Raffles Institution FE 2018 ANS

2018 Y4 FE Physics Exam Answers Qn 1 Qn 2 Qn 3 Qn 4 Qn 5 Qn 6 Qn 7 Qn 8 Qn 9 Qn 10 A A B A D C A C C C Qn 11 Qn 12 Qn 13 Qn 14 Qn 15 Qn 16 Qn 17 Qn 18 Qn 19 Qn 20 B A D D D B D D A B Qn 21 Qn 22 Qn 23 Qn 24 Qn 25 Qn 26 Qn 27 Qn 28 Qn 29 Qn 30 D A D C C A B C C C Qn Suggested Answers 31(a) There is a change in displacement of 19 m in a duration of 1 second. OR The rate of change of displacement with respect to time is 19 m s-1. 31(b) Acceleration = (25 – 20) / [(6 – 3) x 60] = 0.0278 m s-2 31(c) The car is slowing down at a decreasing rate. 31(d) At t = 3.00 min, v = 20.0 m s-1 acceleration = 0.0278 m s-2 (from 31 (b) – ecf) Fresultant = Fengine – Friction - Fair Fair = Fengine – Friction - Fresultant = 1050 – 200 – 1800 x (0.027777778) = 800 N 31(e) From t = 6.00 min to 15.0 min, terminal velocity (v = 25.0 m s-1) is reached as a = 0 m s-2 Fresultant = Fengine – Friction - Fair Fair = Fengine – Friction - Fresultant

= 1450 – 200 - 0 = 1250 N 32(a) Fx(wind) = 5 sin 50° = 3.8302 N Fy(wind) = 5 cos 50° + 1.30 = 4.5139 N Hence, Fwind = √(3.83022 + 4.51392) = 5.92 N (to 3 sf) Direction = tan-1(4.5139 / 3.8302) = 49.7° with respect to the horizontal Scale 1 cm : 0.5 N 33(a) Neutral equilibrium Line of action of the weight always pass through the pivot, allowing the air resistance / kN velocity / m s-1 20 25 1.25 0.8 1 mark – correct shape (quadratic curve) [stated in qn: Fair v2] 1 mark – correct values of 800 N & 1250 N 1 mark – correct values of 20 and 25 m/s 1.30 N 5.92 N 5.00 N 49.7° 1 mark – size of diagram with correct vectors 1 mark – correct magnitude (5.70 N to 6.10 N) 1 mark – correct direction (48° to 51°)

hammer to always stay at its new position when tilted by any angle from its original position. OR Position of C.G neither rise or fall. 33(b) KE2m = mg * (3 – 2) = 0.20 * 9.81 = 1.96 J 34(a) Wavelength = 0.02 m Period T = 3.0 s vdeep = f λ = 1/3.0 * (0.02) = 6.67 x 10-3 m s-1 34(b) Frequency remains constant. Wavelength decreases. 35 1) Gradient of 1st line steeper 2) New t1 less than half the original t1 3) Duration bet. t1 and t2 exactly halved. 36 Electrons on the near side of the rod are repelled to the far end. The attractive force between unlike charges are stronger than the repulsive force between like charges because of the smaller distance apart. 37(a) RL = 3.0 / 2.5 = 1.2 Ω 37(b) 1.50 A 38(a) R1 = 122 / 24 = 6.0 Ω 38(b) I3 = (2/3) (2.0) = 1.33 A ( 3 s.f.) fuse rating = 2 or 3 A 38(c) P4 =(2.02 ) (2.0) = 8.0 W

38(d) P2 = 4.02 / 6.0, P3 = 8.02 / 6.0 = 2.7 W = 10.7 W L2, L4 and L3 38(e) across YZ or across WX 39a North North 39bi X is between P a