Stpm Trials 2009 Phy Q&a (smi Ipoh)

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ST. MICHAEl'S INsTlnmoN, IPOH STPM Trial Examination - 2009 Upper Six Sciences

Pltyslcs 1 (960/1) 1 hr45 min

Instruction_ie> ~didaCM: There are fifty questions in this tesl For each question. four suggested answers are given. You are required to chooee one which is the best answer and marl< ~ on the separate Multiple Choice Answer 5 _ provided.

At1ower ..t the questions. Mar1<s will not be deducted for wrong - . s. Your total score in this leal w~1 be the number Qf questions correctly answered . • If necessary, use the vlJiues of cons/ants provided on page 10.

Thi. question _

consists of 10 printed

1 Using the symbols L, M, T, Nand 0 as the dimensions for length, mass, time, amount of substance and temperab,lre, What is the dimensions of R il the equation pV", = RT? (p It pressure, Vrn is the molar vOlume. R is the gas c:onstaIlt, T is the thermodynamic temperature)

A B

' e"' Mt.'r' 0"

ML7

C D

ML7'2 N'1 e-t ML'" N" 0"

Z A student uses a mia"ometer screw gauge Ie measure the diameter or a baU-bearing. Based on five measurements, the fDlowfng results were obtained minimum diameter 5.28 mm maximum dlameter '"' 5.33 mm mean value = 5.30 mm It is known that thfi actual diameter is 5.24 mm. What Is the most probabte reason why the mean value obtained differs from the actual diamete
=

A

B ~

o

There is I*'8I1ax Ofl'9r. The nu..- 01 readi1gs I;Iken is insutlldenl. The m. ~ aItft:tg tnltr'umel1t is not sens~ enough. There is zero error in ~ measuring instrument

~ A wIlesl suddenly detaIer wIlile n is j!1OVllg horizonlally 8\ 1.. krn h" at a height of 500 m abOVo tho oaf1h. Wh$t is tho horizontal dillance travelled by tho """'" _ ~ Ms tho ground? [T- g to be 10 m s" sod Ignore tho e1Iec:tsqr air resistance and wind!

A

200m

B

300m

C

400m

0

SOOm

4 M and P, _ are·of mass 3.0 kg and 5.0 kg respectively, are connected tog_ with an I n _ string. M lo pulled by a horizontal force of 30 N. 3,0 kg 30

T

''::=::::L P

M

Iftho1l1dion - . MandPwllhthoftoor i04.0 N and 10 N respectively. what is tho tension T In the sVtng? A

4.0 N

B

10,0 N

C

11 .3 N

D

16.0 N

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- 25 A truck of mass 2 600 kg is moving along a horizontal road. When its brake s are applied, heat is produced at a steady rate of 58 .5 kW for 5.0 5 before the truck stops . Estimate the speed of the truck just before the application of the brake s.

A

10.0 m 5. 1

C

o

15.0 m 5. 1

22.5 m 5 "

6 A steel ball of mass 0.005 kg fa lls vertically through a liquid at terminal speed 0.1 m 5", The energy dissipated per second in the liquid by the ball 's motion, in mJ, is

A 0.025

B

5.0

o

10

C

SO

7 The radius of curvature of the North-South Highway at a certain place is 200 m. At what angle I must its surface be banked so that a car moving with a speed of 108 km h- will not experience any lateral strain on the wheel?

A 8

24.6 '

C

o

61 .6 '

65.4 '

To what quantity in linear motion is the moment of inertia of a rotating body analogous?

A 9

B

zero

B

force

momentum

C

acceleration

o mass

The diagram shows a metre rule which is placed on a smooth horizontal table. smooth table

metre rule

t t;!Ii:;!p9"&;;~*'i ~ bl

, impulsive force

When a horizontal impulsiv~ force is applied to one end of the meter rule in a direction perpendicular to its axis, the metre rule will experience

A B C

o

translation only. rotation only. rotation and translation. vibration , rotation and translation .

10 A body of mass horizontal.

m is placed on a rough surface that is inclined at an angle of e to the

If the body remained stationary, the coefficient of static friction J.I is given

A B

~=mg ~=mg\an e

C

).1=

mgkos 9

o

f1=

tane

On

11

1

The escape velocity of a carbon dioxide molecule the Earth's surface is 1.1 x 10" m 5 . , What will be its escape velocity at an altitude of O.4R, where R is the Earth's radius?

A B

10" m 5 . 1 1 1. 1 x 10"ms·

0.9

X

C

o

1.3 x 10" m 5 " 1.5 x 10" m 5"

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-3-

u

12 The graph on the right shows the variation of the gravitational potential energy U of a spacecraft with its distance r from the Earth's centre, The radius of the earth is R

R " O~--------~---+,

- U, - ---f--- ' - ,, -- - - -.:; ,, ,, - U,

If the spacecraft is orbiting the Earth in an orbit of radius r 0 , the total energy of the spacecraft

is

c

·2U ,

13 When the amplitude of a simple harmonic motion changes, which of the following quantities does not change? A

8 14

period velocity

C

D

fix~

A frictionless trolley of mass 1.0 kg is tied to two

sPrings· x~

acceleration restoring force

0= -

points, X and Y, using two stretched

!mIleYI~Y c)

x

The trolley is displaced a small distance (0.050 m) towards Y by a force 5.0 N and then released . oozx, where x is the displacement of the The subsequence motion is represented by the equation a trolley from its equilibrium position. What is the value of 00 2 ?

=.

