2024 RI Prelim H2 Phy Paper 1 Answers

Raffles Institution Year 5-6 Physics Department 1 2024 H2 Physics Preliminary Examination Solution Paper 1 Qn Ans Solution 1 A units of 2 12 2 kg m s kg m smP − −− = = units of 3kg m −= units of ( ) −− − −  ==  12 22 3 kg m s m skg m nn nP units of −= 1m sv For equation to be homogeneous, units of    n P = units of v 2 2 1m s m snn −− = comparing indices of m: 12 1 2nn=  = 2 D 21 16.24 12.78 3.46 mmd d d= − = − = 21 0.03 0.02 0.05 mmd d d = + = + = 0.05100% 100% 1.4451 1.4%3.46 d d   =  = = 3 C From a-t graph: From = 0t to = 1tt , acceleration is constant which implies that the object’s velocity is increasing at a constant rate. From = 1tt to = 2tt , acceleration is decreasing which implies that the object’s velocity is increasing at a decreasing rate. From = 2tt to = 3tt , acceleration is zero which implies that the object’s velocity is constant. Since = dva dt , the gradient of the v-t graph, which gives acceleration, in Option C follows the description above. 4 D Option A: Possible, if lift is decelerating / decreasing in speed on its way up. Option B: Possible, if lift is moving upwards at a constant speed. Option C: Possible, if lift is accelerating / increasing in speed on its way up. Option D: Hence, all the options above are possible, depending on the lift’s acceleration.

Raffles Institution Year 5-6 Physics Department 2 5 C Applying Newton’s second law on the system of both crates, ( )( ) ( ) ( )( ) ( ) − = − + = +  −+== + , 2 100 2.0 3.0 9.81 2.0 3.0 100 2.0 3.0 9.81 10.19 m s2.0 3.0 net both bothF m a a a Applying Newton’s second law on the 2.0 kg crate, ( ) ( ) = − − = = ,2 2 100 2.0 9.81 2.0 10.19 60 N net kg kgF m a T T OR ( ) ( ) = −= = ,3 3 3.0 9.81 3.0 10.19 60 N net kg kgF m a T T 6 B Motorcycle travels in the same direction during the whole duration. Impulse or the change in momentum is the area under the force-time graph. ( )( ) ( )( ) ( )( )11 22 1 400 4.5 1.0 400 2.0 800 4.5 1.5 4.5 1.5 3.0 m s p Fdt v v v  − = − = − − =− = − =  7 D Since cube is floating, there is vertical equilibrium. ( ) ( ) ( )( ) 12 1 1 2 2 1 2 1 1 2 1 1 2 1 2 2 1 1 21 1 2 32 3 2 2 1 cube c U U W V g V g V V g V V V V V V V V VV V V       += + = + + = + − = − = = 8 A Work done by the force to extend the spring is given by the area under force-extension graph i.e. the area bo unded by the graph and the vertical axis of the graph given. This work done goes to increase the potential energy of the spring. The potential energy represented by area P is released upon the removal of the force. The potential energy represented by area Q is retained in the spring that is permanently stretched i.e. the energy used to separate the particles of the spring further apart.

Raffles Institution Year 5-6 Physics Department 3 9 D At constant speed, engine force = resistive force