ACS EXP Prelim 2024 P1 and P2 answers

Y4Exp Physics Prelim 2024 P1 & P2 – Answer scheme Page 1 of 9 Anglo-Chinese School (Independent) PRELIMINARY EXAMINATION 2024 YEAR 4 EXPRESS Physics Answer Scheme Paper 1 1 2 3 4 5 6 7 8 9 10 B C A C B D B D C B 11 12 13 14 15 16 17 18 19 20 D B C A A D C C A B 21 22 23 24 25 26 27 28 29 30 B A B B C A D A D A 31 32 33 34 35 36 37 38 39 40 C B B A C D A A C C Paper 2 Section A Q Part 1 (ai) The velocity-time graph of the car from t = 0 s to t = 1.4 s. Accept students who state 1.4s. A1 (aii) Total displacement = (20 x 1.4) + ½ x (20 + 8.0) x (4.0 – 1.4) Total displacement = 28.0 + 36.4 = 64.4 m Average speed = 64.4 / 4.0 = 16 m/s M1 A1 (b) The velocity-time graph of the car from t = 4.0 s to t = 7.0 s has a negative and constant gradient. (accept 1 mark only) Many different ways of describing the motion: • velocity decreases constantly from 8.0 m/s to 0 m/s. • accept slowing down at a constant rate with a deceleration of 2.67 m/s2./decreasing uniformly from 8.0 m/s to 0 m/s. This means that the car is undergoing a constant deceleration of 2.7 m/s2. B1 B1

Y4Exp Physics Prelim 2024 P1 & P2 – Answer scheme Page 2 of 9 Accept: The car decelerated uniformly from the speed of 8.0 m/s to 0 m/s. At 0 m/s, the car is at rest. Accept: The car decelerated uniformly at a lower rate as compared to time from 1.4 s to 4.0 s. 2 (a) The moment of a force about a pivot is the product of the force F and the perpendicular distance d from the pivot to the line of action of the force. B1 (b) At equilibrium, take moment about the pivot, Sum of anti-clockwise moment = Sum of clockwise moment = 0.472 kg x 10 N/kg x (6.2/100) m = 0.29 Nm 4.72 x 6.2 = 29 Ncm accepted Note: Mass ≠Weight (Hence 0.472 kg≠ 4.72 N) Accept Direction: Downwards B1 M1 A1 (c) ∆F x (1.8/100) = 0.29 (allow e.c.f from (b)) ∆F = 16 N M1 A1 3 (ai) F/(5.0 × 10–5) = 1.2 × 106 Pa F = 60 N M1 A1 (aii) As oil cannot be compressed, the same pressure from piston R will be transmitted equally to piston S. With a larger cross-sectional area for piston S, it will experience a larger force F due to P = F/A. where the PR = PS FR/AR = FS/AS In other words, we can see that PS is a multiplier of PR. This ratio is given by the ratio of AS to AR. This is known as Pascal’s Principle. B1 B1 (b) Since the liquid cannot be compressed, the volume of oil displaced in piston R must be equal to the volume of oil displaced in piston S. AR x dR = AS x dS (5.0 × 10–5 m2) x dR = (1.5 × 10–3 m2) x (0.20/1000 m) x 2 dR = 0.012 m = 1.2 cm = 12 mm Or Principle of Conservation of Energy FR x dR = FS x dS 60 x dR = 1800 x (0.20/1000 m) x 2 dR = 0.012 m {where Fs = P x As = (1.2 × 106 Pa) x (1.5 × 10–3 m2) = 1800N} M1 A1 M1 A1

Y4Exp Physics Prelim 2024 P1 & P2 – Answer scheme Page 3 of 9 60 N : 1800 N 1: 30 Hence (0.20 x 2) x 30 = 12 mm There are two piston S. 4 (a) Source A: Softer due to lower pressure, as the amplitude of a wave is directly proportional to pressure of air. Lower pitch of sound due to lower frequency as shown in the graph as compared to Source B. Or Source B: Louder due to higher pressure, as the amplitude of a wave is directly proportional to pressure of air. Higher pitch of sound due to higher frequency shown in the graph as compared to Source A. B1 B1 B1 B1 B1 B1 B1 B1 (bi) Rarefaction B1 (bii) One wavelength B1 5 (ai) 1m for every two points: -solid line for object (correct height and distance) -dotted for image (correct distance) -correct principal ray 1 drawn -correct principal ray 2 drawn (penalise for missing arrowheads and ruler – max 1m) f = 4.0 cm hi = 3.0 cm M2 A1 A1

