2022 RI Prelims H2 Phy Paper 1 Soln

Raffles Institution Year 5-6 Physics Department 1 2022 H2 Physics Preliminary Examination Solution Paper 1 Qn Ans Solution 1 D 4.5 335.61 15.1 20 (1 s.f.)100v    1(340 20) m svv    v is rounded off to the nearest tens place. 2 A 4 4 4 4 2 34 24 unit of unit of unit of unit of unit of kg m s m kg s Ks m K P AeT Wt AeT Fd tAeT mad tAeT              3 C Take direction to the right and upwards as positive. 3.4cos30 5.0cos60 xx x x su t st u     13.4cos305.0sin60 9.81 7.2240 m s5.0cos60yyyvua t          1 tan , where is the angle that the final velocity makes with the horizontal 7.2240tan 70.9115.0cos60 y x v v        70.911 30 40.911 41       OR     22 12 2 11 2 5.0sin60 2 9.81 3.4sin30 7.2183 m s or 7.2183 m s (reject) yy y y y vu a s v         1 tan , where is the angle that the final velocity makes with the horizontal 7.2183tan 70.8975.0cos60 y x v v        70.897 30 40.897 41      

Raffles Institution Year 5-6 Physics Department 2 4 A Based on the given graph, the direction upwards is positive. At time t, the ball is at its maximum height, and its displacement is +S1 (or +S2). At time 2t, the ball is back at its initial point and is accelerating as it moves downwards (negative velocity). Hence its displacement is zero (S1 = S2) and the gradient of the displacement-time graph is negative. At time 3t, the ball hits the ground since its velocity is –2u and acceleration is constant at –g. Its displacement is –S3 from the initial point. From 2t to 3t, since the ball is accelerating as it moves downwards, its speed increases. Hence the gradient of the displacement-t ime graph should be negative with increasing magnitude. 5 C When P starts to slide: Consider block P: PPfm a Consider block Q: QQfm a PP QQma ma Since PQmm , PQaa  Friction on Block P causes it to decelerate as it moves to the left. Friction on Block Q causes it to accelerate as it moves to the left. 6 A from the time the column is dropped to when it just reaches the surface of the soil, increase in K.E. = decrease in G.P.E 21 02 2 Mv MgH vg H   from the time the column enters the soil to the time when it comes to a stop, by Newton’s second law and taking direction downwards as positive, R dpF dt 2 where is the average resistive force 0 2 21 fi pMg f f t ppMg f t MvMg f t Mg HMv HfM g M g M gtt g t          P Q NP mPg f f NQ mQg direction of motion

Raffles Institution Year 5-6 Physics Department 3 7 B When a quarter of the cube is removed, the C.G. changes to the new geometrical centre and the weight decreases. Upthrust remains unchanged initially. Hence UW and there is a resultant force upwards. Since the upthrust and the weigh