2020 YIJC H1 Physics Prelims P1 Answers

2020 YIJC JC2 H1 Physics Prelim P1 MCQs Solution Ques Ans Ques Ans Ques Ans Ques Ans Ques Ans Ques Ans 1 A 6 B 11 B 16 D 21 D 26 C 2 A 7 C 12 C 17 A 22 B 27 D 3 C 8 C 13 B 18 B 23 A 28 C 4 C 9 D 14 B 19 A 24 A 29 A 5 D 10 B 15 D 20 B 25 A 30 B MCQs Solution S/N Answer Explanation 1 A Q = It SI base units of charge = Ampere seconds Option B : [F] = [m][a] = kg m s-2 Option C : [R]=[P]/[I2] = kg m2 s-3 A-2 Option D : [P]=[W]/[t]= kg m2 s-3 2 A For force of 4 N and 6 N Maximum = 4 + 6 = 10 N Minimum = 6- 4 = 2 N The force will range from 2 N to 10 N depending on the angle between them, hence 1 N is not possible. 3 C Vertical component of velocity on striking ground = vfinal sin = √2𝑔ℎ ----------(1) Horizontal component remain constant = vfinal cos  = v ------------(2) (1)/(2) tan  = √2𝑔ℎ 𝑣 Angle  is largest when h is the largest and v the smallest. 4 C In the presence of air resistance, the drag force which is acting in opposite direction to motion, hence upwards, will increase and causes the net resultant force acting on the object to reduce to zero when it equals and opposte to the weight. Hence the acceleration will decrease and reach zero (state of terminal velocity).

5 D At point A : object just reached the ground for the first time At point B: object just about the leave the ground after first bounce At point C: object reached the maximum height after first bounce At point D: object just reached the ground for the second time 6 B Use s= ut + ½ ut2 (evacuated tube implies no air resistance) L = ½ g(T)2 S = ½ g (T/2)2 =1/4 L = 0.25 L 7 C For a system where no net external force acts, the linear momentum is conserved. For inelastic collision, kinetic energy is not conserved. 8 C Change in momentum = (0.44 x 32) = 14.1 kg m s-1 Since Fave = ∆ (𝑚𝑣) ∆𝑡 = 14.1 9.2 ×10−3 = 1.53 × 103𝑁 9 D This is a statement of Newton’s third law. The force exerted on the man’s feet by the floor is always equal in magnitude but opposite in direction to the force exerted on the lift floor by his feet. 10 B By conservation of momentum 2m (4v) + 3m(-2v) = (5m) Vfinal Vfinal = 2 5 v 11 B Tension in the cord = 900 N Torque on disc = 900 x 0.20 = 180 Nm Torque due to F = F x 1.20 = 180 F = 180/1.2 = 150 N 12 C By Hooke’s law, 𝐹 = 𝑘𝑥 𝑘 = 𝐹 𝑥 𝑘 = 200 0.01 = 2.0 × 104 N m−1 13 B (Note F is the x-axis here instead) Work done is the area of graph bordering the displacement axis.

14 B The loss in gravitational potential energy is equal to the gain in kinetic energy and work done against air resistance. Thus the loss in gravitational potential energy is greater than gain in kinetic energy. 15 D Work done by person= gain in GPE of box + work done against friction = mgh + fd, where d is the distance moved along the slope. = 2001.5 + 1501.5/sin(30°) = 750 J 16 D The total forward force produced, F = f +