Y4Exp Physics Prelim 2024 P1 & P2 – Answer scheme Page 4 of 9 (aii) The image height / size will decrease. (note: it remains upright and virtual) (award e.c.f. based on diagram drawn) B1 (b) X – Ultraviolet Y – Infra-red A1 A1 6 (a) It is the work done by an electrical source to drive a unit charge around a complete circuit. B1 (b) V = IR = (0.50)(12.0) = 6.0 V M1 A1 (ci) Read off the graph the value of I when V = 6.0 V (since it is in parallel with the resistor R) I2 = 2.5 A A1 (cii) I1 = I2 + I3 = 3.0 A (accept ecf) A1 (ciii) V across L = 12.0 V – 6.0 V = 6.0 V Resistance of L = V/I = (6.0 V)/(3.0 A) = 2.0 Ω (1m for method if R=V/I, and use I=3.0 but wrong V) (accept ecf for I1) (accept if correctly use Reff method) M1 A1 (d) Lightbulb L will be less bright. B1 7 (ai) I = P/V = 840/240 = 3.5 A M1 A1 (aii) 5A (4A is not a typical household fuse rating) A1 (b) Electrical Fault Tick The live wire was in contact the neutral wire The live wire was in contact with the ground wire ✔ The neutral wire was in contact with the metal casing The neutral wire was in contact with the ground wire The ground wire was disconnected from the ground ✔ 1m for each correct ✔ For incorrect ✔, max 1m For every 2 incorrect ✔, - 1m A2

Y4Exp Physics Prelim 2024 P1 & P2 – Answer scheme Page 5 of 9 8 (a) 𝐴𝑚95 241 → 𝑁𝑝𝟗𝟑 𝟐𝟑𝟕 + 𝐻𝑒2 4 A2 (bi) No smoke: ionising current when alpha emission hits the detector Smoke: alarm sound when smoke absorbs alpha emission and ionising current is disrupted. B1 B1 (bii) Relatively low penetrating power / relatively high ionising effect Use proper terminology. B1 (c) 1728/432 = 4 half-lives 800/24= 50 Bq M1 A1 (d) 20 counts/min Note: 1 Bq ≠ 1 count/min. Bq is a unit for radioactivity (no. of disintegrations/second) and not detection of radioactivity by GM counter. A1 9 (a) Conduction in liquid molecules The liquid molecules in the hotter region will move faster and collide with the other neighbouring cooler region. Convection currents in the soup The hotter liquid molecules expands, rise up the soup to the surface as they are less dense while the cooler liquid molecules of the soup sinks to the bottom of the glass bowl as they are denser. B1 B1 B1 B1 (b) During heating, the thermal energy from the microwave is transferred to the soup. The liquid particles move faster, hence there is an increase in the particles’ kinetic energy. The increase in average kinetic energy of the particles increases the temperature of the soup. The potential energy of the liquid particles increases with the increase in the average separation of the particles. The sum of the kinetic store and potential store of the particles gives the internal store of the particles. Hence, the internal store of the particles increases. Old Syllabus: Accept students stating: The increase in average kinetic energy of the particles increases the temperature of the soup only without the increase in PE. Hence, the internal store of the particles increases. B1 B1

Y4Exp Physics Prelim 2024 P1 & P2 – Answer scheme Page 6 of 9 New Syllabus: Accept: The potential energy of the liquid particles increases with the increase in the average separation of the particles. Hence, the internal store of the particles increases. (c) Q = mc∆ϑ = (750/1000) kg x 4200 J / (kg ° C) x (80 – 26) ° C Q = 170 100 J ≈170 000 J M1 A1 (d) Energy required to raise the temperature of an object by 1° C is given by: For the soup: Qsoup=msoup× csoup = 0.75× 4200=3150 J/° C For the glass bowl: Qbowl=mbowl× cbowl = 0.6× 750=450 J/° C The glass bowl requires 450 J of energy to increase its temperature by 1° C. The soup requires 3150 J of energy to increase its temperature by 1° C. Since the microwave energy is being transferred to both the soup and the glass bowl at the same rate, the glass bowl, which requires lesser energy per degree of temperature increase, wi