YIJC 2024 JC2 PRELIM H2 Phy P3 Soln

1 ©YIJC 9749/02/YIJC/23 2024 JC2 Prelim Exam H2 Physics Paper 3 Solution 1 (a) (i) Because the speed at B is the same at C, the gain in KE is due to the loss of GPE from A to B only. Gain in KE in the end = Loss in GPE from A to B 21 2 =mv mgh == 2 2(9.81)(2.5)v gh = 7.0 m s-1 C1 A1 (ii) From A to B, the gravitational potential energy drop is converted to kinetic energy of the child. From B to C, the further loss in gravitational potential energy is not converted into kinetic energy but dissipated as work done against contact friction (or thermal energy and sound). Thus, the child’s speed is unchanged. B1 B1 (iii) ( ) ( )( ) ( ) 2 Work done against friction loss of KE from C to D 17.6 54 7.0 02 168 N 3sf CD Fs Fs KE KE F F = = =− =− = OR ( ) ( )( ) 22 2 2 2 0 7.0 2 7.6 average accel., 3.22 m s Average frictional force 52 3.22 168 N ave ave ave ave v u a s a a ma − =+ =+ =− == = M1 A1 (M1) (A1) (b) (i) When the carousel rotates, the chair and its occupant will travel in a straight-line if not for the cable attached to it. A horizontal (centripetal) force is required in order to pull the man towards the centre. This is provided by the horizontal components of the tension on the cable holding the chair. Thus, the cable has to tilt to provide for the needed centripetal force. M1 A1 (ii) All forces must be indicated and labelled in words (not in letters or symbols). Tension on cable and gravitational force (or weight) on person and seat. Deduct mark if FBD includes centripetal or centrifugal force, normal force or any irrelevant forces. B1 weight tension

2 ©YIJC 9749/02/YIJC/23 (iii) Along the y-direction: cosT mg = (balanced) Along the x-direction: 2sinT mr= (unbalanced due to centripetal rotation) ( ) 22 2 1 4.7 8.4sin35 9.518 m tan 9.518tan35 9.81 0.85 rad s r mr r mg g    − = + = == = = C1 C1 A1 2 (a) The first law of thermodynamics states that the increase in internal energy of a system is the sum of the heat (thermal energy) supplied to the system and the work done on the system. B1 (b) (i) 1. ( ) ( ) ( ) 562.79 10 1125 950 10 48.825 48.8 J 3sf onW P V − =−  =−  −   =− =− Max 1m for positive answers. C1 A1 2. ( )( ) ( )( ) ( ) 5 6 5 6 0, since no volume change for Q R 3 2 3 2.10 10 1125 10 2.79 10 1125 102 116.44 116 J 3sf in on on in loss in U Q W W Q U pV QQ −−  = + =→ = =  =   −   =− =− = Max 2m for negative answers, but do not penalize if same type of mistake in 1. C1 C1 A1 (ii) Since there is no heat loss/gained, and there is (positive) work done on the gas through the compression (decrease in volume), the in