Prelim H2P3 MS

weight of book support force on book by table 1 Anderson Serangoon Junior College 2021 H2 Physics Prelim Mark Scheme Paper 3 (80 marks) 1a Taking right as positive, s = ut 4.40 = 5.40 ×t t = 0.8148 s Taking downwards as positive, s = ut + ½at2 h = ½ ×9.81 ×0.8148 2 = 3.2565 ≈ 3.26 m C1 A1 1bi downward pointing arrow labelled weight upward pointing arrow labelled air resistance no credit if magnitude of air resistance exceeds that of weight B1 1bii Air resistance increases (as velocity increases) Weight (or mass) is constant, so resultant force decreases Hence, acceleration decreases B1 B1 B1 1biii At terminal velocity, air resistance equals weight With larger mass, weight is larger. (Greater air resistance), so greater terminal velocity B1 B1 1biv 1. gravitational potential energy to kinetic energy and thermal/internal energy B1 1biv 2. gravitational potential energy to thermal/internal energy B1 2ai Either The support force on book by table and the weight of book are both acting on the book. (N3L states that forces act on different bodies.) They are different types of forces, gravitational and electromagnetic/contact forces. (N3L states that forces must be of the same type.) (So, they are not a pair of action-reaction forces.) Or Reaction force of contact force by table on book is contact force that book exerts on table. Reaction force of weight of book is gravitational force that book exerts on Earth. B1 B1 B1 B1 9749/03/ASRJC/2021PRELIM [Turn Over weight air resistance

2 (So, weight of book and contact force by table on book are not a pair of action-reaction forces.) Credit 1 mark only if student did not mention type of force. 2aii Since book is resting on table, there is no net force on the book, so the two forces are equal and opposite. B1 2b Suppose two colliding bodies A and B (where A and B is an isolated system), By Newton’s third law, force A exert on B, FAB is equal in magnitude and opposite in direction to force B exert on A, FBA. FAB = – FBA Duration of collision is the same for A and B. By Newton’s second law, net force on A, FBA is equal to rate of change of momentum of A. Net force on B, FAB is equal to rate of change of momentum of B. Hence, total (rate of) change of momentum is 0. B1 B1 B1 2c By Conservation of Linear Momentum Sum of initial momentum = Sum of final momentum (28)(88) + (17)(53) = (28)(67)+ (17)v2 v2 = 87.6 m s1 Loss in kinetic energy = total initial kinetic energy – total final kinetic energy = ½ (28)(88)2 + ½ (17)(53)2 – (½ (28)(67)2 + ½ (17)(87.6)2) = 4200 J C1 C1 A1 2d Correct shape – smooth curves for both lines. Line for steel objects has larger peak and smaller duration than line for rubber, with approximately equal area under the two lines. B1 B1 3a base units: kg m s–2 × m = kg m2 s–2 A1 3bi distance of COG from P (= GP) = 17 cos 45° – 4.0 = 8