A

0.25 rad z s ' z

B

1.0 rad z s""2

C

o

100 rad z s' 400 rad z S ' 2

15 Diagrams (a) and (b) show the displacement--distance and displacement·time graphs of a transverse wave. displacementlmm displacementfmm

of\PJa

o f \ J \ J distance/em (a)

.. tlme/ms

(b)

The velocity of the wave is A

16

0.004 m

s·,

B

o

0.4 m s' \

4ms'\

Which of the following statements is not true about stationary waves? A B C D

Particles Particles Particles Particles

between between between between

consecutive consecutive consecutive consecutive

nodes nodes nodes nodes

vibrates vibrates vibrates vibrates

in phase. with the same velocity. with different amplitude. with the same frequency.

17 A spectator is standing by the side of a racing track. When a Formula·1 racing car passes him, he observed that the pitch of the sound produced by the car decreases. This is due to A B

Doppler effect. resonance.

C

o

damping. reflection.

- 4-

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18 A point source produces a sound at the rate of 120 W. What is the intensity level of the sound at a distance of 2.0 m from the source? [threshold of hearing , 10 = 10.12 W m'2j

A

B

30dB

124 dB

C

130 dB

D

141 dB

19 The diagram below shows the extension (x) of a wire varies when it is stretched by a force (F).

F

o LL-_--.l'-_ _ X The shaded region represents the

A S C D

strain in the wire. stress on the wire. strain energy stored in the wire. strain energy per unit volume stored in the wire.

20 The graph shows how the potential energy U between two atoms in a diatomk: moJecule changes with the distance of separation r between them. At a certain temperature T, each atom vibrate~ between two positions which are represented by the points P and Q . The maximum thermal energy of the molecule at temperature T is

A E.

21

B

E, .

E.

C

u

,

o E, Ii. ..

E, + Eo

Q

D

The r.m.s. speed of the molecules in an ideal gas of volume V and pressure p is Co. The gas is

heated until its volume becomes

t V and its pressure becomes 6p. lNhat is the new r.m .s. speed of

the molecules?

A

22

2e.

B 3e.

C

4e.

D

Be.

The graph below represents the distribution of molecular speed in a gas at temperature T . number of molecules, n(v)

z

speed,

v

The quantity z is the

A

modal speed (most probable speed) .

B

root mean square speed. mean square speed. mean speed .

C

o

23 Which of the following equations is change? A

dO=O

B

pV=nRT

C

pV

o

=constant 1 =constant

TV'"

not applicable to an ideal gas undergoing adiabatic

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- 5Two c:onduc:ling biocks are maintained at 8 temperature di1ference of 20 K. and are connected through tW
conducting

bkidcS is filled will lagging.

The rate of heat flow between !he two blocks is A

B

0.13W

0.25W

C

0.31 W

o

0.63W

25 Two Identical metal spheres, X and Y. are given positive charges 01 ax and Qy respectively, wt1efe C it > 'NtJich of the following statements is true about X and Y1

av.

A

B

C

o 28

The elecbic field strength inside both spherea is zero. The eloclric ~al inside both spheres is zero. The eloclric fteld strengiI1 on !he surface of both spheres is zero. The eIocIric potential inside both sphen!s is the same.

The diagnam shows two dlarges - q and + 3q at points A and B respectively. Aa -q

.

".

(AS' BC :CA· 20 em; M is the mid-point of AS )

r/ii-................. :,. C +3q If q • 5

A

iJC. calculate the potential difference belween points C and M .

1350 kV

B

000 kV

C

450 kV

0

4.5 kV

2'1' The flash ligll in a camera consisb of. xeoon diad1arge lube which is connected to a cap8!Clor ""'!ch is charged by. 1 000 V SOUfCe. The ..... _ _ oupplied to the discharge tube is 2 000 W and Ihe du!alion 01 the flash is O.Q.oIO s. Estimate Ihe capac:ilance of lie capacitoI. A

28

B

8O~F

C

160~

0

6O . ooo~

~_ below its critical temP'!f"OlI!te. """'ich
lead becomes •

_ A

B C o

29

40~F

The conduction ele Alons n1OY8 in pains. The vibtatlon of the atoms stopped. The heat content of the materiaf is zero. The number of conduction ~s hed increased tremendously.

A milliammeter can be converted to a voltmeter by connecting

A

o

C D

a resistor of high value in parallel to it. a rc.istor of high value in series with it. & resistor of low value in parallel to it. a. resi_or of tow value in series with it.

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- 630

Six identical resistors, each of value 5 fl, are connected to a 2 V cell whose internal

resistance can be ignored.

~---------------------.X

2V

r

yL-________- - - - - -______

~

The potential difference across X and Y is

A

3. v )

B

c

Jlv 9

o

±v )

31 The diagram shows a 40 em straight conductor carrying a current 0[20 A in a 2.0 T field. The magnetic field acts in the horizontal plane at an angle of 8 with the direction of the current. If the weight ofthe conductor is 8.0 N and the magnetic force is just enough to balance the weight of the conductor, what is the angle (f!

32 The figure shows the trajectories of two charged particles P and Q shot perpendicularly into a region of uniform magnetic field w ith the same velocity. If the radii of the trajectories are the same, identify P and Q.

A B C

0

p a-particle electron electron positron

2IIA

p -

- . --

o - -+_-

Q electron a-particle positron electron

33 The figure shows the side view of a circular coil of N turns and area A in a magnetic field B. Ifthe normal to the plane of the coil makes an angle q with the magnetic field , the total magnetic flux through the coil is

Qrcularcol

,

_---.-:"~L--- B

A

NBA casu

B NBA sinu

C

BAcosu

o BA sinn

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-734 R

1·

d:tW:crrrrr:::Imcrg:ucr& Coil

Soft "'"

In the circuit above, switch S is closed and the current in the coil is increased gradually by adjusting the rheostat. When the current is at its maximum, switch S is opened. The deftection shown on the galvanometer is largest

A 8

C

o

at the instant switch S is closed . while the current in the coil is increasing. when the current is at its maximum. at the instant switch S is opened .

