(SAJC) 2024 H2 Phy Prelim P2 soln_for sharing

1 SAJC 2024 Prelims / 9749 [Turn Over JC2 Prelim (H2 Physics) Paper 2 Solutions 1(a)(i) a = dv/dt v = + a dt = area enclosed by a-t graph  v = area enclosed by F-t graph / m (as F = ma) = [ ½ (1980)(10) ] /1100 = 9.0 m s-1 v = vi - v = 18.0 – 9.0 = 9.0 m s-1 [1 – for p or v using area under graph] [1] (ii) The net force is the braking force [1] (b) A downward slope with speed starting from 18 m s-1 and end at 0 Gradient of curve starts from 0 at t=0 and increases to a max gradient at 10 s. Gradient of curve decreases gradually from max at 10 s to 0 at t=20 s [1] [1] [1] (c) Gradient of the curve decreases non-linearly from a max at t= 0 to 0 at t=20 s. Distance is 0 at t=0 and max at t=20 [1] [1] 2(a) sum / total momentum before (interaction) = sum / total momentum after or sum / total momentum (of a system of interacting objects) is constant [1] if no resultant force / for an isolated system [1] 9

2 SAJC 2024 Prelims / 9749 [Turn Over (b)(i) 3m x 4 = m x v sinϴ (v sinϴ = 12) 2m x 6 = m x v cos ϴ (v cos ϴ = 12) therefore, sin ϴ = cos ϴ or tan ϴ = 1 ϴ = 45° or a method using close triangle of the momentum vectors and trigo [1] [1] [1] (b)(ii) mv x cos 45° = 12m or mv x sin 45° = 12m or (mv)2 = (3m x 4)2 + (2m x 6)2 v = 17 m s–1 or use pyth theorem for momentum [1-ecf] [1-ecf] (c) Chemical energy = ½ mv2 (By conservation of energy) Chemical energy = 0.0050 x 700 or 5.0 x 0.700 = 3.5 3.5 = (0.5 x 3m x 42) + (0.5 x 2m x 62) + (0.5 x m x 172) 0.0050 x 700 or 5.0 x 0.700 = 3.5 = 204m m = 0.017 kg [1] [1-ecf] [1-ecf] 3(a) F = ρgV V = 4 / 3 × π × (2.1 × 10–3)3 = 3.88 × 10–8 m3 ρ = 4.8 × 10–4 / 9.81 × V = 1300 kg m–3 [1] [1] (b)(i) W downwards > U upwards > Fv upwards. total length up = total length down [1] (ii) FV = 7.2 × 10–4 – 4.8 × 10–4 = 2.4 × 10–4 (N) velocity = 2.4 × 10–4 / (17 × 2.1 × 10–3) = 6.7 × 10–3 m s-1 [1] [1]

3 SAJC 2024 Prelims / 9749 [Turn Over 4 (a) From infinity to Mars, |KE gain| = |GPE loss| Thus, | ½ mv2 – 0 | = | msurface - m | = | (− 𝐺𝑀𝑚 𝑅 ) – zero | speed of rock v = 2𝐺𝑀 𝐷 2 where M = 6.4 x1023 kg, D = 6.8 x 106 m = 5.0 x 103 m s-1 [1] [1] (b)(i) ½ m <c2> = 3/2 kT  temperature T = 𝑚<𝑐2> 3𝑘 where mass of 1 atom, m = 4𝑥10−3𝑘𝑔 𝑁𝐴 , <c2> = [Ans in (a) ]2 = 2.5 x 107 m2s-2 = 4.0 x 103 K ● [1 m] for mass of 1 atom, m = 4𝑥10−3𝑘𝑔 𝑁𝐴 ● [1 m, ecf v & m ] for correct caln of T [1] [1] (ii) 1.