 , qp_41 2

5

(a) Some gas, initially at a temperature of 27.2 °C, is heated so that its temperature rises to 38.8 °C. Calculate, in kelvin, to an appropriate number of decimal places, (i)

For Examiner’s Use

the initial temperature of the gas,

initial temperature = ............................................. K [2] (ii)

the rise in temperature.

rise in temperature = ............................................ K [1] (b) The pressure p of an ideal gas is given by the expression p = 13 ρ!c 2" where ρ is the density of the gas. (i)

State the meaning of the symbol !c 2". .................................................................................................................................. .............................................................................................................................. [1]

(ii)

Use the expression to show that the mean kinetic energy <EK> of the atoms of an ideal gas is given by the expression <EK> = 32 kT. Explain any symbols that you use. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [4]

278

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6 (c) Helium-4 may be assumed to behave as an ideal gas. A cylinder has a constant volume of 7.8 × 103 cm3 and contains helium-4 gas at a pressure of 2.1 × 107 Pa and at a temperature of 290 K. Calculate, for the helium gas, (i)

the amount of gas,

amount = ......................................... mol [2] (ii)

the mean kinetic energy of the atoms,

mean kinetic energy = .............................................. J [2] (iii)

the total internal energy.

internal energy = .............................................. J [3]

© UCLES 2010

CEDAR COLLEGE

9702/41/M/J/10

279

For Examiner’s Use

6 Oct/Nov 2010 , Question #2 , qp_42 2

(a) (i)

6

State the basic assumption of the kinetic theory of gases that leads to the conclusion that the potential energy between the atoms of an ideal gas is zero.

For Examiner’s Use

.................................................................................................................................. .............................................................................................................................. [1] (ii)

State what is meant by the internal energy of a substance. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

(iii)

Explain why an increase in internal energy of an ideal gas is directly related to a rise in temperature of the gas. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

(b) A fixed mass of an ideal gas undergoes a cycle PQRP of changes as shown in Fig. 2.1. 10

P

8 volume / 10–4 m3 6

4

2

0

R

Q

0

5

10

15

20

25 30 5 pressure / 10 Pa

Fig. 2.1

280

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7 (i)

State the change in internal energy of the gas during one complete cycle PQRP. change = ............................................. J [1]

(ii)

For Examiner’s Use

Calculate the work done on the gas during the change from P to Q.

work done = .............................................. J [2] (iii)

Some energy changes during the cycle PQRP are shown in Fig. 2.2. change

work done on gas /J

heating supplied to gas / J

increase in internal energy / J

P

Q

.............................

–600

.............................

Q

R

0

+720

.............................

R

P

.............................

+480

.............................

Fig. 2.2 Complete Fig. 2.2 to show all of the energy changes.




281

CEDAR COLLEGE © UCLES 2010

[3]

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9

7 May/June 2011 , Question #4 , qp_42 4

(a) The first law of thermodynamics may be expressed in the form

For Examiner’s Use

ΔU = q + w. Explain the symbols in this expression. + ΔU ................................................................................................................................. + q .................................................................................................................................... + w ................................................................................................................................... [3] (b) (i)

State what is meant by specific latent heat. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................ [3]

(ii)

Use the first law of thermodynamics to explain why the specific latent heat of vaporisation is greater than the specific latent heat of fusion for a particular substance. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................ [3]

282

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8 Oct/Nov 2011 , Question #2 , qp_41 (c) (i)

7

State what is meant by the internal energy of a substance. ..................................................................................................................................

For Examiner’s Use

.................................................................................................................................. .............................................................................................................................. [2] (ii)

Use the equation in (b)(ii) to explain that, for an ideal gas, a change in internal energy ΔU is given by

ΔU ∝ ΔT where ΔT is the change in temperature of the gas. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

CEDAR COLLEGE © UCLES 2011

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[Turn over

9 May/June 2012 , Question #3 , qp_42 3

9

(a) State what is meant by the internal energy of a system. ..........................................................................................................................................