35 In a circuit, an alternating CUlTent of r.m.s. value 1.00 A flows through a resistor. In another circuit, a steady current of magnitude I flows through an identical res istor. If the two resistors dissipate heat at the same rate, the value of I is A

0.71 A

B

C

1.00 A

1.41A

o

2.00 A

36 A sinusoidal voltage of r.m.s. value 0.60 V is connected to a 30 mH inductor. If the peak current flowing through the inductor is 45 mA, what is the frequency of the a.c.?

A

B

18 Hz

C

100 Hz

110 Hz

37 The figure shows the symbol of an operational amplifier.

A

P

B C 0

Q R

S

630 Hz

R P ~--I

Which termina l and label do not correspond?

Terminal

o

s

o~--I'

Label Inverting input Non-inverting input Feedback output Signal output

38 The figure shows an operational amplifier circuit. lOll,.,

,.'",

V.

I

K: r-t:-r: _10V

I

V.

- '-

Which graph represents the variation of the output voltage Va with the input voltage Vi?

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- 8B

A

." - >0

." 2.6 1

0

~

V;IV

_10

Vi/V

-10

V,.IV

C

",,,

0

V.N

D

."

.10

,

-" ,

,,

~5,

V;IV

-10

39

In photoelectric emission, the energy of the photon is used

A

e o

C

as the kinetic energy of the electron on ly. to release the electron from the lattice only. by the electron to produce a new electron called photoelectron to release the electron from the lattice and as the kinetic energy of the electron.

40 A proton and an electron have the same value of the de Broglie wavelength. If the K E. of the proton and the electron -are. respectively, Kp and 1<.. which of the following is correct?

A

Kp

>

C

K,

Kp=K,

I

o K=P K ,

- - -- - - -- --£. ----------£,

41 The fi gure shows some of the energy levels of an atom .

The maximum number of spectral lines produced by the transition of electrons from these four energy levels is

- - -- -- ----£, -----------£,

A

3

B 4

C

5

o

6

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- 942 The graph shows the X-ray spectra I and II produced by an X-ray tube when it is operated at two different voltages.

':

"

," I

Which statement explain s why the minimum wavelength of spectrum II is longer than that of spectrum I? A B C o

43

Wavelength

A higher voltage is used to produce spectrum II. A lower voltage is used to produce spectrum II. An element of smaller atomic number is used as the target material to produce spectrum II. An element of bi gger atomic number is used as the target material to produce spectrum II.

The penetration power o f o f X-rays produced from an X-ray lube can be increased by A B C

o

increasing the length of the X-ray tube . increasing the operating voltage of the X-ray tube . increasing the current flowing through the cathode . changing the target metal to an elem ent o f higher atom ic number.

44 The nucleus of iron l~ Fe is more stable than the nucleus o f bi smuth foll owing is true of the more stable nucleus? A B C

o

4S

z:; Oi . Which of the

Smaller mass per nucleon Lower ratio o f proton to neutron Lowe r binding energy per nucleon Higher binding energy per nucleon I~~ Cd

When a slow neutron is captured by a stationary

nucleus, a

I::Cd nucleus is formed and

a 'Y 4ray photon is emitted. What is the wavelength of the y -ray? [Mass of ~ n = 1.0087 u, mass of ':: Cd "" J 12.9044

A B

2.0SxlO-23 m 6.63 X10- 16 m

U, mass o f I:; Cd "" 113 .9034 uJ

C

o

1.37 X \0-1) m 1.45 xlO-12 m

46 A radioactive sample consists of a nucl)de which emits a-particles at the rate as shown in the graph below.

_01"""""

Determine the decay constant of the nuclide. A

5.5)( 10-3 S· 1

B

9.2x 10 -3 5-'

C

75s

o

125 s

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- 10 47 A radioactive e lement X decays to a radioactive element Y which then decays to element Z. If the sample initially contains X only, which of the foll owing factors influences the ratio the number ory nuclei to the number of X nuclei after a certain time?

or

A B

Ha lf· life of e lement Y Half·life of e le ment Z

C o

"x

48

'io

->

m

Initial number of X nucle i

Surro unding pressure

',Y

+

The equation above represents a nuclear reaction. What do the symbols m, nand Y represent?

A B C

0 49

!I!

!L

X

4

2

2 2 -1

4 4 0

nucleon neutron proton helium nuclide

The antipartic le of the electron is the A

quark

B

meson

neutrino

C

positron

D

50 The strong fo rce is responsible for

A B

bela decay. a lpha decay.

C

o

binding the nucleons together. holding the electrons in the atom.

960 P HYS ICS

Values of constants Speed of light in free space

c

Permeability of free space

3' 00X 1 0'ms~

..

= =

41tx10 , T Hm

•

( 1/(36n)) x 10" F m"

Magnitude of electronic cha rge

e

1.60 x 10 , n C

Planck constant

h

= = = = = = = = = =

1'0

Permittivity of free space

Unified atomic mass constant

u

Rest mass of electron

m. m,

Rest mass of proton

R

Molar gas constant Avogadro constant

L. NA

Boltzmann constant

k

Gravitational constant

G

g

Acceleration of free fall

Prepared by : Lee Ping Kew, August, 2009

Checked by: PRAKAs

8.85 x 10

. 12

6.63 x 10 , 3-4

1 Fm·

Js

2T

kg

, 31

kg

1.66 x 10 · 9.11 x 10

,!

