2021 Prelim H2Phy P1 Ans

1 2021 DHS H2 Physics Prelim Paper 1 Suggested Solutions Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 D B D B C C C D D A Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 B D A A C C B A B A Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 A C B C C D D B A A 1 D Average car tyre radius, R = 0.20 m Cross sectional radius r = 0.10 m Volume = (2πR)(πr2) = 0.039 m3 2 B Units on right side = (unit of k)(A2)(m/m)2 = (unit of k)(A2) Units on left side = W unit of k =W/ A2= (W/A)(A) = V/A = Ω 3 D v increases as the skydiver accelerates vertically downwards. Resultant force decreases (force due to air resistance) causes acceleration to decrease. 4 B v2 = u2 + 2as, 0 = u2 + 2(-9.81)(12.7) v = u + at , 0 = √ 2(9.81)(12.7) + (-9.81)t t = 1.61s, total time = 2t = 2(1.61) = 3.22 s 5 C Gradient = acceleration = (12.0 – 3.0)/15.0 = 9.0/15.0 Resultant force = ma = (2.5)(9.0/15.0) = 1.5 N 6 C perfectly elastic collision : vQ – vp = uP – uQ = 1.2 – (-1.8) = 3.0 -- Eqn 1 COM :

2 (0.30)(1.2) – (0.60)(1.8) = 0.30 vP + 0.60 vQ -- Eqn 2 Solve Eqn 1 and 2, vP = 2.8 m s-1, vQ = 0.20 m s-1 7 C Max pressure the probe can take = atmospheric pressure + pressure due to the liquid On Earth: Max pressure = 100 x 103 + (64)(1000)(9.81) = 727840 Pa On Moon of Saturn: 727840 = 35 x 103 + h (740) (3.6) h = 260 m 8 D When T = (D + W2), the net force on the lift is 0. Hence by Newton’s first law, the lift must have a constant velocity. If it was originally at rest, it will be at rest (constant velocity of 0). 9 D Negative work is done by the upthrust on the cylinder. 10 A At the top of circle, mg + N = mv2/r At minimum speed, N = 0 mg = mv2/r v2 = rg = (1.20) (9.81) v = 3.43 m s -1 11 B A is wrong because magnitude of gravitational force GMm/r2 decreases with increasing distance. In an orbit, gravitational force provides centripetal force, and since centripetal force decreases, so is the centripetal acceleration (Thus D is wrong). From Kepler’s Third Law, T2 r3 T increases as r increases.

3 Hence angular velocity decreases with increasing distance (Thus C is wrong). B is correct, as gravitational potential energy –GMm/r is a scalar and it becomes less negative (hence increases) with increasing distance. 12 D Let the distance from M1 be r from where it experiences no resultant gravitational force. At r, 13 A A is incorrect as the amount of energy absorbs is equal to the latent heat rather than the specific latent heat. 14 A × M1 d r M2

4 15 C As temperature is constant, the internal energy of the gas is unchanged. Hence A is wrong. Work is done by the gas during expansion, hence B is wrong. C is correct as D is wrong as temperature is unchanged thus root-mean-square speed is unchanged. 16 C Positions of the KE = 0 ( at t = 0, T/2 and T ) corresponds to x = amplitude positions. Positions of max KE corresponds to x = 0 positions. Hence C is correct. 17 B Let the height of the cylinder be h. h – 0.835 = / 4 ------(1) h – 0.171 = 3/ 4 ------(2)