ASRJC Prelim H2P3 MS Final (sharing)

1 9749/02/ASRJC/2019PROMO [Turn Over Anderson Serangoon Junior College 2023 H2 Physics Prelim P3 Mark Scheme Paper 3 (80 marks) E – Easy, A – Average, D – Difficult ECF Error carried forward SF Significant figures error M0 No A marks awarded AE Arithmetic error BOD Benefit of doubt ^ More is needed in answer POT Power of ten error CON Contradictory response XP Wrong physics TE Transcription error IR Irrelevant (part) response 1a Gradient is not constant so acceleration changes with speed Hence resultant force changes with speed Resultant force consist of weight and air resistance and weight is constant (hence can be deduced that air resistance varies with speed). A M1 A1 A0 1bi Fnet = mg – kv2 Using N2L, Fnet = ma mg – kv2 = ma m(g – a) = kv2 2 () kvga m−= A B1 B1 A0 1bii At v = 4.0 m s–1 Gradient= 6.0 2.8 3.2 8.00.68 0.28 0.40 − ==− a = 8.0 m s–2 g – a = 9.81 – 8.0 = 1.8 (or 1.81) A M1 A1 A0 1biii From b(i), 2 () kvga m−= , 2 ()g a k vm − = should be a constant for 4.0 m s–1, 22 ( ) 1.8 0.114.0 − ==ga v or 8.0 m s–1, 22 ( ) 9.8 0.158.0 − ==ga v Hence, values of 2 ()ga v − are not the same, suggestion is incorrect. A M1 A1 2a The resultant force acting on a body is the rate of change of momentum of the body and acts in the direction of the change in momentum. E A1 2b Work W is the product of a force F and a displacement s in the direction of the force. Since Power P is the rate of doing work, Hence, P = W/t = Fs/t = Fv A B1 B1 A0

2 9749/02/ASRJC/2019PROMO 2ci At constant speed, no net force acting on lorry. Hence, magnitude of driving force is equal to that of resistive force on lorry. Driving force, F = P/v = 130 000/25 = 5200 N A M1 M1 A0 2cii Friction between tyres of lorry and road exerts a backward force on the road. By Newton’s third law, road surface exert a forward force of equal magnitude on tyres. A B1 B1 2d F - mg sin θ – R = ma F = 36000(9.81 sin 1.4) + 5200 + (36000)(0.15) = 19228 N = 19000 (19200) N A C1 A1 3a Centripetal acceleration: Car on path X and Y has the same acceleration Both experienced the same centripetal force at maximum speed since the centripetal force is provided by the (maximum lateral) friction force. As both cars are identical and hence have the same mass, they will have the same centripetal acceleration. Maximum speed: Car on path Y will have a bigger maximum speed. 2mvF r Frv m = = Since F and m are the same for both cars, vr . Since radius of path Y is bigger than that of X, maximum speed of car on path Y is larger. A M1 A1 M1 A1 3bi For circular motion, centripetal force on moon of mass m is provided by the gravitational force due to Jupiter on moon, 2 2 23 2 2 2 ) 4 ( J J JMmGm GM m mRR RT R M T G R    = = = where ω is the angular velocity of moon. A M1 M1 A0 3bii Since T2  R3, 2 3 2 3 0.676 3.18 2.62 Th Th Am Am Am TR T R T       ==             TAm = 0.506 Earth-days A M1 A1