For Examiner’s Use

.......................................................................................................................................... ...................................................................................................................................... [2] (b) State and explain qualitatively the change, if any, in the internal energy of the following systems: (i)

a lump of ice at 0 °C melts to form liquid water at 0 °C, .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [3]

(ii)

a cylinder containing gas at constant volume is in sunlight so that its temperature rises from 25 °C to 35 °C. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [3]

CEDAR COLLEGE © UCLES 2012

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[Turn over

10 Oct/Nov 2012 , Question #2 , qp_41 2

6

A student suggests that, when an ideal gas is heated from 100 °C to 200 °C, the internal energy of the gas is doubled. (a) (i)

State what is meant by internal energy. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [2]

(ii)

By reference to one of the assumptions of the kinetic theory of gases and your answer in (i), deduce what is meant by the internal energy of an ideal gas. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .............................................................................................................................. [3]

(b) State and explain whether the student’s suggestion is correct. .......................................................................................................................................... .......................................................................................................................................... ...................................................................................................................................... [2]

285

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For Examiner’s Use

11 May/June 2013 , Question #2 , qp_42 2

5

(a) The volume of an ideal gas in a cylinder is 1.80 × 10–3 m3 at a pressure of 2.60 × 105 Pa and a temperature of 297 K, as illustrated in Fig. 2.1.

For Examiner’s Use

ideal gas 1.80 × 10–3 m3 2.60 × 105 Pa 297 K

Fig. 2.1 The thermal energy required to raise the temperature by 1.00 K of 1.00 mol of the gas at constant volume is 12.5 J. The gas is heated at constant volume such that the internal energy of the gas increases by 95.0 J. (i)

Calculate 1. the amount of gas, in mol, in the cylinder,

amount = ........................................... mol [2] 2. the rise in temperature of the gas.

temperature rise = .............................................. K [2]

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[Turn over

6 (ii)

Use your answer in (i) part 2 to show that the final pressure of the gas in the cylinder is 2.95 × 105 Pa.

For Examiner’s Use

[1] (b) The gas is now allowed to expand. No thermal energy enters or leaves the gas. The gas does 120 J of work when expanding against the external pressure. State and explain whether the final temperature of the gas is above or below 297 K. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [3]

© UCLES 2013

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287

12 Oct/Nov 2013 , Question #2 , qp_43

(ii)

7

During the heating, the gas expands, doing 1.5 × 103 J of work. State the first law of thermodynamics. Use the law and your answer in (i) to determine the total energy supplied to the gas.

For Examiner’s Use

.................................................................................................................................. ..................................................................................................................................

total energy = ............................................. J [3]

288

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13 May/June 2014 , Question #3 , qp_41 3

7

The volume of 1.00 kg of water in the liquid state at 100 °C is 1.00 × 10−3 m3. The volume of 1.00 kg of water vapour at 100 °C and atmospheric pressure 1.01 × 105 Pa is 1.69 m3. (a) Show that the work done against the atmosphere when 1.00 kg of liquid water becomes water vapour is 1.71 × 105 J.

[2] (b) (i)

The first law of thermodynamics may be given by the expression ΔU = + q + w where ΔU is the increase in internal energy of the system. State what is meant by 1.

+ q,

...................................................................................................................................... [1] 2.

+ w.

...................................................................................................................................... [1] (ii)

The specific latent heat of vaporisation of water at 100 °C is 2.26 × 106 J kg−1. A mass of 1.00 kg of liquid water becomes water vapour at 100 °C. Determine, using your answer in (a), the increase in internal energy of this mass of water during vaporisation.

increase in internal energy = ..................................................... J [2]

© UCLES 2014

CEDAR COLLEGE

9702/41/M/J/14

[Turn over

289

(ii)

1 (d)

6

7

1

(a)

(b) 22 13

(c)

8

(a)

ratio = 2.0 (allow 1 s.f.) advantage: e.g. easy to change the voltage Page 1 Mark Scheme disadvantage: e.g. cables require greater insulation A LEVEL – NOVEMBER 2005 ....... rectification – with some justification

