ASRJC 2024 JC2 H2 Physics Prelim P2 MS

1 9749/02/ASRJC/2024PRELIM [Turn Over Anderson Serangoon Junior College 2024 H2 Physics Preliminary Examination Mark Scheme Paper 2 (80 marks) 1a The net momentum of a system remains constant provided no external resultant force acts on the system. B1 B1 1bi (total initial momentum in the direction at right−angles to the direction of the initial path of ball A = 0) (taking ↑ as positive) 0 = 4.0 × 6.0 sin θ – 12 × 3.5 sin 30° θ = 61 ° C1 A1 1bii Considering momentum along the direction of the initial path of ball A: By Conservation of linear momentum, total initial momentum = total final momentum 4.0 × v = 4.0 × 6.0 cos 61° + 12 × 3.5 cos 30° v = 12 m s−1 C1 A1 1biii total initial k.e. = ½ (4)(12)2 = 288 J total final k.e. = ½ (4)(6.0)2 + ½ (12)(3.5)2 = 145.5 J Since the total kinetic energy before and after collision is different, the collision is inelastic. M1 A1 2ai take any two sets of coordinates to determine a constant value (F/x) F/x constant hence obeys Hooke’s law M1 A1 2aii 1 2 4.5 1.5 250 Nm(1.8 0.6) 10k − − −== − EP = area under graph or ½Fx or ½kx2 = 0.5 × 4.5 × 1.8 × 10–2 or 0.5 × 250 × (1.8 × 10–2)2 = 0.041 J (0.0405 J) C1 A1 2b Loss in KE of cart = Gain in EPE of spring KE = ½mv2 = 0.0405 v = (2 × 0.0405 /1.7)½ = 0.22 m s–1 C1 A1

2 9749/02/ASRJC/2024PRELIM 3a Intensity is proportional to (amplitude)2 Since amplitude is A cos θ, the graph of amplitude is a cosine function. Hence I = kA2 cos2  (Malus’ Law) Shape: M1 Axes values: A1 M1 A1 3bi I = kA2 cos2 ϕ , and Io = kA2 Intensity after passing through polaroid R = Io cos2 30o = 0.75 Io Intensity after passing through polaroid Q = (0.75 Io) cos2 (90°− 30°) = 0.19 Io C1 A1 3bii Working below is not required: Intensity after passing through polaroid R, IR = Io cos2 ϕ Intensity after passing through polaroid Q = IR cos2 (90° - ϕ ) = Io cos2 ϕ sin2 ϕ = Io (0.5 sin 2ϕ)2 where sin 2θ = 2 sin θ cos θ = 0.25 Io sin2 2ϕ Shape: sine-square graph with 4 cycles, starting with zero labelling of every 45° correct maximum value – 0.25 Io M1 A1 B1 3c Longitudinal waves cannot be polarised because the oscillation of particles in the wave is parallel to its direction of energy transfer. A1 θ/° intensity 0.25Io ϕ/o 0 30 45 90 180 270 360 P Q R ϕ 90– ϕ 0.19Io

3 9749/02/ASRJC/2024PRELIM [Turn Over 4a progressive waves transfer/propagate energy and energy in a stationary waves is localised/stored, OR amplitude constant for progressive wave and varies (from max/an