<p> S7 MOCK EXAMINATION 2001 –2002 AL PHYSICS PAPER I</p><p>SOLUTIONS 1. (a) An isochronous oscillation is one in which the period of oscillation is independent of its amplitude. 1</p><p>(b) (i) An ellipse - v 2  2 (A2  x 2 ) 1 (ii) A straight line with a negative slope - ˙x˙  2 x 1 1 1 (iii) A parabola concave downwards - K  mv 2  m2 (A2  x 2 ) 1 2 2 (iv) A parabola concave upwards with a positive y-intercept such that 1 U  kx 2  constant 2 since the spring has elastic p.e. at equilibrium 1</p><p>(c) Since velocity at P = 0, OP is the amplitude of oscillation. Let OP = A and apply the equation v 2  2 (A2  x 2 ) , at point Q, 42  2[A2  (A  2) 2 ]………(1) ½ Similarly, at point R, (3) 2  2 [A2  (A 10)2 ] ………(2) ½</p><p>Divide equation (1) by equation (2), we have A  5.51 cm 1</p><p>Substitute the value of A  5.51 cm into equation (1), we have   0.9417 rad s 1 . ½ 2 The period of oscillation T   6.67 s ½ </p><p> m (d) The period T  2 would remain unchanged because k and m are the same. 1 k</p><p>(e) (i) Let m1 and m2 be the masses of objects M and N respectively.</p><p>2 k  2  k 2      ½ m  T  m</p><p>When object M oscillates alone,</p><p>2  2  k    ………(3)  6.67  m1</p><p>- 1 - 1. (e) (i) (Continued) When objects M and N oscillate together,</p><p>2  2  k    ………(4)  8.75  m1  0.05</p><p>Divide equation (4) by equation (3),</p><p>2  6.67  m    1 ½  8.75  m1  0.05</p><p>Therefore, the mass of object M, m1  0.07 kg 1</p><p>(ii) Substitute m1  0.07 kg into equation (3),</p><p> force constant of the spring k  0.062 N m 1 1 </p><p>12</p><p>2. (a) (i) Let  be the angular acceleration of the flywheel in the first 10 revolutions.</p><p>2 2 2   f  i ½</p><p>2  0.2    12.5103 1      2.04 rad s 2 210  2</p><p>Let T be the tension in the string. mg  T  ma  mr ½ T  m(g  r)</p><p>T  0.5(10 12.5103  2.04)  4.99 N 1</p><p>(ii) Let  be the angular deceleration of the flywheel in the last 50 revolutions. 2  0.2    12.5103 1      0.41 rad s 2 2 50 2</p><p>(iii) Let I be the moment of inertia of the flywheel and axle about the axis of rotation. Resultant torque acting on the flywheel and axle in the first 10 revolutions is Tr    I where  is the frictional couple at the bearings ½ Moreover,   I ½</p><p>- 2 - 2. (a) (iii) (Continued) 4.9912.5103 Therefore, Tr  I(  )  I   0.026 kg m 2 1 2.04  0.41</p><p>(b) (i) The principle states that angular momentum of a system is conserved if there is no resultant torque applied on it. 1 5 (ii) (I) Initial angular momentum, L  I   I   2 ½ A A 2</p><p>Final angular momentum, L  (I A  I B )  (I A  I B ) 1 2 ½ By conservation of angular momentum, L  L I 3 B  1 I A 2 1 1 (II) Initial k.e., K  I 2  I (5) 2 ½ 2 A 2 A</p><p>1 2 1  5  2 Final k.e. K  (I A  I B )   I A 2 ½ 2 2  2  The percentage decrease in the k.e. of the system is K  K  100%  60% 1 K The energy is used for overcoming friction between the discs. Finally, it is converted into heat. 1 12</p><p>3. (a) Negative 1</p><p>(b) I = nAQv 1 I 16 v = = = 1.0 mm s 1 1 nAQ (1.01029 )(2.0103 )(0.5103 )(1.61019 )</p><p>(c) When Hall voltage is attained, electric force = magnetic force. F = BQv 1 = (0.1)(1.6  1019)(1.0  103) = 1.6  10  23 N 1</p><p>F 1.61023 (d) (i) E = = = 1.0  10  4 N C 1 1 Q 1.61019 V (ii) E = 1 d V = Ed = (1.0  104)(2.0  103) = 0.2  V 1</p><p>- 3 - 3. (Continued) (e) Magnetic field should be applied in a direction perpendicular to the North-South direction of the Earth’s magnetic field. 1 (f) Use stronger applied magnetic field Use larger current through the slice Use semiconductor with smaller number of charge carriers per cubic metre Use a thinner slice ANY THREE of the above 3 12</p><p>4. (a) V All vector are drawn correctly 1 L VL and VC are equal in length ½ V and VR are equal in length ½ </p><p>V V R I</p><p>V C</p><p>(b) (i) signal generator is connected correctly 1 Two CRO's are connected correctly 1 </p><p>(ii) CRO 2 reads the p.d. across R. </p><p>Current through the circuit can be found by VR/R. 1 (iii) At resonance, 1 f0 = 2 LC 1 1 9 LC = 2 2 = 2 2 = 4.05  10 s 1 4 f0 4 2500</p><p>- 4 - 4. (b) (iii) (Continued) 1 And C = 2 2 , using the given values of L, the corresponding value of C 4 f0 L can be calculated.