A1 B1

Syllabus 9702

B1

(a) GM / R2 = Rω2 …………….…………………...…..………………….. (i) 3.0 A (allow 1 s.f.) A1 ω = 2π / (24 × 3600) ………………………………..……..…………… (ii) 3.0 A (allow –11 1 s.f.) A1 6.67 × 10 × 6.0 × 1024 = R3 × ω2 Total 3 22 ………………………………………………………… 0 R =- 7.57 × + 107 (-1 for each error) B2 R = 4.23 × 10 m ……………………………………………………….. + + 0 (-1 for each error) B2

+ + 0 (-1 for each error) B2 (b)(i) ∆Φ = GM/Re – GM/Ro …...………………………………………….….. Total = (6.67 × 10–11 × 6.0 × 1024) ( 1 / 6.4 × 106 – 1 / 4.2 × 107) λ = h/p or λ = h/mv M1 7 –1 = 5.31 × 10 J kg …………………………………………………. with λ , h and (or mv) p identified A1 ∆EP = 5.31 × 107 × 650 Mark …………………………………………………. Page 4 Scheme Syllabus 10 × 10LEVEL J …………………………………………………….. A/AS EXAMINATIONS - JUNE 2004 9702 1= 3.45 2 mv E = C1 (c) e.g.22satellite will already have some speed in the correct direction … = p /2m or v = √(2E/m), hence M1 (a) work = ρ∆V × T ………………………..……………….. h/√(2mE) A0 (a) λ(i)=obeys the done law pV = constant 5 = 55T x 10………………………………………………. x (150 – 40) x 10-6 at all values of p, V and = 605 J E = qV C1 5 -9 2 -31–2 + 400) – (1020 (ii)n x =10energy wasted (2500 605) = 1275 J )2 (b) (0.4 (2.9 × 3.1x=×10 10 (8.31 ×-19 290) ) x×210 x 9.11 x) /1.6 x 10 x V+=…..………………..………... (6.63 x 10-34 C1 efficiency = 1625/2900 =V 3.73 molscores ………………………………………………………………. V(iii) = 9.4 (2 s.f. 2/3) A1 = 0.56 or 56% Total 3 .4 290 S shown at the peak B1 (c) at new pressure, n = 3.73 × × n compression/expansion are both adiabatic (b) similarity: e.g.

mol relative ….………………………………………. (i) Kr and U on right of peak=in4.23 correct positions B1 Total change = 0.50 (ii)1 binding energy of mol U-235……………………………………………………….… = 2.8649 x 10-10 J number of strokes = 0.50 / 0.012 x= 10 42-10(must round up for mark) ……. J binding energy of Ba-144 = 1.9211 Option T - Telecommunications -10 C2 binding energy of Kr-90 = 1.2478 x 10 J (a) energy correct words orJsymbols …..…………………………...….. 1433 (a) 10 statement, lg(P1/P ) or 10-11lg(P A1 release = 23.04 x 10 2/P1) (-1 if 1 or 2 s.f.) C1 2 E = mc2 (b)(i) w =10p∆V ………………………………………………………………….. 2 (b) = 14 8dB x 10-28 kg (ignore s.f.) A1 m = (3.04 xlg(25.4/1.0) 10-11)/3.0 5 x 10 ) = 3.38 =above 1.03the × 10 × (2.96level × 10–2 – 1.87 × 10–5) reference (iii) e.g. neutrons are single particles, = (–) 3050 J …………………..……………….…………..…………… neutrons have no binding energy per nucleon B1 (c) (i) loss of signal power/energy 4 Total 14/3.2 (ii)(ii)q =length 4.05 ×=10 J ………………………………………………………….

(b)

Paper 4

[2]

ANSWERS

2 .9 300 difference: e.g. in petrol engine, energy input at constant volume

C1 C1

[2] [9] M1 A0

[3]

[6] C1 [6] C1 C1 Paper A1 06

[4]

B1

[1]

C1M1 [2] M1A1

[2]

[2]

A0 A1C1 C1A1 [3] A1 [5] [7]

15

(c)

number of molecules = NA

[2]

B1

[1] C1 [7] C1 A1 B1B1 [1] [3]

……………………………………………….

