2024 ASRJC JC2 H1 Physics Prelim P1 MS

1 8867/01/ASRJC/2024Prelim [Turn Over Anderson Serangoon Junior College 2024 JC2 H1 Physics Preliminary Examination Mark Scheme Paper 1 (30 marks) 1 2 3 4 5 6 7 8 9 10 D A B D C B C A B D 11 12 13 14 15 16 17 18 19 20 C A B A A C C A C A 21 22 23 24 25 26 27 28 29 30 D C A D A B D D B D 1 D 2 2 3 intensity units of I I      E At kg ms m m s kg s 2 A Z = X + (–Y) as shown by the vector triangle below. This is equivalent to option A. 3 B Velocity is a vector; distance is a scalar 4 D Taking downwards as positive, 21 2 21 210.0 0 (9.81) 1.43 s s ut at t t      2 2 2 1 2 0 2(9.81 )(10.0) 14.0 m s v u as v v       X Y Z

2 8867/01/ASRJC/2024PRELIM 5 C Constant speed up to t1 means s increases at a constant rate, hence a straight line with positive gradient. As speed decreases at constant rate, distance travelled increases at a decreasing rate (decreasing gradient) and reaches a constant (zero gradient) when speed is zero. 6 B The vertical component of acceleration is the acceleration of free fall which is a constant in the absence of air resistance. 7 C Action-reaction must be of the same type. Weight is the gravitational force on the man due to the Earth. 8 A Consider the 2 masses as a system Weight of 2 kg mass  friction between 8 kg mass and plane = (mass of 2 masses) × a a =  . . . . . 2 0 9 81 5 0 8 0 2 0   = 1.5 m s-2 9 B Fnet = change in momentum / change in time = 0.2 (10.0 – (–20.0)) / 0.10 = 60 N 10 D Consider vertical equilibrium when there were three springs and weight W 3kx = W k = 3 W x Consider vertical equilibrium when there were two springs and weight 2W 2kx’ = 2W kx' = W x’ = 3 ( )3 W W xWk x   11 C Two equal and opposite parallel forces whose lines of action do not coincide. 12 A Weight must be vertical. The three forces form a closed triangle. 13 B total work done by man = Work done against friction + gain in GPE, since KE constant

3 8867/01/ASRJC/2024Prelim [Turn Over Work done against friction = total work done by man – gain in GPE = 1500 – mgh = 1500 – 5.0 x 9.81 x 12 = 911. 4 J Average friction = work done against friction / distance travelled along plane = 911.4 / ( 12 sin 30  ) = 37.9 ≈ 38 N 14 A By Conservation of Energy, loss in KE = Gain in GPE.  ∆E = ∆mgh = mg∆h Hence E varies linearly with height, i.e. a straight line. Since y is vertical displacement, at maximum height (largest y value) E = 0. Hence the answer is A. 15 A The centripetal acceleration at Singapore is:   2 2 2 6 2 2 2 2 6.4 10 3.4 1024 60 60a r r T           