1.67 x 10'27 kg

mar I

8.31 J K ., 6.02 x 10

23

1.3Bx10 ·

mol -'

23

JK-t

6.67 x 10 . 11 N m 9.B1ms -

2

kg · 2

2

Endorsed by/?~

a-w'1:..U..

c;u", Kmrtm $aifU

SML St. Midwd

LPK@C:IPIIYSICSFORM6lEXAMl09STPMT,ia[ PtJfMrf.doc:

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ST. MICHAEL'S INSTITUTION, IPOH STPM Trial Examination - 2009 Upper Six Sciences

Full Name: _ _ __ _ __ _ _ _ _ __

Class No.:

Physics 2 (960/2) 2 hr 30 min

Class: Upper Six Science _ _

(as in the Class Register)

Instructions to candidates: Answer all the questions in Seclion A in the spaces provided. All working must be shown. For calculations, relevanl values ofconstants in the Data Booklet· must be used. For numerical answers, units must be quoted wherever they are appropriate. Answer any (our questions from Section B. For this section, write your answers on the answer sheets (test pad) provided. Begin each answer on afresh sheet o/paper, and arrange your answers in numerical order. Tie your answer sheets to this question booklet (from page I to page 6 only) . • ljnecessary, use the values of constants provided on page 10 in this question booklet.

This question booklet consists of 10 printed pages

Section A Answer all the questions in this section. 1 (a) Define a vector quantity. ... ........ ............... ... . ...... .. . ................... .....

........... ........ .... ...•..................•......... .•.........

(b) A r!gid wire frame is in the form of a cube. A force of 20 N is applied along a diagonal from the origin 0 to a point A which is diagonally opposite to O. Refer to the Figure on the right. Three forces of equal magnitude, each parallel to a side of the cube, are then applied at point O. Determine the magnitude and direction of these three forces so that they will balance the 20 N force.

[1 J

y

20N

,,

,, f---

OV"'''-~--T----~

,,

, ,,

,

x

-------y z

[4J

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- 22 The archerfish hunts by dislodging an unsuspecting insect from its resting place with a stream of water expelled from the fish's mouth. Suppose the archerfish squirts water with an initial speed of 2.5 m 5"1 at a beetle on a leaf 4.0 em above .the water's surface, as shown below.

(a) If the fish aim in such a way that the stream of water is movirlg horizontally when it hits the beetle. what is the launch angle, 9?

[2]

(b)

How much time does the beetle have to react?

[2]

(e) What is the horizontal distance, d, between the fish an~ the beetle when the water is launched?

( I]

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-33

Figure (a) shows the bone and muscle structure ofa person's arm holding a 5.0 kg mass in

equilibrium. The forearm is horizontal and is at right angles to the upper arm. Figure (b) shows the equivalent mechanical system. fM is the force exerted by the biceps muscle and FJ e lbow joint.

is the force at the 5.0 kg

E

p

Er-+-~r--------

N

20N

5.0om·_ om

om

35 em

35cm

Figure (a)

Figure (b)

(a) (i) Suggest a reason why a 20 N fo rce has been inc luded in this system.

(1)

(i i) State the condition(s) that must be met by the forces when the arm is in equilibrium .

[2)

(b) Calculate the magnitude orlhe force FM .

[ I)

(e) In many athletes, the distance between the e lbow joint (E) and the muscle attachment (P) is greater than 5.0 em. Explain why this is an advantage in lifting and throwing events .

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

960/2

(1)

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- 44

(a) State the first law of thermodynamics.

/1)

(b) Explain the following observations: (i) when gas from an aerosol can is suddenly released, the container becomes cooler.

(ii) while pumping up a bicycle tyre, the pump barrel becomes warmer.

(2)

[2)

5 An ideal solenoid consists of 1000 turns of wire per em wound around an air-filled cylindrical structure. The solenoid is of2.0 em long and cross-sectional area 1.8 cm2 , A current of2 .0 A passes through the wire. (a) Calculate the magnetic flux in the centre of the solenoid.

II I

(b) Calculate the self-inductance of the solenoid.

[2J

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-5 6

A sinusoidal a.c. supply is applied to the circuit as shown below . .-----,D~-----, A

R

a.c . ....,

L--------------J a (a) (i) Write down the equation representing the potential difference between A and B.

[I [

(ii) On Graph I below, sketch the variation with time of the potential difference between A and B for two cyc les. (b) A diode is then connected to the circuit at point D. On Graph II below, sketch the variation with time or the potential difference between A and B for two cycles. (e) (i) Describe ho w you would modify the circuit in o rder to smoothen the p.d. across AB.

[I] [ 11 [11

(ii) After the modification , sketch the variation with time of the potential difference between A and B for two cyc les on Graph III below. [IJ

v

v

v

01-----+

o

o

Graph I

Graph II

Graph III

7 (a) What is a photon? [1 J

(b) A p.d. (V) is applied across a photocell as shown in Figure (a). The photocell is then illuminated with monochromatic light of wavelength 365 nm. The current in the circuit is measured for various values of V. The results are shown in Figure (b).

, v

V,rl.,..

d.c. ,uppty

• -, Figure (al

•

,

' wv

Figure (bl

(i) Deduce the value of the stopping potential. (1 J

960/2

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·6· (ii) Calculate the maximum kinetic energy of the photoelectrons emitted_

[1 )

(iii) Determine the work function of the cathode materiaL

[2)

8 The initial number of atoms in a 4.0 g radioactive element is 8.0 x 1021 • The half-life of the e lement is 5 hours. (a) Calculate the number of atoms which decay in 12 hours.