(i) energy broadcast frequency = kHz ) = 37500 / (6.02 × 50 1023 8 3 3.0 x 10 = 50 x–20 10 x λ = 6.2 × 10 J (accept 1 sig.fig.) …………………………..…. λ = 6000 m = 7.0 kHz (a) (i)(ii)ω =bandwidth 2πf ………………………………………………………....………….. (iii) maximum frequency = 3.5 kHz = 2π × 1400 Total = 8800(or radorbit) s–1 is………………………………………………………….. (a) period 24 hours equatorial (orbit) (ii) a0 =(satellite (–)ω2x0orbits) ……………………………..………………………………… from west to east 2 –3 =©(8800) × 0.080 × 10 University–2of Cambridge International Examinations 2004 =allow 62002 m s → 40 …………………….……………………………………. (b) (i) GHz GHz (ii) prevent swamping of the (low power) signal received from Earth

(b) (c)

straight line through origin with negative gradient …….…………….... end points of line correctly labelled advantage: e.g. fewer satellites…………………………………….. required

aerials point is fixed direction/no tracking required (c) (i) zero displacement ………………………………………………………… (any sensible suggestion, 1 mark) disadvantage: e.g. noticeable time delay in messages (ii) v = ωx0 ……………………………………………………………………. reception difficult at Poles –3 (any sensible suggestion, 1 mark) = 8800 × 0.080 × 10 –1 Total ……………………………………………………………. = 0.70 m s

CEDAR COLLEGE

[3]

[1]

A1C1 [2] [2]

A1

A1

[2]

[1] B1 [8] C1B1

[1]

A1

(b)

4

[2]

[1] B1

[3] Total (iii) ∆U = 4.05 × 104 – 3050 = 37500 J …no e.c.f. from (a)………………… A1 [6] (a) amplitude of the carrier M1 penalise 2 sig.fig. once onlywave varies in synchrony with the displacement of the information signal A1 [2] = 4.4 km

16

[2]

© University of Cambridge International Examinations 2005

[1]

C1 C1 C1A1 A1 A1C1 A1 [5] [7] B1A1 B1 B1C1 [3] B1A1 B1 [2]

[2]

[2]

[2]

M1 A1

[2]

B1B1

[1]

C1 [2] A1 [7]

B1

290

[2]

y = 3 mm = 28.3 × 10–3 √(112 – 32) = 0.30 m s–1 (allow 1 s.f.)

Page 2 4 4

(a) ∆U = q + w

C1 C1 A1

Mark Scheme: Teachers’ version GCE AS/A LEVEL – May/June 2010 (allow correct word equation)

Syllabus 9702

[3]

Paper 41 B1 [1]

Section A 1

5

5 2

(b) either kinetic energy constant because temperature constant (a) angle (subtended) at centre of circle potential energy constant because no intermolecular forces (by) arc equal in length to radius so no change in internal energy or kinetic energy and potential energy both constant (M1) so no change in internal energy (A1) (b) (i) point S shown below C reason for either constant k.e. or constant p.e. given (A1)

M1 B1 M1 B1 A1

[2] [3]

B1

[1]

(ii) (max) force / tension = weight + centripetal force centripetal force = mrω2 (a) change/loss in kinetic energy2= change/gain in electric potential energy 15 = 3.0/9.8 × 0.85 × ω 2 × ½mv2 = q2 / 4πε–10 r ω = 7.6 rad s 2 × ½ × 2 × 1.67 × 10–27 × v2 = (1.6 × 10–19 )2 / (4π × 8.85 × 10–12 × 1.1 × 10–14 ) 6 × 10 m s–1 or 27.2 + 273.2 (a) v(i)= 2.5 27.2 + 273.15 300.4 K

C1 C1 B1 C1 C1 A1

[4]

M1 A0 C1

[3]

A1

[2]

C1 A1

[1]

2 (b) (ii) pV =11.6 ½Nm and pV = NkT 3 ½ m = 2 kT (award 1 mark of first two if not used)

C1

3

½ × 2 × 21.67 × 10–27 × (2.5 × 106 )2 = 2 × 1.38 × 10–23 × T > 8isKthe) mean / average square speed (b) T(i)= 5(
C1 A1 B1

