2020 RI Prelims H2 Phy Paper 1 Soln

© Raffles Institution 9749/01   2020 Raffles Institution Preliminary Examinations – H2 Physics Paper 1 Suggested Solutions 1 B    12 12 22 33 1 1222 2 2 33 122 J s N m kgunits of m s kg m s m s kg m s m kg m s mm G c               2 C  1212 2 3 4 14 3 2 2Pa b c d a b c d   32 42 134% 2% 6% 2 5%42 1% 2% 9% 10% 22% Pab c d Pa b cd         3 B The gradient of the displacement-time graph gives the velocity. Car M is accelerating as it has an increasing velocity (since gradient increases). Car N is not accelerating as it has a const ant velocity (since gradient is constant). Statement is incorrect. 4 D p mu Since horizontal velocity u remains unchanged, cosuv  . horizontal direction: Since horizontal velocity remains unchanged, initial momentum = final momentum change in momentum = 0 vertical direction: initial momentum = 0 final momentum sinmv  (downwards) change in momentum sin 0 sin tan tancos umv m mu p       (downwards) 5 D Frictional force on the boxes by the belt, in the forward direction, is needed to change their horizontal state of motion from rest to the final speed of 0.80 m s–1. By Newton’s third law, these boxes exert an equal and opposite force (in the backward direction) on the belt which slows it down. An external force that is equal and opposite (in the forward direction) to this force needs to be applied to the belt to keep it moving at constant speed. horizontal force applied,   0.80 2.5 2.0 NdmFv dt 

2 © Raffles Institution 9749/01   6 C Since the pyramid is floating in the water, this means the upthrust on the pyramid is equal and opposite to the weight of the pyrami d. (Hence option B is correct since the pyramid remains floating even when a liquid of higher density is added.) Upthrust is given by the weight of the water displaced by the submerged part of the pyramid. When the pyramid is inverted and floating, it w ill sink deeper in order to displace the same volume of water to achieve the same upthrust. This means the pressure at the submerged apex is higher than that at the submerged base when the pyramid is upright. Hence the pressure difference between the surface of the water and the lowest point of the pyramid is greater when it is inverted. (This explains why option C is incorrect and why option A is correct.) Option D is correct as proven below: ('w' for water, 'o' for object) constant for water and object of the same density ww oo o w wo Vg Vg V V      7 C decrease in G.P.E. increase in K.E.  energy lost to surroundings energy lost to surroundings  decrease in G.P.E. – increase in K.E.     22 22 11 22 110.80 9.81 10 4.0 1.222 72.7 73 J mg h mv mu        8 C Since body is moving up the track with constant velocity, ther