DHS 2022 Prelim P2 Guide

1 2022 DHS H2 Physics Prelim Paper 2 Suggested Solutions 1 (a) vx = (6.02 – 4.82)1/2 M1 = 3.6 m s–1 A0 or 6.0 sin  = 4.8, so  = 53.1° and vx = 6.0 cos 53.1° M1 = 3.6 m s–1 A0 (b) (i) straight line from (0, 4.8) to (0.49, 0) A1 straight line continues with same slope to (0.98, –4.8), labelled Y A1 (ii) a horizontal line A1 from (0, 3.6) to (0.98, 3.6), labelled X A1 (c) s = ut + ½at2 = (4.8  0.49) + (½  –9.81  0.492) C1 = 1.2 m A1 or s = ½(u + v)t = ½  (4.8 + 0)  0.49 C1 = 1.2 m A1 or v2 = u2 + 2as s = 4.82 / (2  9.81) C1 = 1.2 m A1 Y X time / s

2 (d) kinetic energy at maximum height change in gravitational potential energy = 1 2⁄ (mvx2) mgh = 1 2⁄ (3.62) 9.81(1.2) C1 = 0.56 A1 or kinetic energy at maximum height change in gravitational potential energy = 1 2⁄ (mvx2) 1 2⁄ (mvY 2) = 1 2⁄ (3.62) 1 2⁄ (4.82) C1 = 0.56 A1 (e) With air resistance, the resultant force is larger than weight, resulting in larger deceleration, hence the actual time is shorter. B1 2 (a) energy (stored) / work done represented by area under graph B1 energy = ½ (180) (4.0 x 10-2) C1 = 3.6 J A1 (b) (i) either momentum before release is zero M1 so sum of momenta of trolleys after release is zero A1 or force = rate of change of momentum M1 force on trolleys equal and opposite A1 or impulse = change in momentum M1 impulse on each trolley is equal and opposite A1 (ii) 1. M1V1 = M2V2 B1 2. E = ½ M1V12 + ½ M2V22 B1 (iii) 1. 2 2 2 2 1 2 1 ()2 () 2 2 KE mv m mvm mv m p m = = = = 2. As trolley B has a smaller mass, it has a larger kinetic energy because momentum is the same/ constant for both trolleys. B1 M1 A0

3 3 (a) Phase difference = 2t T = 2(2.5−1.5) 3 = 2 3 = 120o A1 (b) At t = 0.75 ms, resultant displacement = sY + sZ = 4.0 – 1.0 = 3.0 µm A1 (c) I  A2 C1 Thus I = k (4)2 and IZ = k (2)2 = 0.25I A1 (d) v = f =  T C1 so  = vT = 330 (3.0 x 10−3) = 0.99 m A1 4 (a) (i) Current in R4 or R1 = 0.30 + 0.30 = 0.60 A C1 R = V / I M1 = 2.4 / 0.60 = 4.0 Ω A0 or p.d. across R3 or R2 = 2.4 / 2 = 1.2 V C1 R = V / I M1 = 1.2 / 0.30 = 4.0 Ω A0 (ii) E = 2.4 + 2.4 + 1.2 C1 = 6.0 V A1 or total resistance = 10 Ω C1 E = 10 × 0.60 = 6.0 V A1 (b) total resistance in