[1] [4]

(ii) ρ = Nm/V with N explained 2 > so, pV = 1/3 Nm = ½m = 3/2 kT (any sensible comment, /1<E mark)

B1 B1 B1 B1

[4]

6

(if T < 10 K, should comment that too low for fusion to occur)

6

(c) (i) pV = nRT 7 × 7.8 loss × 10–3 = n × 8.3 flux × 290 2.1 × 10prevent (a) (i) either of magnetic n = 68 mol or improves flux linkage with secondary (ii) mean kinetic = 3/2 kT eddyenergy current (losses) (ii) reduces = 1.38 × 10–23 × 290 (in ×core) reduces losses of energy3/2 = 6.0 × 10–21 J

3

B1

[1]

C1 B1 A1

[2] [1]

B1 B1 C1 A1

[2] [2]

realisatione.m.f. that total internal energy is the (b) (iii) (i) (induced) proportional to / equal to total kinetic energy M1 C1 –21 × 68 × flux 6.02(linkage) × 1023 energy = 6.0 ×of10 rate of change (magnetic) A1 C1 [2] = 2.46 × 105 J A1 [3] (ii) changing current in primary gives rise to (1) changing flux in core (1) (a) (i) flux to-and-fro / backward and forward two limits) B1 [1] links with the secondary coil motion (between(1) changing flux in secondary coil, inducing e.m.f. (1) (ii) no energy loss or gain / no external force acting / constant energy / constant amplitude B1 [1] © UCLES 2008

(iii) acceleration directed towards a fixed point acceleration proportional to distance from the fixed point / displacement

(b) acceleration is constant (magnitude) so cannot be s.h.m.

B1 B1

[2]

M1 A1

[2]

291

CEDAR COLLEGE

© UCLES 2010

(iii) graph: g = 0 at least ⅔ distance to Moon gE and gM in opposite directions correct curvature (by eye) and gE > gM at surface 6 2

B1 M1 A1

[3]

B1

[1]

(ii) sum of kinetic and potential energy of atoms / molecules due to random motion

M1 A1

[2]

(iii) (random) kinetic energy increases with temperature no potential energy (so increase in temperature increases internal energy)

M1

(a) (i) no forces (of attraction or repulsion) between atoms / molecules / particles

(b) (i) zero (ii) work done = p∆V = 4.0 × 105 × 6 × 10–4 = 240 J (ignore any sign)

A1

[2]

A1

[1]

C1 A1

[2]

B3

[3]

(iii) change

work done / J

heating / J

increase in internal energy / J

P→Q Q→R R→P

+240 0 –840

–600 +720 +480

–360 +720 –360

(correct signs essential) (each horizontal line correct, 1 mark – max 3 )

Page 3

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Mark Scheme: Teachers’ version GCE AS/A LEVEL – May/June 2011

Syllabus 9702

(a) +∆U: increase in internal energy © UCLES 2010 +q : thermal energy / heat supplied to the system +w: work done on the system

B1 B1 B1

[3]

(b) (i) (thermal) energy required to change the state of a substance per unit mass without any change of temperature

M1 A1 A1

[3]

M1 M1 A1

[3]

M1 A1

[2]

B1 B1 B1

[1] [1] [1]

C1 292 A1

[2]

(ii) when evaporating greater change in separation of atoms/molecules greater change in volume identifies each difference correctly with ∆U and w

5

Paper 42

(a) (i) (induced) e.m.f. proportional to rate of change of (magnetic) flux (linkage) / rate of flux cutting (ii) 1. moving magnet causes change of flux linkage 2. speed of magnet varies so varying rate of change of flux 3. magnet changes direction of motion (so current changes direction)

(b) period = 0.75 s CEDAR COLLEGE frequency = 1.33 Hz

(c) graph: smooth correctly shaped curve with peak at f0

M1

(ii) either pV = NkT or pV = nRT and links n and k and <EK> = ½m 3 Page 3 Mark Scheme: Teachers’ version clear algebra leading to <EK> = kT GCE AS/A LEVEL – 2May/June 2012

