2022 JVS 4N Physics ANS

1 [Turn over Answer Scheme Paper 1 [20 marks] 1 2 3 4 5 6 7 8 9 10 B C D D B C C A B A 11 12 13 14 15 16 17 18 19 20 C C C D B B D B A B Paper 2 [30 marks] Section A – 14 marks 1a 1b 1c mark at the lowest point max KE = max GPE = 0.8 X 10 X 3 [1] = 24 J [1] 1 1 1 3 marks 2a 2b Distance = 400 m Average speed = 250 / 35 = 7.14 m/s (3 sf) 1 1 1 3 marks JURONGVILLE SECONDARY SCHOOL PRELIMINARY EXAMINATION 2022 Secondary 4 Normal (Academic) Science Physics

2 [Turn over 3a 3b Clockwise moments = (480000 x 20) = 9600000 Nm CW = ACW W x 3m = 9600000 Nm W = 3200000 N 1 1 1 3 marks 4a 4b Time = 95 s (accept 94 – 96 s) The intermolecular bonds are being formed during freezing and only the internal potential energy of the molecules decreases. [1] The temperature remains constant as there is no change in the kinetic energy of the molecules. [1] 1 1 1 3 marks 5a 5b Amplitude = 2.4 – 2.5 cm Wavelength = 7.8 – 7.9 cm 1 1 2 marks

3 [Turn over Section B – 16 marks 6a 6b (i) plots [1] line of best fit (curve) [1] (ii) 3.4 min to 3.6 min [1] (i) Pot L [1] as black and dull object is a good emitter of radiation [1] so the kettle will lose heat the fastest. (ii) Pot J [1] as white and shiny object is a poor emitter of radiation [1] and the cork mat will further reduce heat loss through conduction [1] to the metal table top. 2 1 2 3

4 [Turn over 8 marks 7ai 7aii 7b 7c P – X-ray Q – visible light R – infrared radiation S – microwave Note: every two correct answers = 1m 𝑓 = 𝑣 𝜆 𝑓 = 3 × 108 700 × 10−9 𝑓 = 4.28 x 1014 Hz Sound needs a medium to travel. [1] No air in space (vacuum). [1] Distance = 1 500 x 0.4 = 600 m 2 1 1 1 1 1 1 8 marks 8ai 8aii 8aiii 8bi 8bii Rxy = ( ¼ + ¼ ) -1 = 2  Reff = 2+4 = 6  I = 10 / 6 = 1.67 A (3 sf) V = (10/6) x (2) = 3.33 V (3 sf) 1.2 x (4/60) x 30 = 2.4 kWh 2.4 (0.25) = $0.60 1 1 1 1 1 1 1 1 8 marks

5 [Turn over