Stpm Physics 2007 P2 Ans

1 Gain in KE = Loss of PE 1 1 mv 2 + Iω 2 = mgh 2 2

1 1 1 1 1 = + = + C C1 C 2 1.5 2 C = 0.86 μF (b) For capacitors in series, the charge stored in each capacitor is equal to the total charge stored. Charge stored in each capacitor = CV = 0.86 μF x 24 =20.64 μC

5 (a)

2

1 1 1  v  mv 2 +  mr 2   = mgh 2 22  r  1 2 1 2 Critically v + v = gh damped 2 4 4 gh 4 × 9.81 ×1 v= = = 3.62 m s −1 3 3

2 (a)

Displacement

(c) When a dielectric is added to each capacitor, the capacitance increases. The potential difference across the capacitor is constant. Therefore, the charge stored in the capacitor increases.

Overdamped

6 (a) Number of turns, N = 1000 x 2 = 2000

Underdamped

Magnetic flux, 0 Time

φ=

(b) Resonance occurs when a system undergoing forced oscillation oscillates with the maximum amplitude when the driving frequency is slightly less than the natural frequency of the system. 3 (a) (i) A single slit diffraction pattern is formed. (ii) The fringe separation of the interference pattern increases. λD [Use the formula y = : when a is decreased, y increases] a (b) The lens is placed on a flat piece of glass Thin air flim and a thin film interference pattern is formed by the reflected light. A uniform fringe separation will confirm the flatness of the lens surface. 4 (a) Stress =

F F 200 = = = 2.55 ×10 6 N m −2 2 2 − 3 A πr π × (5.0 ×10 )

Fl Ae Fl Fl 200 × 0.25 e= = = = 6.56 ×10 −6 m 2 AE π r E π × 5.0 ×10 −3 2 × 9.7 ×10 10 24 V 24 V

(b) E =

(

)

1.5 μF

2.0 μF

µ AN 2 I  µ NI  φ = NBA = N  0  A = 0 l  l  −7 −4 2 4π ×10 ×1.8 ×10 × 2000 × 2.0

C

2.0 ×10

(b) Self-inductance, L =

−2

= 9.05 ×10 −2 Wb

∆φ µ0 AN 2 4π × 10 −7 × 1.8 × 10−4 × 20002 = = ∆I l 2.0 × 10− 2

L = 4.52 x 10–2 H

7 (a) Energy absorbed by the atom = energy of a photon of the light beam =

hc

λ

=

6.63 ×10 −34 × 3.00 ×10 8 = 1.81 ×10 −15 J −9 0.110 ×10

(b) KE of the electron, E = 180 eV = 180 x 1.60 x 10–19 J 1 mv 2 = E . 2 Therefore, v =

2E = m

2 ×180 ×1.60 ×10 −19 = 7.95 ×10 6 m s −1 −31 9.11 ×10

8 (a) Mass defect, m = (1.0087 + 10.0130) – (7.0160 + 4.0026) = 0.0031 u Total kinetic energy released, E = mc2 E = 0.0031 x 1.66 x10–27 x (3.00 x 108)2 E = 4.63 x 10–13 J The total kinetic energy released consists of the KE of the Lithium atom and the KE of the Helium atom. Kinetic energy of the helium atom

2 1 × ( 4.0026 ×1.66 ×10 −27 ) × (9.10 ×10 6 ) = 2.75 ×10 −13 J 2 Kinetic energy of the lithium atom = 4.63 x 10–13 – 2.75 x 10–13 = 1.88 x 10–13 J ALTERNATIVE SOLUTION Using E = mc2, the KE equivalent to 1 u is, = 1.0 x 1.66 x 10–27 x (3.00 x 108)2 = 1.494 x 10–10 J In a nuclear reaction, mass-energy is conserved. The conservation equation in terms of mass is:

x v cos θ x Substitute t = into (2) v cos θ 2  x  g x  y = ( v sin θ )  −    v cos θ  2  v cos θ  From (1), t =

=

(Rest mass of n) + (Mass equivalent to KE of n) + (Rest mass of B) + (Mass equivalent to KE of B)

gx 2 2v 2 cos 2 θ (b) (i) The time taken for the bullet to strike the jeep is equal to the time taken for the bullet to drop through a vertical height of 15 m. The initial vertical component of the velocity of the bullet = 0 1 2 Using the equation, s = ut + gt , with u = 0, s = 15 m and g = 9.81 m s–2 2 y = x tan θ −

(Rest mass of Li) + (Mass equivalent to KE of Li) + (Rest mass of He) + (Mass equivalent to KE of He)

=

t=

The conservation equation in terms of kinetic energy is: (KE equivalent to rest mass of n) + (KE of n) + (KE equivalent to rest mass of B) + (KE of B)

=

(KE equivalent to rest mass of Li) + (KE of Li) + (KE equivalent to rest mass of He) + (KE of He)