8

13 9

M1 Paper A1 42

Syllabus 9702

(c) (i) graph: horizontal line, y-intercept = 7.0 mJ with end-points of line at –2.8 cmkinetic energy of molecules/atoms/particles B1 +2.8 cm and (c) (i) sum of potential energy and M1 reference to random (distribution) A1 (ii) graph: reasonable curve B1 with maximum (0,7.0) end-points of line at (–2.8, 0) (ii) no intermolecular forcesatso no potential energy B1 0) energy is (change in) kinetic energy and this B1 is (changeand in) (+2.8, internal proportional to (change in ) T B1 (iii) graph: inverted version of (ii) M1 with intersections at (–2.0, 3.5) and (+2.0, 3.5) A1 (Allow marks in (iii), but not in (ii), if graphs K & P are not labelled) Page 2 Mark Scheme Syllabus Paper GCE AS/A LEVEL – October/November 2012 9702 41 B1 (d) gravitational potential energy Section A (a) is potential proportional to the product the masses and (a) force sum of energy and kineticofenergy of atoms/molecules/particles inversely to the square of the separation referenceproportional to random (distribution) either point masses or separation >> size of masses

M1 A1

[2]

[1] [2]

[2] [2] [2]

[1]

M1 A1

[2] [2]

B1 M1 A1

B1 M1 A0

[2] [3]

½ × 4.00 × 1014 × 620 × ({7.30 × 106}–1 is increasing (ii) 1. either molecules/atoms/particles move faster/ 7 = 9.26 ×energy 10 J (ignore anywith signtemperature in answer) (increases) or kinetic increases B1 8 J if evidence that E separately for each r) M1 (allow 1.0 × 10 K evaluated no change in potential energy, kinetic energy increases internal energy increases A1 2. ∆EP = 4.00 × 1014 × 620 × ({7.30 × 106}–1 – {7.34 × 106}–1) = 1.85 × 108 J (ignore any sign in answer) 8 J) 1.8 orenergy 1.9 × 10 (a) (i) as r (allow decreases, decreases/work got out (due to) M1 mass is negatively charged A1 attraction so point (iii) either (7.30 × 106)–1 – (7.34 × 106)–1 or ∆EK is positive / EK increased increased (ii) speed electrichas potential energy = charge × electric potential B1 electric field strength is potential gradient B1 field strength = gradient of potential energy graph/charge A0 (a) (i) sum of potential energy and kinetic energy of atoms / molecules / particles reference to random (b) tangent drawn at (4.0, 14.5) B1 (ii) no intermolecular A2 gradient = 3.6 × 10–24 forces potential (for <no ±0.3 allow 2energy marks, for < ±0.6 allow 1 mark) –24 –19random motion) of molecules internal kinetic energy (of field strengthenergy = (3.6 ×is 10 ) / (1.6 × 10 ) –5 motion –1 (reference to random here then allow back credit from gradient value)to (i) if M1 scored) A1 = 2.3 × 10 V m (allow ecf –5 –1 (one point solution for gradient leading to 2.3 × 10 Vm scores 1 mark only)

C1 A1

[2]

© University of Cambridge International Examinations 2011

(b) (i) as lattice structure is ‘broken’/bonds broken/forces between (b) (i) gravitational force provides the centripetal force molecules reduced (not molecules separate) 2 2 2 /r = GMm/r and E = ½mv mv K no change in kinetic energy, potential energy increases hence K = GMm/2r internalEenergy increases

4

10 2

3

[3] C1 A1

[2]

[2] M1 A1

[2]

[2] M1 A1 B1 B1 B1

[2]

[3] [4]

(b) kinetic energy ∝ thermodynamic temperature either temperature in Celsius, not kelvin so incorrect or temperature in kelvin is not doubled

B1 B1

[2]

(a) temperature of the spheres is the same no (net) transfer of energy between the spheres

B1 B1

[2]

(b) (i) power = m × c × ∆θ where m is mass per second 3800 = m × 4.2 × (42 – 18) m = 38 g s–1

C1 C1 A1

[3]

