2023 RI H2 Physics Prelims P1 Answers

1 © Raffles Institution [Turn over 2023 Preliminary Examinations H2 Physics Paper 1 Solutions 1 B When  is zero, F2 = F (largest), F1 = 0 (smallest). When  is 90, F2 = 0 (smallest), F1 = F (largest). 2 C 22 2yyv u gy=− At the highest point, vy = 0. Hence, 202 yu gy=− 2yu gy= Since the maximum height reached is the same for all three paths, uy is the same too. Option A: Sy = uyt + at2, since all three paths end on ground level, time of flight is the same. Path Z has the longest range. Sx = uxt, path Z has the highest horizontal component and the largest initial speed 22() xyu u u=+ . Option B: Sy = uyt + at2, since all three paths end on ground level, time of flight is the same. Option D: Path X has the shortest range. Sx = uxt, path X has the lowest horizontal component. 3 A By Newton’s second law, taking the direction of F1 as positive, 12 12 netF ma F F ma FFa m = −= −= Since object was at rest, F2 is initially zero ( ) 2 2Fv and acceleration is maximum initially. Acceleration then decreases as F2 increases with increasing speed as object accelerates. Acceleration becomes zero eventually when F1 = F2 and object travels at constant (maximum) speed henceforth. 4 A At equilibrium, the lines of action of three forces must meet at a point. The arrows in the vector triangle must form a closed loop. 5 C ( )( )( ) ( )( )( ) 3 3 0.120 750 9.81 0.060 1000 9.81 1.4715 10 1.47 10 Pa o o w wP h g h g=+ =+ = = 6 D Work done is area under the force extension graph. This would be the work done to stretch the spring from x1 to x2. ceiling wall string Q bar P weight tension Force of wall on bar weight tension Force of wall on bar

2 © Raffles Institution 7 C ( ) ( ) ( ) P PQ Q PQ total gain in energy total loss in energ y gain in GPE gain in KE loss in GPE W.D. b y friction 1.5sin30 0 gain in KE 1.5 0 1.5PQm g m g f = + = + − + = − − ( ) ( ) ( ) ( )( )( ) ( )( ) ( )( )( ) PQgain in KE 1.5 1.5 1.5sin30 4.0 9.81 1.5 2.5 1.5 3.0 9.81 1.5sin30 33.0375 33 J QPm g f m g= − −  = − −  = = 8 D 2 2 2 1.2 0.040 9.81 0.30 1 0.238862 mvT mg r mv mv += +  = = By conservation of energy, ( ) 22 2 2 11 22 10.23886 0.040 9.81 2 0.30 0.040 2 23.715 mv mgh mu u u += +    =   = 2 223.715 79.1 m s0.30 ua r −= = = 9 B Graph G1 for path from S to Q means that the two stars have different masses (since there is no point where resultant force is zero). Graph G 3 shows that star X has a smaller mass than star Y (the values of g at start is lesser than at the end). The point of g = 0 is closer to X. Since the spacecraft is accelerating, it is possible for the time duration from P to point of g = 0 to be approximately equal to the time duration from point of g = 0 to R. 10 C Given Q = – 2.0 J kg−1 R Q R PQ 1 PR 4 1 (2.0) 0.5J4    == = − = − Work d