(b) Determine the mass of the radioactive element

960/2

~ft

after 24 hours.

[4)

[II

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- 7-

Section B

Answer any four questions in this seclion

9

[2]

(a) Describe briefly the characteristics of a projectile motion.

(b) A ball-bearing o f mass 50 g rolls off the edge of a horizontal platform at a height 2.4 m and strikes the fl oor at a horizontal distance of3.2 m from the edge o f the table.

(i) Ca lculate the time taken by the ball-bearing to reach the fl oor. (ii) Determine the speed of the ball-bearing just before it falls. (iii) Calculate the magnitude and direction of the velocity of the ball-bearing just befort: it hils

fu_

[2] [2] ~]

(iv) Dctennine the average power of the ball-bearing.

14]

10 (a) (i) Stale Newton's law of universal gravitation . (ii) Define gravitational field strength. (iii) Use your answer to part (i) and (ii) to show that the magnitude of lhe gravitatio nal fi eld strength at the earth's surface (Eo) is given by the expression

where M is the mass of

£ :It: GM

•

[2] [I] [2]

R'

the earth, R is the radius of the earth and G is the gravitational constant. (b) A communication satellite is in an orbit such that its period of revolution around the earth is 24 hours . [2]

(i) Explain the significance of this period . (ii) What is the term given to this particular orbit?

(iii) Show that the radius of the orbit,

Ro, is given by R• •

[ I]

VGMT

4n '

2

where T is the period

[ 2]

of the orbit, and G and M have the same meaning as in (a)(iii). (c) Estimate the minimum kinetic energy which must be given to a satellite o f mass 2500 kg at the earth's surface for it to reach a point which is at a distance Ro from the centre of the eanh. State two assumptions made in your calculations.

[M · 6.0 x 10" kg, R · 6.4 x \0' m]

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-8 11 (a) (i) State two assumptions of an ideal gas. (ii) State two physical conditions under which a real gas behaves as an ideal gas. (iii) A 0.45 m ) tank contains 8.0 kg of butane gas. Assuming that the gas behaves as an ideal gas, calculate irs pressure at 27°c'

[Molecular mass of butane is 58 g mor

l

(3]

1

(iv) Butane gas normally behaves as a real gas. The aelUal pressure of the butane gas is higher than the calculated value in (a)(iii). Give a reason .

(b) (i) What is meant by the degrees of freedom of a gas molecule? (ii) Write an expression relating the total kinetic energy E of a gas molecule to the number of degrees offreedomf Explain any other symbols used. (iii) The escape ve locity of Mars is 5.0 x 10) m 5. 1• If the temperature of Mars is 300 K, determine whether oxygen gas can exist o n the planet. [Molecular mass of oxygen is 32 g mol'l

(21 [2]

[ I) [I) [2)

[4)

1

2

12 (a) Two thin conducting plates have an area of 0.50 m each. They are placed para lle l to each other and 20 mm apart. One plate is maintained at +75 V while the other at - 75 V by a d.c. supply. (i) Determine the amount of charge stored on each plate. (ii) Calculate the energy per unit vo lume stored in the electric fi eld between the plates.

[3] [3]

(b) The fi gure shows a simple circuit of the photographic flash used in a camera.

The capacitance of the capacitor is 40.0 .... F, and the resistance of the resistor is 45.0 k!l. (i) Explain how the capacitor functions in this application. (ii) Calculate the time required to charge the capacitor to 63% so that a good flash can be

obtained. (iii) Suggest a way to reduce the charging time of the capacitor.

960/2

[41

[4] [I]

papercollection

- 9IJ

Figure (a) shows how two op-amp are connected together in series. Both are supplied with

± 9.0 V. Figure (b) shows how the voltage gain of a non-inverting amplifier de pends on freq uency. Voltage gain

input

+

0----1+

output

200'

22kU 2k

P 200

UcU

1 kU

OV

OV

20 2

Figure (a)

30

3

300

31t

30k lOOk

3M

frequency' Hz Figure (b)

(a)

(i) Name the type o f feedback c ircu it used. (ii) Explain the meaning of this type offeedback. (iii) State two advantages of this type o f feedback.

[ I] [Ij [2]

(b)

(i) Calculate the voltage gain of each individual amplifie r. (ii ) Deduce the voltage gain o f the whole system.

[ I] [I]

(e)

A steady vo ltage of - 3.0 mV is applied to Ihe input of the system. (i) Determine the voltage at the output. (ii) Calculate the e lectric potential at po int P.

[ I]

(d)

[21

A s in uso idal a.c. of peak voltage 30 mY is then applied to the input. On the same axes, sketch the graph of (i) the input voltage against time. (ii) the output voltage aga inst time.

[4 1

In you r graph . label each line carefully and labe l all the important values o n the axes. (e)

Using the graph in Figu re (b), estimate the bandwidth of the whole system .

[21

14 (a) Give the meaning of eacn of the following terms.