3 9749/02/ASRJC/2019PROMO [Turn Over 3biii Since the orbital period of Thebe is not the same as the rotation period of Jupiter, So not in stationary orbit. Or orbital period of Thebe = 0.676 Earth-days, not the same as a Jupiter-day which is approximately 0.417 Earth-days, so not (possible to be) in stationary orbit. D M1 A1 (M1) (A1) 4a Change in depth of water. E B1 4b L = /4 v = f So, v = f (4L) = 480 × 4 × 0.18 = 345.6 m s–1 = 346 (350) m s–1 A C1 A1 4ci A node at the water surface and an antinode at the open end of the air column Length of air column equals ¾ of wavelength of the stationary wave, with total of 2 nodes and 2 antinodes labelled correctly Correct sinusoidal shape (only award when both M marks are scored) A M1 M1 A1 4cii Vibrate along axis of the tube, Vibrate with maximum amplitude. A B1 B1 4ciii L = ¾  so  = 4/3 L = ’/3 where ’ is the wavelength in part (b) Since v = f, f = 3 f’ where f’ is the frequency in part (b) f = 3 × 480 = 1440 Hz D C1 A1 5a As temperature (of the filament) increases with V the resistance of the lamp increases hence V/I ratio increases. A M1 A1 bi From I-V graph, find the common current when the individual p.d. of each lamp sum up to 12 V. I = 0.40 A (± 0.01 A) A B1 B1

4 9749/02/ASRJC/2019PROMO bii Current is the same, but p.d. across lamp B is higher than that of lamp A OR resistance of lamp B is higher than resistance of lamp A Using P= IV, lamp B will have greater power dissipation. A M1 A1 c From bi, VA = 4.0 V , VB = 8.0 V For zero current in the galvanometer G, k(XJ) = 4V where k = 12/0.80 XJ = (4/ 12) x 0.80 = 0.27 (0.267) m OR For zero current in the galvanometer G XJ = [4/(4+8)] x 0.80 = 0.27( 0.267) m A C1 A1 (C1) (A1) 6a Each coil acts as a current-carrying conductor in the magnetic field of the neighbouring turn Magnetic field (due to current) in one loop cuts / is normal to the current in second loop. By Fleming’s left-hand rule, there will be a (magnetic) force on the second loop towards the first loop. (By Newton’s third law, this gives rise to mutual attraction between the loops.) Hence, separation decreases. A M1 M1 A0 6bi Mass exert a downward force on the frame. / mass exert a (clockwise) moment on the frame For frame to have no net moment about pivot (or for the frame to be in rotational equilibrium), magnetic force must act downwards on near side of frame. A B1 A1 6bii F = BIL 24 × 10-3 = B × 4.5 × 4.6 × 10-2 B = 0.116 T = 0.12 T A C1 A1 7a When the rate of decay of Xe-143 is faster than that of Cs-143, the number of Cs-143 increases. When the rate of decay of Xe-143 decreases, as number of Xe-143 decreases, till it is below the rate of decay of Cs-143, the number of Cs-143 decreases. Or When the rate of decay of Cs increases, as number of Cs increases, till it is above the rate of decay of Xe, the number of Cs-143 decreases A B1 A1 (A1) 7b Due to the spontaneous nature of radioactivity decay, the rate of decay is unaffected by external environment such as surrounding temperature. A A1 7c The number of radiations does not just come from Xenon, but also from Caesium. The count rate may include background radiation. A B1 B1 7d β-particles have high speeds/energy D B1

5 9749/02/ASRJC/2019PROMO [Turn Over when charged particle is accelerated/slowed down electromagnetic radiation is produced β-particles stopped in the lead produce X-ray radiation which escapes. M1 A1 B1 8ai Heat lost by water = heat gained by mercury mwatercwaterTwater = mmercurycmercuryTmercury (18.7) (4.18) (37.4 – T) = (6.94) (0.140) (T – 23.0) T = 37.2 °C A C1 C1 A1 8aii The glass of the thermometer OR the beaker containing the water have negligible heat capacity or gained negligble heat. A B1 8aiii Use a liquid with a lower (specific) heat capacity (than mercury), or Use a smaller mass of mercury. A B1 8aiv It depends on properties of a real substance, and 0 °C is not absolute zero D B1 B1 8av Ideal gas E B1 8bi Internal energy is the sum of the kinetic energy due to the random motion of the molecules and the potential energy Ep due to intermolecular forc