2020 ASRJC H1 Physics Prelims P2 Answers

1 8867/02/ASRJC/2020Prelim Anderson Serangoon Junior College 2020 H1 Physics Prelim Solutions Paper 2 (80 marks) 1ai EITHER By conservation of energy, Loss in k.e. = Gain in g.p.e. ½ m (2.02 – v2) = mg∆h ½ (2.02 – v2) = 9.81 (0.13) v = 1.204 = 1.2 m s−1 OR Taking direction up the slope as positive, acceleration along slope, aslope = − g sin 30° Using v2 = u2 + 2as, we have v2 = 2.02 + 2(− g sin 30°)(0.13/sin 30°) v = 1.204 = 1.2 m s−1 1aii At the top of the ramp, horizontal component of velocity = v cos 30° = 1.2 cos 30° vertical component of velocity = v sin 30° = 1.2 sin 30° Using sy = uyt + ½ ay t2, and taking downward as positive, 0.13 = (−1.2 sin 30°)t + ½ (9.81)t2 t = 0.2351 s d = (1.2 cos 30°)(0.2351) = 0.2443 = 0.24 m 1bi Considering momentum in the direction at right−angles to the direction of the initial path of ball A: By Conservation of linear momentum, EITHER total initial momentum = total final momentum OR total initial momentum in the direction at right−angles to the direction of the initial path of ball A = 0 0 = 4.0 × 6.0 sin θ – 12 × 3.5 sin 30° θ = 61.04 = 61 ° 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 61.04° + 12 × 3.5 cos 30° v = 11.998 = 12 m s−1 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.

2 8867/02/ASRJC/2020Prelim 2ai • arrows correctly labelled and form a closed loop • angles labelled correctly 2aii EITHER use sine rule: T sin (90° − 25°) = W sin[(90° − 18°)+25°] ➔ T = 474.8 = 470 N OR considering equilibrium of forces: ∑FY = 0 ➔ R sin 25° + T cos 18° = W = 520 ∑FX = 0 ➔ R cos 25° = T sin 18° Solving, T = 474.8 = 470 N 2bi For the oil droplet to be equilibrium, it means that the electric force is upwards. Since this is a negatively charged droplet, the direction of electric field is downwards. Since electric field points from high to low potential, metal plate A is of a higher potential. 2bii Since oil droplet is in equilibrium, weight = electric force mg = qE 2.6 × 10−14(9.81) = (8 × 1.60 × 10−19)(E) E = 1.99  105 N C−1 3a elastic energy stored = ½ (1400)(0.10)2 = 7.0 J 3bi Using conservation of energy, Loss in EPE of spring = Gain in GPE of mass + WD against friction 007.0 2 9.81 3.0=   + hh Solving the equation, h0 = 0.309 = 0.31 m 3bii As the mass loses energy due to frictional forces on its way up and down, the elastic potential energy when the mass next comes to a stop will be lesser than at the start. Hence, the compression in the spring will be less than 10 cm. W T 18° R 25°

3 8867/02/ASRJC/2020Prelim 4a The dir