Let the kinetic of the Lithium atom = EL [(1.0087 x 1.494 x 10–10) + 0 + (10.0130 x 1.494 x 10–10) + 0] = [(7.0160 x 1.494 x 10–10) + EL + (4.0026 x 1.494 x 10–10) + 2.75 x 10–13] EL = 1.88 x 10–13 J (b) The reaction energy is the energy released in the form of the total kinetic energy of the decay products = 2.75 x 10–13 + 1.88 x 10–13 = 4.63 x 10–13 J 9 (a) The force in the object = – mg. Therefore, the acceleration of the object = –g. Let the time taken for the object to move from P to Q = t Horizontal displacement, x = ( v cos θ ).t …………..……. (1) 1 2 Vertical displacement, y = ( v sin θ ).t + ( − g ) t 2 Q gt 2 ………..... (2) y = ( v sin θ ).t − 2 v sin θ v y P

v cos θ x

2s = g

2 ×15 = 1.75 s 9.81

(ii) Distance travelled by the jeep = 10 m s–1 x 1.75 s = 17.5 m Distance from the security post = 700 m – 17.5 m = 682.5 m (iii) The bullet travels a horizontal distance of 682.5 m in 1.75 s 682 .5 = 390 m s −1 Initial speed of the bullet = 1.75 (iv) Vertical component of the velocity of the bullet = 0 + gt = 9.81 x 1.75 = 17.17 m s–1 2 2 Speed of the bullet = 390 +17 .17 = 390 .4 m s −1 θ 17.17 m s–1

390 m s–1

 17 .17  0  = 2.52 390  

θ = tan −1 

The speed of the bullet is 390.4 m s–1 at an angel 2.520 below the horizontal. 10 (a) Doppler effect is the apparent change in the observed frequency of a wave due to the relative motion of the source of the wave and the observer. (b) y = 2.5 ×10 −5 sin 2π ( 500 t −1.4 x ) . Compare with y = a sin (ωt − kx ) ω = 2π ( 500 ) = 1000 π s −1 k = 2π (1.4 ) = 2.8π m −1 ω 1000 π = 357 m s −1 Speed, v = = k 2.8π

ω = 2πf . Therefore frequency, f =

ω 1000 π = = 500 Hz 2π 2π

v = 2 m s–1

Wall

Detector

(c) (i)

Speed of sound waves = v Speed of source of sound = uS Frequency of sound given out by the source The detector receives sound reflected from the wall and sound directly from the car. Consider the Car and the Wall The car is a source moving towards the wall.  v  Apparent frequency of sound received by the wall, f 1 =   v −u   f0 S    357  f1 =   × 500 = 503 Hz  357 − 2  Frequency of sound reflected by the wall and received by the detector = 503 Hz Consider the Car and the Detector The car is a source moving away from the detector.  v  Apparent frequency of sound received by the detector, f 2 =  v +u   f0 S    357  f2 =   × 500 = 497 Hz  357 + 2  Frequency of sound from the car received directly by the detector = 497 Hz (ii) Beat frequency = f1 – f2 = 503 – 497 = 6 Hz (iii) The reflected travels in the opposite direction. The waves undergo a phase change of π due to reflection at the wall. Equation of the reflected wave is y = 2.5 ×10 −5 sin 2π ( 500 t +1.4 x − 0.5) (iv) Intensity is directly proportional to (amplitude)2 Since the approaching wave and reflected wave have equal amplitudes,

the ratio of the intensities of the approaching wave and reflected wave is 1 : 1. pV T = 11 (a) Using pV = nRT , we have nR 5 −3 3 ×10 × 2.0 ×10 At A, TA = = 361 K 0.2 × 8.31 TB = TA = 361 K since A to B is an isothermal expansion. V VB = C From B to C, pressure is constant. Therefore, TB TC

TC =

VC 2.0 ×10 −3 × 361 TB = = 60 K VB 12 .0 ×10 −3

(b) From A to B, work done by the gas, WAB = nRT ln

VB VA

WAB = 0.2 ×8.31 ×361 × ln

12 .0 2.0

WAB = 1075 J From A to B, temperature is constant. Therefore, pAVA = pBVB V  2 .0  5 4 p B = A pA =   × 3 × 10 = 5 × 10 Pa VB 12 . 0   From B to C, work done on the gas, WBC = p(VC – VB) WBC = 5 x 104 (2.0 x 10–3 – 12.0 x 10–3) WBC = –500 J From C to A, no work is done because the volume is constant. Net work done by the gas = 1075 – 500 = 575 J (c) Use the First law of thermodynamics, Q = ∆U +W From A to B, ∆U = 0 . Therefore, Q = W = 1075 J From B to C, pressure is constant. Therefore, Q = nC p, m ∆T 5 5 × 0.2 × 8.31 × ( 60 − 361 ) 3  Q = n R + R ∆T = nR ∆T = = −1251 J 2 2 2  From C to B, volume is constant. Therefore, Q = nC v, m ∆T 3 3 × 0.2 × 8.31 × ( 361 − 60 ) 3  Q = n R ∆T = nR ∆T = = 750 J 2 2 2  Net heat absorbed by the gas = 1075 – 1251 + 750 = 574 J

