AJC_H2_PHY_P1_Soln

1 9749/01/AJC2017/Prelim [Turn Over 2017 AJC JC2 H2 Physics Prelim Solutions Paper 1 (30 marks) Answer 1 2 3 4 5 6 7 8 9 10 D A B B C D B D C D 11 12 13 14 15 16 17 18 19 20 A C B D A C C C A C 21 22 23 24 25 26 27 28 29 30 A D A C A D B B B A No Answer & Solution 1 Ans: D 2000 pV = 2000 x 10-12 V = 2 x 10-9 V = 2 x 10-15 x 106 V = 2 x 10-15 x M V p.d. = work done per unit charge, hence V is equivalent to J C-1 2 Ans : A For the projectile motion, horizontal speed is constant, v. Vertical speed is zero initially and acceleration is g, downwards. Let the height of the table be h. Using s = ut + ½ at2 h = 0 + ½ gT2 T = g h2 T is independent of mass. Range, D = vT. Since both v and T are independent of mass, D will remain the same. 3 Ans : B The scale reads the contact force between the sack and scale. Since the scale reading is larger than the mass, the upward contact force must be larger than the downward weight. The sack experiences a net force and acceleration upwards. (12 – 10) x 9.81 = 10 x acceleration acceleration = 1.962 ≈ 2.0 m s -2. 4 Ans : B For elastic collision, total KE is conserved OR relative speed of approach equals to relative speed of separation. Total KE before impact = ½ mv 2 + ½ mv2 = mv2 Thus, total KE after impact = mv2. Answer C is wrong because during collision, KE is not conserved. It is converted to elastic PE then back to KE. Answer D is wrong because total momentum before impact = mv + m(-v) = 0

2 9646/01/AJC2016/Prelim Answer A is wrong because when the spheres stick together, it is already a perfectly inelastic collision. Also note that relative speed of approach = u 1 – u2 = v – (-v) = 2v (non-zero). If the spheres stick together, they will share the same speed and relative speed of separation will be zero. 5 Ans : C Work done is minimum if the force applied is just enough to overcome the barrel’s weight on its way up the plank, with no change in speed. The barrel gains GPE without any gain in KE. Minimum work = Gain in GPE = mgh = 50 x 9.81 x 1.6 = 784.8 J ≈ 780 J Alternative method : Work done = Force x distance = mg sin θ x d = 50 x 9.81 x (1.6/3.4) x 3.4 = 784. 8 J 6 Ans: D Upthrust = weight of air displaced (smallest force as density of air is very small). Viscous drag is proportional to speed which increases as ball falls. At terminal speed, downward weight of ball is a constant which is balanced by upward drag and upthrust. Hence, weight > viscous drag > upthrust. 7 Ans: B For equilibrium, all three forces must meet at a common point, which is a point below trapdoor along the line of action of W. Hence option C and D are wrong as direction of H is wrong. Tension must act away from the door, hen