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 Ju