[II

(i) nucleon number (ii) proton number (iii) binding energy

[Ij [2]

(b) The structure o f the nucle us was investigated by the experiment shown in the Figure below.

protons intense radiation

beryllium

960/2

solid containing

hydrogen atoms

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- 10 Alpha particles were fired at a piece of beryllium and an intense radiation was found to be emitted from the beryllium. When this radiation entered a hydrogen-rich solid, protons were knocked forward from the solid . The initial reaction in the beryllium is

tHe

+

tBe

dn

->

IgC

+

In the above equation, which symbol is used to represent the alpha particles? (ii) What information does the symbol give about the alpha particles? (iii) Suggest, with a reason, which particle is responsible for knocking a proton oul of the solid containing hydrogen atoms. (iv) The intense radiation was originally thought to be y-rays . Why does the existence of the knocked-forward protons make this impossible? The masses of the particles in part (b) arc as follows. (i)

(c)

i He

:Be

+

4.00260 u

---+

9.01212 u

1.00867 u

Values of constants c Po

Permittivity of free space

Co

Magnitude of electronic charge

e

Planck constant

h

Unified atomic mass constant

u

Rest mass of proton

m, m,

Molar gas constant

R

Avogadro constant Gravitational constant

L. NA k G

Acceleration of free fall

g

Rest mass of electron

Boltzmann constant

Prepared by: Lee Ping Kew,

= = = = = = = = = = = = =

7

4nx10· Hm- 1 8.85x 10 · 12 F m· 1 (1/(36n)) x 10. 9 F m- 1 1.60 x 10

- 19

C

6.63.x 10 -34 J s 1.66x 10

·27

kg

9.11 x 10-

31

kg

1.67 x 10 .

21

kg

8.31 J K -1 mol·' 6.02 x 10 23 mold 1.38x10 · 23 JK·1 6.67 x 10 ·11 N m 2 kg 9.81 m S-2

Checked by:

·2

Endorsed by:

..

960/2

ms

3.00 x 10

August, 2009

LPK@C:PHYSfCSFORM6lEXAMI09 STPMTri8J Pspor2.doc

[2J

12.0000 u

960 PHYSICS

Permeability of free space

[2J

+

(i) Calcu late the mass defect. in kilogram, in the reaction. (ii) Hence. find the energy equivalence of this mass defect. Express your answer in MeV.

Speed of Itght in free space

[1 J [ 1)

'._'-

[2J [31

papercollection

S.M.K. ST. MICHAEL IPOH STPM TRIALS 2009 PHYSICS PAPER 1 (ANSWERS) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

C D C B C B B D C D A A A C D B A B C B B A C D A

26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

C C A B A D D A D B B C A D B D B B D C B A C D C

papercollection

ST. MICHAEL'S INSTITUTION, IPOH STPM Trial Examination - 2009 Upper Six Sciences

Physics 2 (960/2)

2 hr 30min

(Marl
Secti on A 1

(a)

... . possessed both magnitude and direction.

(b)

The 3 forces are in the negative C-x, negative O-y and negative O-z directions. .. (1 m)

1

Y 20 N

,

Lei their magnitude be p.

,,

f:O -

, •• • ,•

Using algebra,

J~2 + p2 +p2)=20 W

(2m)

= 20 p= 11 .5 N

z

~ . ' , -

--

I

'

••• , , ,

·

,

'X

4

-------y,

(1m)

5

TOTAL

2

a

Consider vertical motion: ';= u 2 + 2as When water reaches target. v = O. 0= (vo sin 9)2 + 2(-g)h .. = (2.5 sin 9)2 + 2(-9.81)(0.040)

..

.I'in

2 0=

(15)' 9 = 20 .8

I~

0

2

.

Or:

,-_(u."} -2

..

1=

2.5sin20.So 9.8 1

= 0.090

c

4 .0 an 2.5 sin e ' 0 - ... "- ....... ..... "d..._ .......... ......... ,-_...... 2.5 cos e (diagram optional)

2(9.8 1)(0.040)

Consider vertical motion: v=u+al when waler reaches target, v = O. .. 0= (vosin 0) + ato = 2.5 sin 20.a + (-9.81)1

..

•,,

=0.1256

.. b

2.5 ms"

(~l UH

_ (

2

2(0.040) ) 25sin20.8° +0

= 0.090 s

5

Consider horizontal motion: d = (vocos9)t = (2.5 cos 20.8j (0.090) = 0.210 m

1

TOTAL

- 1-

5

a (l)

3

(ii)

b

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This is to lake into account of the weight o f the forearm.

1

1. The resultant force in any direction is zero. 2. The (~s ultant moment about any pOint is zero.

1 1

.

Taking moment about E.

eO) { " )

,

{")

F" " 2 + (5.0X9.81 100 100 100

1

FM "" 403N

c

The fOfce (FloC) exerted by the athletes witl have a greater moment about the elbow E.

1

TOTAL 5

4

(a)

The heat supplied to a system equals to the sum of the increase in its internal energy and the external work done ID: the system.

(b)

(i)

Work is done ~ the gas using part o f ils internal energy.

1

Since the gas expands adiabatically, heat does not flow inlo the sys tem causing its internal energy to decrease.

1

(ii) The work done on the gas is changed into internal energy.

1

Since the gas is compressed adiabatically, heat does not flow oul of the system causing its internal energy to increase.

1

TO TAL

5

(a)

1

Magn etic flu x density, B "" 1J.,n1 7 2 = (4rr x 10. )(1000 x 10 )(2 .0) = 0.2513 T

5

converl 1000 em" 10 m"

1

Magnetic flu x, <%> = BA = (0 .2513»(1 .8 x 10-4) = 4.523 x 10-5 4.52 x 10-liWb

=

2

• OR

Magneliejlux, f/> ... BA - (P",I)A ... (~~ X Iq')(I000x Ui) (2.0)(1.8x Ifr ) - ~. 52 x Wb

ur

(b)

Magnetic flux linkage, <%>' = N¢

= (n~ 4> = (2000 x 2)(4.523 x '0") . :c

NB: n is in em" ,

lis in em .. o.t

0.09046 'Ml

Also ¢' = II L = <%>'/1

1

= 0.0904612.0 = 4.52

X

1

10.2 H TOTAL

-2 -

5

6

.-

V'" Vo Sin wi

(a)

(i)

(e)

Connect a capacitor parallel to R.

