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 and Q 39bii upwards 39c Neutral point shifts to right Magnetic field of the solenoid without the iron core is much weaker 1 mark if student has “shifts to right” and “magnetic field is weaker” 40a Solenoid experiences a change in flux linkage / Cutting of field lines by the solenoid 40b Drop the magnet at greater height / throw the magnet through the coil / increase the number of turns per unit length/ use a stronger magnet 40c Induced voltage on both sides The second peak is higher than the first peak The time taken for the second peak is less than that of first peak 41(a) kg m-1 s-2 (b)(i) E = v2 X ρ = (5000)2 x 8000 = 2.0 x 1011 kg m-1 s-2 (e.c.f for wrong units in 41(a))

(b)(ii) v = f  =(5 x 103) / (3 x 106) = 1.67 x 10-3 m (c)(i) time = (6.5 div )(0.5) = 3.25 µs thickness = (3.25 x 10-6) /2 x 5000 = 0.00813 m (to 3 s.f.) (c)(ii) Decrease y-gain setting. Higher energy or intensity or louder ultrasound source. Reduce attenuation of source by using a waveguide. Any two of the above. (d) draw and label source and detector on opposite sides of the steel sample