2023 RI H2 Physics Prelims P2 Answers

1 © Raffles Institution [Turn over 2023 RI Preliminary Examinations H2 Physics Paper 2 Solutions 1 (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. 2 (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 2mu + 2 1 2mu = ( ) 2 A 12 2mv + ( ) 2 B 2 2 2 AB3 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, ( ) ( ) ( ) A A 2 2 cos 90 2 cos ... 1 m u mu mv uv   − = =  = Considering vertical motion, taking up as positive, ( ) AB BA 0 2 sin sin 2 sin ... 2 mv mv vv   =− = (1)2 + (2)2, ( ) ( ) ( ) ( ) ( ) 2 2 2 2 B A A 2 2 2 BA 2 cos 2 sin 4 ... 3 u v v v u v v + = + +=

2 © Raffles Institution [Turn over From equation in (b)(i), ( ) 2 2 2 AB 22 2 B A 32 3 ... 42 u v v uvv =+ −= Substituting (4) into (3), ( ) 2 2 2 BA 22 B 22 B 22 B B 5 5 5 1 4 34 2 62 35 5 3 5 3.5 10 4.5185 10 4.5 10 m s3 u v v uv uv vu vu − +=  −=  =− = = =  =  =  Alternatively, Let the horizontal and vertical components of the velocity of particle A after collision be vA,x and vA,y respectively, and the vertical (downward) velocity of particle B after collision be vB. By principle of conservation of linear momentum, Considering horizontal motion, taking right as positive, ( ) ( ) ( ) A, B, A, A, 2 2 0 2 ... *2 xx x x m u mu mv v uv uv − = = = = Considering vertical motion, taking up as positive, ( ) A, B B A, B A, 02 2 ... ^2 y y y mv mv vv vv =− = = By Pythagorean theorem, ( ) ( )( ) 2 2 2 A A, A, 22 2 B A 2 2 2 BA from * and ^22 4 xyv v v vuv u v v =+ =+   += which is identical to equation (3). (b) (ii) 2. where u, vB and v are the magnitudes of the vectors. vB u  v vB u v Alternatively, 

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