[(\

.

-

0

f'I

TV ~aPh

,

T

2T

(a)(ii)

V

oV\ f\

,

I

(al (b)(;) (ii)

(iii)

1

(b)

0

.

",. . ,',. ~ ,

2T

T

Graph Il

(c)(ii)

2T

.... a packet (quantity) of electromagnetic energy.

1.60 X 10' J

1 1

5

1 . .. 1

1.0 V E m&>! - eV, :; ( 1.S0 X 10" ")(1.0)

1

Graph I II

TOTAL

7

--1

V

V

papercollection --

11 1.0 eV

1

E...... - hf-Wo Wo =hf-E", .. ::: he_ E

A.

=

max

(6.63xIO- 34 X3.0x I0 8 )

365)( 10-9 1 = 3.85 X 10. • J

- (1.6OxlO- 19 )

II 2.4 1 eV

.2

TOTAl

8

(a)

Number of halt· lites, n '" 1215

2.4

Number of radioactive atoms left, N

T

1

=N o( Xi f

= B.O X 1021 ~Y:i 1 = 1.516x10

)2.4

1

Number of radioa ctive atoms decayed ::: N o - N

=8.0 X 102 1 -211.516 X 10 21 = 6.484 X 10 =6.48 10

.

X

(bl

Number of radioactive atoms left after 24 days, N

21

1

No( Xi )

= 8.0 X 1021 ~Y.i

)2<415

= 2.872 x 10 21

1

= =

At the beginning : B.O x 10 atoms 4 .0 9 20 21 After 24 days: 2.872 x 10 = [ (2.872 x 1021) I (8.0 x 10 ») X4. 0 0.14 4 9

1

OR After 24 dnys, mass left, m - m.,( v.r - 4.0x(v.i'-' - O.144g TOTAL

·3·

5 .-

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Section B 9

(a)

(b)

1. Horizontal component: M otion under constant speed . Vertical component: Motion under uniform acceleration (a ;: g, acceleration o f free-fall). 2. The trajectory (path of projectile) is a parabola .

1 1

(i) Consider vertica1 motion: Use:

..

s:::ut+ y.at 2 2.4 = (0) + Y.i (9.B 1)

obtained:

e

,,~

<Subs'. s '" 1.4, u - O. a - 9.8 >

t = 0.69949 = 0 ,699 5

"., II" i ,

~

3.2 m

2

", _____1

.-- -.

(ii) Consider horizontal m otion; Use:

..

x = ut (3 .2) = v. (0.69949)

Obtained;

v.

<Subs/.

x

=

11. t .. 0.69949>

=4.5747 = 4.57 m 5 ·'

2

(iji) Consider vertical molion: Use:

..

v = u+at = (D) of- (9.Bl){ 0.69949) = 6.8619 m s·'

1 <Subsf. u .. 0, a .. 1i.8/, I

Vy

magnitude of velocity,

V = ~(VI2 +v/

=

0.69949>

)

= J(4.5747' + 6.86 19' ) = B.247

2

= 8.24 m s " Let th e angle between v and the horizontal = e .

o=tan--(::) =tan . ,(6.8619 ) 4.5747

2

= 56 .309

= 56.3° (Iv)

E power =-

,

Imv2

,

=-'--

just before ;/ hilS the ground, the balf-bearing possesses K.E.

= t (0.050X8.247)'

2 1

0.69949 = 2.430B

1

::: 2.43W TOTAL

-4-

15

,.

papercollection

a

(i) Newton's law --- the gravitational force between two bodies is directly proportional 10 the product o f their masses and inversely proportional to the square afthe distance between them.

2

(or, in equQ{ionjormprovided meaning of al/ the symbols are given).

.

-

(ii) Gravitational fiel d strength --- gravitational force acting on a unit mass .

1

(iii) Consider a body of ma ss m on the Earth's

surlace.

;?

Gravitational force, F = G AI; R F

E.,=m

By definition,

(G~~l =

m

M

2

m GM

(diagram optional)

- fi2 b

(i)

1. lis period of revolution equals to the period of rotation of the Earth. 2. Its position relative to a fixed point on Earth does not change.

1 1 ~

(ii) Geosynchronous /I geostationary /I parking orbit

[-;-. 1

(iii) 'v\Ihile in orbit, centripetal force = gravitational force

..

, Mm maJ Ro '" G - -,

..

Mm m(2JtyR T · =G ,

1

R. R.

GMT2

..

Ro =V 4,'

1

I--

(c)

radius of orbit,

GMT2

Ro

=V 4, '

(6.67.xlO IIX6.0xl024X24x60x60)2 = 3, = 4.23

X

7

4,'

1

10 m

where Uo = P .E. at distance R o, U1

6,K=U o - U, ,

=P.E. at Earth's surface.

Mm ) = _GMm_(_G R. R = GMm(

-'~_ '1 R R.

= (6.67 x

IO~IIX6. 0

x 10 24X 2500{

I 6.4 x 10

=1.33 X 1011 J

6-

1 4.23 x 10

7)

1 1

1 1

Assumptions: 1. No air resistance 2. Speed of the satellite at level Ro is zero.

,

\

\

5.

TOTAL

15-

papercollection

11

(a)

(i) Assumptions

1. The internal energy of the ideal gas consists only of the total kinetic energy of the gas molecules. 2 . There is no force between the ga s molecules except during collisions.

.

