YIJC 2024 JC2 PRELIM H1 Phy P2 Solutions

1 ©YIJC 8867/02/YIJC/24 2024 YIJC JC2 Prelim Exam H1 Physics Paper 2 Solution 1 (a) 1. Initial speed or velocity is zero 2. (non-zero magnitude of) acceleration is constant (uniform) and in a straight line B1 B1 (b) (i) magnitude of acceleration at t = 8.0 s is less than that at t = 14.0 s (due to gradient) direction of acceleration at t = 8.0 s is opposite that at t = 14.0 s (due to sign of gradient) B1 B1 (ii) a = gradient or a = (v-u)/t or a = v/t <a> = (0– (-10)) / 16 a = 0.625 m s-2 C1 A1 (iii) Net displacement (final position) from X = area under v-t graph s = ½ (-10)(4) + ½ (20)(8) = - 20 + 80 = 60 m C1 A1 2 (a) The principle of conservation of momentum states that the total momentum of a system is conserved throughout a collision if there is no net external force acting on the system. B1 (b) (i)             1 , assume is to the right. 0.100 2.0 0.200 0 0.100 0.200 1.3 0.60 m s 2sf ve v v v       Magnitude = 0.60 m s-1 Since v is negative, the direction is to the left. C1 A1 B1 (ii)     1 1 relative speed of approach 2.0 0 2.0 m s relative speed of separation 1.3 0.6 1.9 m s          Since the relative speed of approach is not equal to the relative speed of separation, the collision is inelastic. OR               2 2 2 2 1 1total initial KE 0.100 2.0 0.200 02 2 0.20 J 1 1total final KE 0.100 0.6 0.200 1.32 2 0.187 J 3sf       Since the total initial kinetic energy is not equal to the total final kinetic energy, the collision is inelastic. M1 M1 A1 (M1) (M1) (A1)

2 ©YIJC 8867/02/YIJC/24 (c) By Newton’s third law, the force that block A exerts on block B is equal in magnitude and opposite in direction to the force that block B exerts on block A. By Newton’s second law, the (net) force (on block A) is the rate of change of momentum (of block A.) Hence, the rate of change of momentum of block A is equal in magnitude and opposite in direction to the rate of change of momentum of block B. M1 M1 A1 3 (a) (i) The sum of forces in all directions is zero, and The sum of moments about any axis is zero. B1 B1 (ii) Taking moments about B, the weight of the table top exerts an anticlockwise moment. In order for the table to be in equilibrium, there must be an upward vertical (frictional) component of force acting at the hinge to produce a clockwise moment about B which is equal to the anticlockwise moment of the weight of the table top. OR If there is only a horizontal normal contact force at hinge A, there will be no moment provided by this force and the table will not be in equilibrium. B1 B1 (B1) (b) Taking moments about A,         130 0.30 sin 20 0.60 190 N 3sf T T    C1 A1 (c)  190sin20 130 6