ASRJC 2022 Prelim P2 Guide

1 9749/02/ASRJC/2022PRELIMS [Turn Over Anderson Serangoon Junior College 2022 JC2 H2 Physics Prelim Mark Scheme Paper 2 1a According to Newton’s 3rd Law, F12 = −F21 … (1) Taking the time of collision to be t, Using Newton’s 2nd Law, eqn (1) can be rewritten as, 1(2pp )tt  =− OR ()2 2 2 2 1 1 1 1m v - m u m v - m u tt =− 2 2 2 2 1 1 1 1m v m u m v m u− =− + OR 1 1 2 2 1 1 2 2m v m v m u m u+ = + B1 M1 A1 1bi By conservation of momentum, the gain in momentum of body Q must be equal to the loss in momentum of body P. Since total momentum is conserved, when Q gains momentum (positive gradient), P loses momentum (negative gradient). Hence, the gradients of the graphs must have opposite sign. OR Gradient of the graph is the net force acting on the object. Since the 2 forces are action- reaction pair, hence they must be in the opposite direction. B1 1bii When there is no net external forces acting on the system. B1 1biii Total initial momentum = 20 + 16 = 36 kg m s-1 Total final momentum = 24 + 12 = 36 kg m s-1 Total momentum is conserved. A1 1biv Force acting on body P = rate of change of momentum of body P = 3 12 - 16 (300 - 150) 10 − = −26.7 N = −27 N Magnitude of force = 27 N C1 A1 1bv Impulse = F t = 26.7 × 150 × 10-3 = 4.0 N s OR Impulse = change in momentum = 24 – 20 = 4.0 N s A1 1bvi Relative speed of approach = 116 20 3.0 m s2.0 4.0 pQuu −− = − = Relative speed of separation = 124 12 0 m s4.0 2.0 QPvv −− = − = RSS is not equal to RSA, and RSS is zero, so the collision is perfectly inelastic. B1 B1 A1 2ai Constant amplitude B1 2aii Period = 0.75 s ω = 2π / T ω = 8.4 rad s–1 C1 A1

2 9749/02/ASRJC/2022PRELIMS 2aiii Either use of gradient or v = ω y0 v = 0.168 m s–1 (allow ±0.02 for construction) C1 A1 2bi f = 1/T = 1/0.75 = 1.3 Hz B1 2bii at ½ f0 ‘pulse’ / energy is provided to mass on alternate/some oscillations so ‘pulses’ / energy build up the amplitude (don’t accept maximum amplitude) M1 A1 3ai Transverse A1 3aii =  = 12 8 4 distance = speed 4.4 10 3.0 10 1.5 10 s t A1 3aiii power received by dish aerial = intensity received  area of dish   − = =  = 2 12 2 23 output (source) power 280 4 25 280 4 (4.4 10 ) 2.9 10 W P r C1 C1 A1 3aiv The actual power is greater because the transmitting aerial is directing power towards the Earth instead of radiating uniformly in all directions as assumed in (a)(iii). A1 3bi v = f  = 0.040  1.5  = 0.060 m C1 A1 3bii path difference = (44 – 29) / 6.0 = 2.5  either waves have path difference = (n + ½)  or waves have phase difference = π so destructive interference M1 M1 A1 3ci Incident sound wave fr