2023 RI Prelim P2 AS

1 © Raffles Institution [Turn over 2023 RI Preliminary Examinations H1 Physics Paper 2 Solutions 1 (a) 2 2 2 2 2 2 0.8504 4 9.858 9.86 m s 36.9 20 LT g Lg T                       (b) 2 2 0.2 0.00129.858 36.9 0.850 0.118 0.1 m s g T L g T L g g                     g = (9.9  0.1) m s2 2 (a) To hit the coconut horizontally, vertical component of the velocity = 0. Initial vertical velocity = 20 sin . Using v2 = u2 + 2as, and setting v to 0 m s1 and s = 18.0  2.2 = 15.8 m: 0 = (20sin )2 – 2(9.81)(15.8) sin = 0.880 =>  = 61.7 (b) By v = u + at, 0 = 20 (0.880) – 9.81t t = 1.79 s (c) Horizontal displacement = 20  cos(61.7)  1.79 = 17.0 m (d) The downward acceleration will be larger and hence the same initial vertical velocity will be reduced to zero over a shorter vertical displacement. Hence,  has to be larger so that the initial vertical velocity is larger. Since initial horizontal velocity is smaller and it will decrease due to the air resistance in the horizontal direction, the horizontal displacement will be lower. 3 (a) product of mass and (linear) velocity (b) (i) Since collision is elastic, total kinetic energy is the same before and after collision. 1 2  2 12 2m u  2 1 2mu    2 A 12 2m v    2 B 2 2 2 A B3 2 shown mv u v v  (ii) 1. Let the scattering angle of B be  respectively. By principle of conservation of linear momentum, Considering horizontal motion, taking right as positive,    A2 2 cos 90m u mu mv      

2 © Raffles Institution  A2 cos ... 1u v  OR A,x A cos (1') 2 uv v    Considering vertical motion, taking up as positive, A B0 2 sin sinmv mv     B A 2 sin ... 2v v  OR B A,y A sin (2') 2 vv v    (1)2  (2)2,           2 2 2 2 B A A 2 2 2 B A 2 cos 2 sin 4 ... 3 u v v v u v v       OR By Pythagorean theorem        2 2 2 A A, A, 22 2 B A 2 2 2 B A from 1' and 2'2 2 4 ... 3 x yv v v vuv u v v               From equation in (b)(i),   2 2 2 2 2 2 B A B A 33 2 ... 4 2 u vu v v v     Substituting (4) into (3),   2 2 2 2 B B 2 2 B 5 5 5 1 B 34 2 6 2 5 5 3.5 10 4.5185 10 4.5 10 m s3 3 u vu v u v v u                   (b) (ii) 2. where u, vB and v are the magnitudes of the vectors. By Pythagorean theorem,   2 2 B 2 2 5 5 5 1 5 3 8 8 3.5 10 5.7155 10 5.7 10 m s3 3 v u v u u u                   B 1 1 B tan 5tan tan 52 3 v u v u          52 below horizontal vB u  v vB u v Alternatively, 

3 © Raffles Institution [Turn over    27 5 22 1 1.7 10 5.7155 10 9.7 10 kg