Given, V = 240 sin 120 π t . Compare with, V = V0 sin 2π f t Therefore, V0 = 240 V and f = 60 Hz 1 1 1 = = Reactance, X = = 8.84 x 103 Ω ωC 2πfC 2π × 60 × 0.3 ×10 −6 V0 240 = = 170 V (ii) rms voltage, Vrms = 2 2

12 (a) (i)

rms current, I rms

W W and X are connected to the AC supply. Y and Z are the output 1 4 terminals. Input In the first half cycle, diodes 1 and 3 are forward Z Y bias; diodes 2 and 4 are AC reversed bias. 2 3 The current flows through the load, R from A to B. In the next half cycle, X diodes 2 and 4 are forward bias; diodes 1 and 3 are reversed bias. R The current flows through the load, R in the B A same direction A to B. Therefore, in both half cycles, the current flows in the same direction to produce a potential difference across the load. Output (ii) Smoothing is achieved by connecting a capacitor across the load. Varying DC When the output voltage increases, the capacitor is charged. When the output voltage decreases, the capacitor discharges through the load to maintain a steady voltage across the load. The capacitance should be large enough so that the time constant of the discharge C

Smoothed output

Time

B

2.

There are certain allowed or permissible orbits for which the energy of the electron is constant, that is, the electron does not transmit energy in the form of electromagnetic radiation. Energy is only transmitted when there is a transition of the electron from an orbit with higher energy to another orbit with lower energy. In an allowed orbit, the angular momentum of the electron are integral

h , where h = Planck’s constant. 2π (b) The electrostatic attraction between the nucleus (+e) and an electron (–e) acts as the centripetal force for the electron to orbit around the nucleus.

e2 mv 2 e2 = r = , that is, …….....................[1] r 4π εo r 2 4πmε 0 v 2 From Bohr’s postulate, angular momentum of the electron, nh mvr  = ..……………….... [2] , n = 1, 2, 3, ……. 2π nh . 2πmr nh Substitute v = into [1], 2πmr

From [2],

v=

e 2  2πmr  r=   4πmε 0  nh 

e 2 4π 2 m 2 r 2 . Therefore, 4πmε 0 n 2 h 2 1 mv 2 = 8.64 ×10 −20 . (c) (i) 2 r=

v=

r=

2

h 2ε o n 2 π me 2

2 × 8.64 ×10 −20 = 4.36 ×10 5 m s −1 9.11 ×10 −31

(ii) From [1] , r =

e2 4πmε 0 v 2

(

)

2

e2 1.60 × 10 −19 r= = 4πmε 0 v 2 4π × 9.11 × 10 −31 × 8.85 × 10 −12 × 4.36 × 10 5

R 0

1.

multiples of

V 170 = rms = = 1.92 × 10 −2 A 3 X 8.84 × 10

(b) (i) A full wave rectifier consists of 4 diodes connected in the form of a bridge circuit.

Output voltage

13 (a) Bohr's postulates

A

(

r = 1.33 x 10–9 m From the answer to (b), r =

h 2ε o n 2 πme 2 r 2 n = , we have π me 2 h 2ε 0

)

2

π × 9.11 ×10 −31 × (1.60 ×10 −19 ) ×1.33 ×10 −9 2

n = 2

(6.63 ×10 )

−34 2

× 8.85 ×10 −12

n=5 The allowed orbit is n = 5. (iii) The radius of the orbit = 1.33 x 10–9 m as calculated in (c)(ii) 4 9 12 1 14 (a) (i) 2 α + 4 Be → 6 C + 0 X (ii) The α-particle is the helium nucleus. X is a neutron. (iii) 1. Carries no charge. 2. Moves at high speed

(b) Using E = mc2 and 1 u = 1.66 x 10–27 kg E = (1 x 1.66 x 10–27) x (3.00 x 108)2 J 1 J is equivalent to 1.60 x 10–19 eV of energy.

(

1 × 1.66 × 10 −27 × 3.00 × 10 8 E= 1.60 × 10 −19

)

2

eV

E = 9.34 x 108 eV = 934 x 106 eV = 934 MeV The energy equivalent to a mass of 1 u is 934 MeV. (c) Let the element be X. When the ions of X and carbon-12 enter the magnetic field in the deflection chamber of the mass spectrometer, the magnetic force acts as the centripetal force. The ions move in a semicircle of radius r. mv 2 qvB = r The radius of the circular path, r =

mv qB

Assuming v, q and B to be constant, r ∝ m rX m 26 .2 m X = X . That is, Therefore, = and mX = 14 rC mC 22 .4 12 The element could possibly be nitrogen. Assumptions: The ions of the element and carbon-12 carry the same charge and move with the same speed in the mass spectrometer.