2025 RI H2 Physics Prelims P3 Section A Answers

2025 Preliminary Examination H2 Physics Paper 3 Solutions 1 (a) (i) 2 2 1 1 2 2 2 21 1 2 1240 180 0 (9.81)2 3.497 s 1240 150 0 (9.81)2 4.284 s time 4.284 3.497 0.787 s s ut at t t t t tt =+ − = + = − = + = = − = − = (ii) The time taken will be smaller/shorter as the average speed will be higher and the distance travelled remains the same. (iii) 22 2 1 2 0 2(9.81)(240) 68.6 m s v u as v v − =+ =+ = (b) B1 − Correct shape between 0 to tA with zero gradient at t = 0 B1 − Constant gradient between t = tA and t = tB (no kink at tA) 2 (a) The principle of conservation of momentum states that when bodies in a system interact, the total momentum of the system remains constant, provided no net external force acts on it. (b) Taking velocity to the right as positive. By the principle of conservation of momentum for bodies A and B, ( )( ) ( )( ) ( ) ( )60 11 90 5 60 90 1.5 3.5 eq (1) if A A B B A A B B AB AB pp m u m u m V m V VV VV = + = + + − = + += s / m t / s 0 0 tB tA

2 © Raffles Institution Since collision is elastic, ( ) ( )11 5.0 16 eq (2) A B B A BA BA u u V V VV VV − = − − − = − −= eq (1) − 1.5  eq (2): 1 2.5 20.5 8.2 m s (shown) A A V V − =− =− (c) (i) ( ) ( ) ( )max max area under graph 1 0.40 60 1 8.22 2760 N F t m v u F F − = −   = − −  = (ii) By adding soft padding to the walls, so that the duration of collision can be lengthened (to reduce the maximum force experienced by the skaters when they hit the walls). 3 (a) (i) Elastic potential energy of bow = area under graph = 1 210 0.60 63.0 J2  = By conservation of energy, assuming that loss in elastic potential energy = gain in kinetic energy of the arrow 32 1 163.0 32 102 62.7 m s v v − − =    = (ii) The force required to hold the compound bow at maximum draw is lower. This allows easier aiming / more accurate aiming / archers to hold the bow steady for longer periods without tiring their muscles. Alternative answer: For the same draw, the compound bow has greater elastic potential energy and is able to shot an arrow with greater speed. (b) The elastic potential energy of the bow, designed to be transferred to the arrow, instead dissipates as vibrations and noise within the bow's components. This can cause broken strings and damage the bow. (c) (i) By conservation of energy, loss in kinetic energy = gain in gravitational potential energy ( ) 3 2 3 , , , 1 32 10 52 32 10 9.81 8.0 1.52 43.264 2.040 41.2 J K final K final K final E E E −−    − =    − −= =

3 © Raffles Institution [Turn over (ii) ( ) ( ) 2 2 2 1 2 18.0 1.5 52 sin15 9.81 2 4.905 13.459 6.5 0 0.6256 s or 2.118 s yyS u t at tt tt t =+ − =  + − − + = = 52 cos15 0.6256 31.4 m xxS u t= =   = 4 (a) (i) (ii) Star S and planet P attract each other by Newton’s law of gravitation. By Newton’s third law, the force of attraction on star S by planet P is equal in magnitude and opposite in direction to the force of attraction on P by S. (The magnitude of this gravitational force of attraction is 2 SPGM M r .) Since the only force acting on each star is this gravitational force of attraction, the gravitational force provides the centripetal force on each star. Hence the magnitude of the centripetal force on each star is the same. (iii) Since both S and P orbit at the same angular velocity , ( ) ( ) 22 SP S P 0.12 0.12 M r M r r r  = = (b) (i) 81 period 1500 days 22 1500 24 3600 4.85 10 rads T T  −− = ==  = (ii) 9S S 8 70 1.44 10 m 4.848 10 vr  −= = =   S O P

