NJC_2025_H2_Physics_Prelim_P2_Ans

2025 H2 Physics P2 Suggested Solution 1(a)(i) From energy conservation, gain in g.p.e. = loss in k.e. 𝑚𝑔(10.0) = 1 2 𝑚𝑢2 − 1 2 𝑚(5.00)2 𝑢 = 14.87 ≈ 14.9 m s-1. M1 (a)(ii) 14.873 𝑐𝑜𝑠 𝑐𝑜𝑠 𝜃 = 5.00 𝜃 = 70.4° Or Vertical motion: 𝑢𝑦 = 14.9 𝑠𝑖𝑛 𝑠𝑖𝑛 𝑞 ↑, 𝑎𝑦 = 9.81 ↓, 𝑠𝑦 = 10.0 ↑, 𝑣𝑦 = 0 From 𝑣2 = 𝑢2 + 2𝑎𝑠 taking ↑ +ve: 0 = 𝑢𝑦2 + 2(−9.81)(10.0) 𝑢𝑦 = √196.2 = 14.873 𝑠𝑖𝑛 𝑠𝑖𝑛 𝜃 𝜃 = 70.4° M1 A1 M1 A1 (a)(iii) Consider vertical motion: 𝑢𝑦 = 14.873 𝑠𝑖𝑛 𝑠𝑖𝑛 70.4° ↑, 𝑎𝑦 = 9.81 ↓, 𝑠𝑦 = 0, 𝑡 =? From 𝑠𝑦 = 𝑢𝑦𝑡 + 1 2 𝑎𝑦𝑡2 and taking ↑+ve: 0 = (14.873 𝑠𝑖𝑛 𝑠𝑖𝑛 70.4° )𝑡 + 1 2 (−9.81)𝑡2 0 = {(14.873 𝑠𝑖𝑛 𝑠𝑖𝑛 70.4° ) + 1 2 (−9.81)𝑡} (𝑡) 𝑡 = 0 (reject) or 𝑡 = 2(14.873𝑠𝑖𝑛𝑠𝑖𝑛 70.4° ) 9.81 = 2.86 s or 0 = (14.873 𝑠𝑖𝑛 𝑠𝑖𝑛 70.4° ) − 9.81𝑡′ 𝑡 = 2𝑡′ = 2.86 s M1 A1 (a)(iv) Rate of change of momentum = resultant force = weight = 𝑚𝑔 = 8.00 x 10-3 x 9.81 = 0.0785 N A1 (b)

2(a) No resultant external force acting on the two colliding / interacting objects. B1 (b)(i) Force on B = rate of change of momentum of B = (24 − 20) × 103 3.0 − 1.5 = 2667 ≈ 2700 N M1 A1 (b)(ii) Gradient is equal to the rate of change of momentum of the truck and this is also equal to the impact force exerted on the trucks according to Newton’s 2nd law Since the impact forces between the two trucks are an action-reaction pair and and are of equal magnitude but acts in opposite directions, the gradients have the same magnitude but opposite signs. B1 B1 (b)(iii) Impulse acting on each truck (during the collision) is equal to the change in momentum of the truck. Since the impulse on the two trucks are equal and opposite, it follows that the gain in momentum of one truck must be equal to the lost in momentum of the other truck. Hence total momentum is conserved. B1 B1 (b)(iv) Total kinetic energy of the two trucks before collision = (20,000)2 2×4000 + (16,000)2 2×2000 = 114 kJ Total kinetic energy after collision = (24,000)2 2×4000 + (12,000)2 2×2000 = 108 kJ [C1- for calculation of KE before and after collision] Change in kinetic energy = − 6000 J (Perfectly) Inelastic collision A1 B1

3(a) 𝑐𝑜𝑠 𝑐𝑜𝑠 𝜃 = 0.60−0.08 0.60 𝜃 = 29.93 ≈ 30° M1 (b) B1 (c) Using principle of moments and taking moments about P: 700 × 0.60 𝑠𝑖𝑛 𝑠𝑖𝑛 30 = 𝐹(1.20 − 0.08) 𝐹 = 187.5 ≈ 188 N = 190 N M1 (d) Sum forces vertically: 𝑅𝑦 = 700 N Sum forces horizontally: 𝑅𝑥 = 188 N 𝑅 = √7002 + 1902 = 725 = 730 N (Accept between 725 to 730) C1 C1 A1

4(i) They are an action-reaction pair or they are equal in magnitude and opposite in direction. B1 (ii) 𝑀𝑟1 ( 2𝜋 𝑇 ) 2 = 2𝑀𝑟2 ( 2𝜋 𝑇 ) 2 𝑟1 𝑟2 = 2𝑀 𝑀 = 2 B1 (iii) Since 𝑟1 𝑟2 = 2 and 𝑟1 + 𝑟2 = 3.0 × 1012 𝑟1 = 2 1+2 × 3.0 × 1012 = 2.0 × 1012 m C1 A1 (iv) Consider forces acting on M. From Newton’s 2