CEDAR COLLEGE

© University of Cambridge International Examinations 2012

(ii) some thermal energy is lost to the surroundings so rate is an overestimate

293

M1 A1

[2]

either reference to 22 point masses or separation >> ‘size’ of masses = 7.57 × 10 R = 4.2 × 107 m (b) gravitational provides force (missingforce out 3600 givesthe 1.8centripetal × 105 m and scores 2/3 marks) 2 2 GMm / R = mRω where m is the mass of the planet 112 GM (a) =(i)R3ω 1.2 pV = nRT 1.80 × 10–3 × 2.60 × 105 = n × 8.31 × 297 n = 0.19 mol (c) ω = 2π / T either M2. MSun = (Rstar / RSun)3 × (TSun / Tstar)2 = mc∆T star / ∆q 3 2 Mstar = 4 ×=(½) × 2.0 × 10×30∆T 95.0 0.190 × 12.5 = × 10 ∆T3.2 = 40 K31 kg 2 3 (allow for2 correct answer with clear logic shown) or Mstar = (2π) 2Rmarks star / GT = {(2π)2 × (6.0 × 1011)3} / {6.67 × 10–11 × (2 × 365 × 24 × 3600)2} (ii) p/T == constant 3.2 × 1031 kg (2.6 × 105) / 297 = p / (297 + 40) 2.95 ×and 105 potential Pa 2 (a) (i) sum pof=kinetic energies of the molecules reference to random distribution

A1

B1 M1 A1 A0

[2] C1 A1 [3]

[3] C1 A1 [2]

C1 C1 B1 A1 A1[3] [2] (C1) (C1) (A1) M1 M1 A0 [1] A1 [2]

(b) for change internal energy is 120forces J / 25 J (ii) ideal in gas, no intermolecular M1 B1 internal energy decreases / ∆U is negative / kinetic energy of molecules decreases so no potential energy (only kinetic) A1 M1[2] so temperature lower A1 [3]

(b) (i) either change in kinetic energy = 3/2 × 1.38 × 10–23 × 1.0 × 6.02 × 1023 × 180 C1 = 2240 J A1 or R = kNA energy = 3/2 × 1.0 × 8.31 × 180 (C1) = 2240 J (A1)

[2]

12

(ii) increase in internal energy = heat supplied + work done on system 2240 = energy supplied – 1500 energy supplied = 3740 J

3

(a) work done bringing unit positive charge from infinity (to the point) Page 3 Mark Scheme GCE AS/A LEVEL – May/June 2014

Syllabus 9702

–3 13 (i) either either change both potentials are= positive / same sign 3(b) (a) in volume (1.69 – 1.00 × 10 ) so same sign or liquid volume << volume of vapour or done gradients are 5 5 work = 1.01 × 10positive × 1.69 &= negative 1.71 × 10(so (J)fields in opposite directions) so same sign © Cambridge International Examinations 2013

M1 A1 M1[2] (M1) A1 [2] (A1) B1

2. work done on the system (iii) allow value of x between 10 nm and 13 nm

A1

(a) kinetic (energy)/KE/EK

[3]

M1 A1 [2] Paper 41

(b) (i) 1. heating of system/thermal energy supplied to the system (ii) the individual potentials are summed

(ii) ∆U = (2.26 × 106) – (1.71 × 105 ) (allow 0.42 V → 0.44 V) (iv) V = 0.43 V = 2.09 × 106 J (3 s.f. needed) energy = 2 × 1.6 × 10–19 × 0.43 = 1.4 × 10–19 J 4

B1 C1 A1

B1

[1]

[1] B1

[1]

[1]

C1 M1 A1 [2] A1 A1 [3] B1 [1]

© Cambridge International Examinations 2013

(b) either change in energy = 0.60 mJ or max E proportional to (amplitude)2 /equivalent numerical working new amplitude is 1.3 cm change in amplitude = 0.2 cm

(a) graph: straight line at constant potential = V0 from x = 0 to x = r curve with decreasing gradient CEDAR COLLEGE passing through (2r, 0.50V0 ) and (4r, 0.25V0 ) 5

(b) graph: straight line at E = 0 from x = 0 to x = r

B1 B1 B1

[3]

B1 M1 294 A1

[3]

B1