4 © Raffles Institution (iii) ( ) SS P P 9 10 S P S 0.12 0.12 11 1.444 10 9.333 1.35 10 m0.12 rr rr r r r =  = + = + =  =  (iv) Gravitational force between S and P provides the centripetal force for their circular motion. ( ) ( ) ( ) ( ) ( ) 2 S2 SP 2 2 S P S 2210 9 8 11 25 0.12 0.12 1.348 10 1.444 10 4.848 10 0.12 6.67 10 7.71 10 kg MM G Mr rr r r rM G M M   − − = + +=      =  = (c) Note Star S will appear to be moving in simple harmonic motion. 1500 t / days v / m s−1 0 t = 0 and 1500 days t = 750 days 1.44  109 m t = 375 days and 1125 days 1.44  109 m v = 0 v = 0 vmax

5 © Raffles Institution [Turn over 5 (a) The first law of thermodynamics states that the increase in internal energy of a system is equal to the sum of the heat supplied to the system and the work done on the system, and the internal energy of a system depends only on its state. (b) (i) (ii) Since the gas expands, the area under the graph gives the work done by the gas. (or the negative work done on the gas) Since internal energy is proportional to temperature, 0U= as there is no change in temperature, on on by 0 OR QW Q W Q W =+ =− =+ Hence, the amount of heat supplied to the gas is equal to the area under the curve AB. (c) For experiment 1, the volume remains constant and work done on gas = 0 Hence, increase in internal energy is equal to heat supplied. For experiment 2, the gas expands and work done on gas is negative. Hence, increase in internal energy in experiment 2 is less than that of experiment 1. (increase in internal energy in experiment 2 is equal to heat supplied minus work done by gas) 0 100 200 300 400 500 0.0 0.5 1.0 1.5 2.0 2.5 t / s v / m s−1 A P / kPa V /  10−3 m3

6 © Raffles Institution Since internal energy is proportional to temperature, increase in temperature of experiment 2 is smaller than that in experiment 1 and gas in experiment 1 will have a higher final temperature. OR For experiment 1, all the heat supplied was transferred into increasing the microscopic kinetic energy of the gas, whereas For experiment 2, the heat supplied was transferred into increasing the microscopic kinetic energy of the gas as well as to do work against external pressure (the piston). Since the amount of heat supplied is the same, the gain in kinetic energy for experiement 1 is higher and hence the gas in experiment 1 will have a higher final temperature. 6 (a) The electric potential at a point in an electric field is defined as the work done per unit positive charge by an external force in bringing a small test charge from infinity to that point. (b) (ii) The electric field strength is the negative of the gradient of the potential - distance graph in Fig. 6.2. ( ) 2 1 1200 100 5.0 10 0 26000 N C dVE dx − − =− −−=− − = Alternative answer: − − ==  = 2 1 1300 5.0 10 26000 N C VE d 0 V / V x / cm 100 –1200 5.0

7 © Raffles Institution [Turn over (iii) ( )( ) ( )( ) ( )( ) ( ) − − − − ==  = = =   =+ = +   = 19 10 2 27 22 10 2 10 1.60 10 26000 1.913 10 m s 131 1.66 10 Since the electric field is uniform, the acceleration is also uniform. Using kinematics equation, 2 0 2 1.913 10 5.0 10 2 1.913 10 5 EF qE ma qEa m v u as v ( ) − −  = 2 1 .0 10 43700 m s Alternative Answer: By principle of conservation of energy, gain in kinetic energy = loss of electric potential energy ( ) ( )( ) ( ) ( ) −− − − = −  =  − −  = 2 27 2 19 1 1 02 1 131 1.66 10 1.60 10 100 12002 43700 m s ABmv q V V v v (c) The positively charged xenon ions need to be neutralised so that they will not be attracted back towards the negatively charged plate B, which will reduce / cancel the thrust on the spacecraft created by the eject