2024 RI Prelim H2 Phy Paper 2 Answers

Raffles Institution Year 5-6 Physics Department 1 2024 H2 Physics Preliminary Examination Solution and Comments Paper 2 1 (a) (i) Take direction up the slope as positive. ( ) 22 0 2 0 0 2 0 7.0 2 9.81sin30 4.9949 4.99 m v u as s s =+ = + −  == OR By the principle of conservation of energy, increase in G.P.E. = decrease in K.E. ( ) 2 2 2 0 2 1 0, where is the max. vertical height fr om ground2 2 7.0 2(9.81) sin30 7.0 sin302(9.81) 4.9949 4.99 m mg h mv h vh g hs  = −  = = =  =   == (ii) *Both graphs aligned at the same total energy and 0s and clearly labelled. 1.  ( ) ( ) 22 2 11 2 sin22 1 sin2 K K E mv m u g s E mu mg s = = + −  = −  Graph of EK-s is a straight line with negative gradient and vertical intercept 21 2 mu . 2. ( ) ( ) ( ) sin sin P P E mg h mg s E mg s =  =  = Graph of EP-s is a straight line through the origin with positive gradient. energy EP EK s / m 0 0

Raffles Institution Year 5-6 Physics Department 2 (b) (i) ( ) 22 22 1 2 4.014.0 2 9.81sin30 sin30 10.841 10.8 m s (shown) v u as v v − =+ = + −    == OR decrease in K.E. = increase in G.P.E. ( ) ( ) ( )( ) 22 2 2 2 1 11 22 2 14.0 2 9.81 4.0 10.841 10.8 m s (shown) mu mv mg h v u g h v − − =  = −  = − == (ii) Take directions to the right and upwards as positive. Time of flight after the ball leaves the top of the slope to the ground: ( ) ( ) 21 2 214.0 10.8sin30 9.81 2 1.6080 s or 0.50713 s (NA) y y ys u t a t tt t =+ − =  + − =− Horizontal distance travelled: ( )( ) 21 2 10.8cos30 1.6080 0 15.040 15.0 m x x xs u t a t=+ =  + == *State both values of t and reject the negative one.

Raffles Institution Year 5-6 Physics Department 3 2 (a) The initial total momentum of both balls is not zero. Since there is no net external force acting on the balls as a system, by the principle of conservation of momentum, the total momentum of both balls must remain unchanged and cannot be zero. Hence, the balls could not be stationary at the same time. (b) By the principle of conservation of momentum, 4.0 ( 1.0) 3.0 ----- (1) A A B B A A B B A B A B AB AB m u m u m v m v u u v v vv vv + = + + = + + − = + += Since collision is elastic, ( 1.0) 4.0 5.0 ----- (2) B A A B AB AB u u v v vv vv − = − − − = − − =− 1 (1) (2) 2 3.0 ( 5.0) 4.0 m s (shown) B B v v − − = − − = (c) By Newton’s second law, the average force on ball B by ball A is ( )( ) , 0.50 4.0 1.0 0.25 10 N B net B pF t =  −− = = By Newton’s third law, the average force on ball A by ball B has the same magnitude of 10 N. OR From equation (1) or (2) in part (b), vA = 3.0 − vB = 3.0 − 4.0 = −1.0 m s−1. By Newton’s second law, the average force on ball A by ball B is ( )( ) , 0.50 1.0 4.0 0.25 10 N A net A pF t =  −− = =

Raffles Institution Year 5-6 Physics Department 4 (d)