2

3. Collisions between the gas molecules are perfectly elastic. 4. The volume of the gas molecules is negligible compared to volume of the gas. (any two) (ii)

1. Allow pressure II when the pressure approaches zero. 2 2. At high temperature.

(iii) For an ideal gas,

Thus, pressure,

pV p

=(;

)RT

1

mRT

~--

MV _ (8 .0 X8 .31 X27 + 273 )

-

"~S8xl? ~ XO.4S )

=7.64 x 10

m, M must in kg (SI unit)

1

Pa

(iv) At this high pressure of 7 .64 x 10 Pa, butane gas deviates from the behaviour of the ideal gas /I The volume of the gas molecules is nol negligible.

(b)

1

1

(;) Independent modes of motion the gas molecule II

IndeQendenl modes of acqutring kinetic energy by the gas molecule.

1

, .L kT

(ii)

1

2

where k is Boltzmann's constant and T is the thermodynamic temperature of the gas. (iii) Translational K.E . :c

1

f mv 2 = t kT

..

v : J 3:r

using: m

== J 3Rr

=

(MIN.J and kNA

~

R

At

.

3(8.3 1 X300 V. J2 r iO - I '" 483 m s'\

1

=

1

3

This velocity is less than the escape velocity of Mars (i.e. 5.0 x 10 m S · I ) II oxygen molecules can not escape from the gravitational pull of Mars

1

Thus. oxygen can exist on Mars.

1 TOTAL

·6·

15

papercollection

12

(a)

(;)

Q ~ CV

~ ( £;/)v ~ (8 085XW")(0050) )[75 _ (- 75)J 20xI O-)

= 3.3 186 x 10-e =3.32x 10"C

1

(ii) energy per unit volume =

energy stored in the capacitor . volume _ of _ the _electric _ field

=tQV

/'

.......

1 0

::K

Ad 0 0 +<3.3 186xI 0· ' )(15 0)

V BY <; d-o>

(0.50)( 20x I0·' )

1

1

= 2.4889 x 10"'" = 2.49 x 10'" J m.J

(b)

.(i) When the switch is opened, the capacitor is charged by the battery.

Energy is stored in the (electric field between the plates of the) capacitor. 4

When the switch is closed. the capacitor discharges (through the flash bulb). The energy stored (in the capacitor) is used to produce a flash (in the flash bulb) . (ii) During charging, p.d. across capacitor, V

Vo(1

e

)

lM1en it is 63% charged, V = 0 B3Vo -<'C. 0 63Vo = Vo(1 -e ) 0 .63 = (1 - e~) e-VCR

1

= 0 .37

etICR = 110.37

[ ~ CR ln(1 ) 0.37

~ (40 0XIO· ' )(45 0X IO' ) ln( _ I _ ) 0

2

0

0.37

= 1.7896 = 1.795

1

(iii) use a resistor of resistance smaller than 45 .0 kO .

1

TOTAl

·7-

15

papercollection

13

(al (II Negative feedback

1

(ii) A fraction of the output is fed (applied) to the inverting input.

1

(iii) 1, The voltage gain is independent of the op-amp characteristics. 2. The amplification is constant over a wider frequency range.

,

3 . Distortion of th e output is reduced.

2 (Any two)

(bl

(II

Rf + R,

A=

1

22k + 1 Ik

=

= 23

(ii)

(cl

(231 = 529

1

(II 1

Vi= AVo

:: (S29)(- 3.0 x 10..:1) = -1.S89V

Ik x( - 1.589) Ik + 22k

Vp =

(ii)

1 1

= - 0.0691 V

(dl

,

='- 1j-;;':+

,'.lL. ,

_ / -1-\

,"

--

-

-

-

,

-

1

,m" ,

,~, .3 mV

- -I 1 -4~v-

,, -

'-

,

.

1

- "','

....

,

••

,-

,~-- + -\--y--,-..

_L,J

I' __

"l' V-, I ' T,

,,

, , ,

,

I

~"

-~fI£t - I' \- 1, - ' -,•

,III ,

,-

r' 1"7"

" "::

.

/i'

!. ' •

'-

,

I'"

-

,

,

,

,

/ , 2

\ ,

I

\ 1-1

_

~

,

-,

1 ,• • , \11

2

~ ,iJ

,

~

,,

I I r t~± =1==l= -

ii

,

1--

- ~-.

\

-

I

~

[-l--,-

-r', ---r-+I

_ ~,

NB: If there's no saturation. peak voltage of output = A (VJpeak = 529x30mV = 15.9 V

(el

2

3 Hz to 9.5 x 10· Hz

TOTAL

,8,

15

papercollection

I.

(a)

(i) Tolal number of protons and neutrons in a nucleus . (ii) Number of protons in a nucleus. (iii) The amount of energy required to completely separate all the nucleons in the

nucleus_

,

1 1

1 1

2

2

1

1

1

1

,

(b)

(;)

4

2He (ii)

an alpha particles consist of 2 protons and 2 neutrons.

(iii) the neutrons

1 2

reason: the C-12 nucleus is too large to penetrate the solid .

1

(iv) Th e e mitted protons have momentum in the forwa rd direction.

1

Thus, it requires thai the incident particles have at least the same momentum. V-rays will have much smaller momentum ( ie,

(e)

2 1

h P = A. )

(;)

t.m = (4 .00260u + 9.0 1212u) - (1 .00867u + 12.00000u) = 0.00605u

1

= (O . 000605~1 . 66 x 10 -21) kg = 1.DOx 10 · kg

(ii)

E

2

1

me

=( 1.00 x 10 -.2')(3.00

X

10 11) 2

= 9 .00 x 10 _13 J 9.00 x 10 _ Il eV = (1.60 x \0 - '., )

9.00 x 10- 13

= ( 1.60

x 10

19 )(106 )

1

J

1

MeV

= 5.63 MeV

1 TOTAL

· 9·

15