2025 RI H2 Physics Prelims P1 Answers

2025 Preliminary Examination H2 Physics Solution Paper 1 1 A The vector sum of the two components is not R. 2 A The d will increase initially as Car Y moving slower and Car X. After that, Car Y will move with speed greater than Car X and the relative speed is increasing . Hence gradient is getter steeper. 3 C Net force up the slope ( ) ( )630 15 30 sin30 15 30ga− + = + 29.095 m sa −= For the 15 kg crate, net force up the slope 15 sin30 15R g a− = ( )15 15 sin30 15 9.095 9.81sin30 210 NR a g= + = +  = 4 B Energy stored during the process is the area under the force-extension graph 5 A For translational equilibrium, the net force has to be zero . Since the forces act in the opposite direction on the same object and have the same magnitude, the net force acting on the object is zero. Option B: Even though the two forces act in opposite directions, they are spaced apart from each other which results in a torque or rotation. Hence, the object is not in equilibrium. Option C: The direction of the torque of a couple depends on the direction of the two forces and is independent of the position of the pivot. Option D: The pair of forces act on the same object, so they are not an action-reaction pair of forces. 6 D dUF dx=− Magnitude of force = gradient of graph = 1.400 0.3125 6.0 N0.380 0.200 − =− 7 D Drag force, D = 250 N Resolving forces along the slope 1.0sin 250 900 9.81 985.75 N12F D mg  = + = +   =  985.75 24 23.7 kWP Fv= =  = Alternative method: P = rate of work done against drag + rate of increase in GPE = (250  24) + 900  9.81  2 = 23.7 kW ( )( ) ( ) ( ) ( ) ( )( ) 1 2 2 1 1 2 2 1 1 2 2 1 2 1 2 1 2 T T e e T T x x x x T T x x = + − = + − − −  = + − N F W D

2 © Raffles Institution 8 B At the top of the circle, 2 22 (0.10)(6.0) (0.10)(9.81) 6.2 N0.50 mvT mg r mvT mg r += = − = − = 9 B ( )   =− =−     =−      =− 3 2 1 volume density 14 3 4 3 MGGrr Gr r Gr 10 B ( ) ( ) ( ) 32 32 32 22 2 3 3 22 2 3 3 33 3 3 3 in the same time , 3 3 90 467.7 467.7 360 107.7 Y Y Y Y Y Y Y Y Y X X X X X X X X X Y Y X XY X YY XX Y Mm MMG m R G GR R R Mm MMG m R G GR R R MG RR M RG R t            = → = → = = → = → = = = = ==  =  =  − =  Alternative method: ( ) ( ) 23 32 32 3X Y Tr T k r T k r  = = 2 33 5.196 X Y XY T T TT  =  = X takes ¼ period to move thought 90. 1 1.2994 XYTT= 0.299 360 108Y =  =  11 C ( ) ( ) 2 2 2 0 22 22 0 2 2 2 2 1 2 1 2 1 3.0 2.1 m KE m x x PE m x xxKE PE x xx x   =− = − == −= = 12 D The child on a swing is pushed at regular intervals matching the swing's natural frequency, and the amplitude of the swing increases each time. This is an example of resonance where the driving frequency matches the natural frequency of the system.

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