</p><p>Given values of L (in mH) Calculated values of C (in F) 10 0.41 40 0.10 1 100 0.041 150 0.027 It can be seen that L = 40 mH and C = 0.10 F are the suitable values. The corresponding resonant frequency = 2516.5 Hz which is closest to 2500 Hz. 1</p><p>(iv) Ir.m.s. reaches a maximum value at resonant frequency. 1</p><p>Vr.m.s. Ir.m.s. = R</p><p>Vr.m.s. 6 R = = 3 = 120  1 I r.m.s. 50 10 (v) correct shape 1 higher peak for smaller resistance 1</p><p>12</p><p>5. (a) (i) When saturation just occurs. all VCC drops a cross 2 k,</p><p> ICRL = VCC</p><p>VCC 6 IC = = = 3 mA ½ RL 2000 3 IC 310 and IB = = 27.27 A ½  110</p><p> VIN = IB RB + VBE ½ = (27.27  106)(10  103) + 0.6 = 0.873 V ½ </p><p>- 5 - 5. (a) (Continued) (ii) V /V OUT </p><p>6 correct shape ½ correct values in x-axis and y axis ½ </p><p>V /V 0 IN 0.6 0.873</p><p>(b) At proper biasing, the input voltage is adjusted so that the amplifier works in the linear region of the input-output voltage characteristics graph. ½ V The output voltage is at the middle of the linear region , i.e. CC . ½ 2</p><p>VCC (c) (i) When the amplifier is in proper biasing condition, VOUT = . ½ 2</p><p>VCC IC RL = 2</p><p>VCC 6 IC = = 3 = 1.5 mA ½ 2RL 2(210 ) 3 IC 1.510 (ii) IB = = 13.64 A ½  110</p><p> VIN = IB RB + VBE ½ = (13.64  106)(10  103) + 0.6 = 0.736 V ½ </p><p>R1  0.736 =  6 R1  R2</p><p>R1 = 6.99 k  ½ (iii) They act as a potential divider ½ which divide the supply voltage into smaller values. ½ </p><p>- 6 - 5. (Continued) (d) (i) (correct shape  ½ ; correct max. & min. voltages  ½) 1 (ii) (correct shape  ½ ; correct max. & min. voltages  ½ ; show distortion  ½ ;</p><p> correct corresponding values of VIN when distortion  ½ ) 2</p><p>Input voltage /V Input voltage /V 0.2 0.1 0.137 time /s time /s 0 0  0.1 0.137  0.2 Output voltage at X /V Output voltage at X /V</p><p>6 6 5.2</p><p>3 3</p><p>0.8 time /s time /s 0 0</p><p>(iii) They are used to filter out the d.c. components in the a.c. signals. ½ 1 I agree with the student. The reactance of a capacitor X = C 2fC will be smaller and hence the opposition to a.c. signal is lowered. ½ 12</p><p>6. (a) The sound waves arrive at A in phase (phase difference = 0), 1 so they interfere constructively. 1</p><p>(b) The phase difference increases from zero as the man walk from A to B. 1 The loudness decreases due to less constructive interference and 1 it become a minimum when the sound waves are vibrating 1800 out of phase. 1</p><p>(c) v = f v 330  = = = 0.165 m 1  2000</p><p>- 7 - 6. (Continued) D (d) s = 1 a (0.165)(5.0) = = 1.375 m 1 0.6</p><p>(e) The positions of maxima and minima reverse 1 because the loudspeakers are now vibrating out of phase. 1</p><p>(f) Since the emission of light from the light bulbs is in a random manner. The light waves may not have constant phase relation, i.e. they may not be coherent. 1 Separation of two light source is 0.5 m which is too large. It should be comparable to the wavelength of light. 1 12 1 7. (a) (i) Since intensity I  1 r 2</p><p>I r 2 2 = 1 I 2 1 r2</p><p> r 2 0.52 I = 1 I = ( )(2.4 104) = 6 10  5 W m 2 1 2 2 1 2   r2 1 I (ii) h = 10 log ( 2 ) I0 6.0105 = 10 log ( ) = 78 dB 1 1.01012</p><p> r 2 I (iii) 0 = 2 2 I r2 0</p><p>5 I 2 6.010 r0 = r2 = (1) = 7746.0 m 1 12 I0 1.010 (iv) Intensity is defined as the rate of flow of energy through a unit area perpendicular to the direction of travel of the sound at the place in question. 1 Intensity level of a sound source is its intensity relative to some agreed ‘zero’ intensity level. 