2018 RI H2 Physics P1 solutions

© Raffles Institution 9749/01   2018 Raffles Institution Preliminary Examinations – H2 Physics Paper 1 Answers 1 C 6 B 11 C 16 D 21 B 26 D 2 D 7 D 12 B 17 C 22 B 27 B 3 C 8 C 13 C 18 D 23 C 28 A 4 A 9 A 14 A 19 D 24 B 29 D 5 D 10 A 15 B 20 A 25 C 30 A Paper 1 Suggested Solutions 1 C ker ker 70 20 50 g 112 g Liq Total Bea Liq Total Bea mm m mm m           3 20 . 6 50 10.0 20 . 6 5.0 0.5 g cm (1 s.f.)50 10.0 Liq Liq Liq Liq Liq Liq m V mV mV             2 D After first bounce, 22 2 00 0 79 3 1 1 31 16 16 4 2 2 4 where is the speed just before the ball hits the ground left initial initialKE KE KE mv m v v                 If the height of the second triangle is now 0 3 4 v , time duration of the second triangle is 3 0.754 tt (due to similar triangles). Since air resistance is negligible, the time taken by the ball to move up after the first bounce is the same as the time taken to move down to the ground again. OR Since air resistance is negligible, the gradient of the velocit y-time graphs will all be the same with the value of g. 0 0 22 2 00 10 1 0 79 3 1 1 31 16 16 4 2 2 4 3speed after bounce, 4 31 31 3 0.7544 4 left initial initial v gv g tt KE KE KE mv m v vv vtv g tt tgg g                               

2 © Raffles Institution 9749/01   3 C Horizontal motion: 4.0 x x sv t st vv   Vertical motion: 2 2 1 1 2 14 . 01.5 1.1 2 14 m s ysg t g v v      4 A This is a completely inelastic collision. Hence, total mechanical energy is not conserved. 5 D By Newton’s second law, net force, 3 5 41 54 5 FFm m a FaF mm       Consider the forces acting on Y: 355 43 55 5 XY XY FFFm m FF FFm m     6 B There cannot be any normal reaction when the sphere is in equil ibrium because it will be the only force with a horizontal component that is not balanced. 7 D Taking moments about the point on the base of the block through which R acts,       225.0cos30 5.0sin30 10 2 10 1.0 8.0 10 0.0473 m 4.7 cm x x       8 C Work done to stretch it 10 cm:  2 1 1 0.10 4.02 800 N m k k    Work done to stretch it 20 cm:    21 800 0.20 162  Additional work required = 16 – 4 = 12 J 9 A 2 2 1 cos 40 --- (1) sin40 --- (2)1.2sin40 (2) tan 40 2.5 m s(1) 1.2 sin 40 Tm g mvT v vg       

3 © Raffles Institution 9749/01   10 A The equations for potential energy and kinetic energy are given by : p GMmE R and 2 k GMmE R As R increases, pE increases (becomes less negative) and kE decreases. 11 C 2 3 3 2 , , 12 0 2 4and since = 3 42 83 3 For a body to escape a planet’s gravitational influence from its surface: escape escape escape escape X xx xx escape Y y Y YY GMm GMmv vRR MR GR GRv R vR v R R vR R              