1</p><p>- 8 - 7. (Continued) (b) (i)</p><p>1</p><p>At A the light source is moving towards the observer, distances between wavefronts are compressed. Wavelength of light is then decreased. ½ At B the light source is moving away from the observer, distances between wavefronts are extended. Wavelength of light is then increased. ½ (ii) At A, moving source towards the stationary observer, c fA = ( ) f c  v c c = ( ) f ………………………… (1) 1 599.9109 c  v At B, moving source away from the stationary observer, c fB = ( ) f c  v c c = ( ) f ………………………… (2) 1 600.1109 c  v Solving (1) and (2), v = 5  10 4 m s 1 1 (iii) circumference of orbit = orbital speed  Period 2a = v  T vT (5104 )(3.6106 ) a = = = 2.86  10 10 m 1 2 2 12</p><p>8. (a) (i) The Young modulus of a material is a measure of its elasticity. It is defined as F Stress the ratio E   A . 1 Strain e l</p><p>(ii) Let the Young moduli of wires P and Q be EP and EQ respectively.</p><p>- 9 - 200 3 2 [ ( 210 ) ] 10 2 EP   6.37 10 N m 1 5103 20 8. (a) (ii) (Continued) 1 E  E  3.181010 N m-2 1 Q 2 P (iii) Let point A be at a distance of x away from the end of the block and T be the tension in wire P. 1 By the result in (a)(ii), the tension in wire Q = T 1 2 Take moment about the centre of mass of the block, 2T (10  x)  T(12  x 10) 1</p><p> x  6 cm 1</p><p>(b) (i) Energy required = Area under the curve 1 1 1 U  (141010 )[(3.00  2.88) 1010 ]  (210 10 )[(3.03  3.00) 1010 ] 2 2</p><p>U  0.811020 J 1</p><p>(ii) At the near neighbourhood of the equilibrium separation in the F-r graph, the curve is approximately a straight line. Therefore, F  r  F  kr that is the Hooke’s law. 1 (iii) k equals to the slope of the graph at the equilibrium separation, i.e.</p><p>10 r0  3.0010 m . 1</p><p>91010 k   75 N m 1 1 (3.00  2.88) 1010 Hence, the Young modulus for the solid is</p><p> k 75 11 2 E   10  2.510 N m 1 r0 3.0010 12</p><p>9. (a) When electromagnetic radiation of high enough frequency falls on metal surfaces, electrons are ejected from them. The phenomenon is known as the photoelectric effect. 1 It implies that light has a wave-particle dual nature. 1</p><p>- 10 - (b) The work function of a metal is the minimum amount of energy to liberate an electron from it. 1</p><p>- 11 - 9. (b) (Continued) The energy of the incident photon is partly used for liberating the electron from the metal and partly converted into the k.e. of the emitted electron in a photoelectric emission. ½ 1 hf    m v 2 ½ 2 e max which is the photoelectric equation.</p><p>(c) (i) Number of photoelectrons emitted per second, Power of incident light n   Quantum Yield ½ Energy carried by each photon 104  I  1.61019  0.16 ½ 4 (1.61019 )</p><p> I  4106 A 1</p><p>(ii) Photoelectrons are emitted from the lower plate to neutralize the positive charges on the upper plate when the light is turned on. 1 Q CV 50106  2 I    t   25 s 1 t t 4106</p><p>(iii) By the photoelectric equation, the work function of the lower plate is</p><p>  hf  K max  4 1.5  2.5 eV 1</p><p>Hence, the threshold frequency of the incident light is  2.51.61019 hf    f    6.031014 Hz 1 0 0 h 6.631034</p><p>(iv) When the p.d. between the plates reaches the stopping potential, no photoelectrons can be emitted. 1 The upper plate and the lower plate should be negatively charged and positively charged respectively at that moment such that ½ the p.d. between them is 1.5 V . ½ 12</p><p>10. (a) (i) -radiation has a short range in air. 1 It cannot leave the smoke detector. 1</p><p>(ii) No, the detector will not work because -radiation does not cause as much ionization as -radiation. 1</p><p>- 12 - 10. (a) (ii) (Continued) And it is more penetrative so that smoke will not be effective enough to reduce the ionization current. 1</p><p>(b) (i) Half-life – the time taken for radioactive source to disintegrate by one half of its initial number of atoms 1 Decay constant – the probability of decay per unit time for a radioactive source 1</p><p> ln 2 0.6931 10 1 (ii) (I)     2.301510 min ½ T1 5730 365 24 60 2</p><p>1 N  6.02 10 23  1.31012  6.5221010 ½ 0 12</p><p>A0  N 0</p><p>10 10 A0  2.301510  6.52210  15 decays per minute 1</p><p>7.5 (II) Activity of 1 g of carbon =  3.75 decays per minute ½ 2</p><p>Using A  A0 exp(t) , ½</p><p> t  2T1  11460 years 1 2</p><p>(III) Carbon exists in all living things. 1 The half-life of carbon is comparable with the time span of human history. 1 12</p><p>- END -</p><p>- 13 -</p>