2025 ZHSS Physics Prelim Paper 1 and 2 MS

1 Paper 1 1 C 11 B 21 A 31 B 2 A 12 D 22 B 32 C 3 B 13 D 23 C 33 B 4 C 14 C 24 B 34 B 5 A 15 D 25 C 35 A 6 D 16 B 26 A 36 A 7 C 17 A 27 C 37 D 8 B 18 C 28 B 38 C 9 D 19 B 29 A 39 B 10 D 20 B 30 A 40 B

2 Paper 2 Section A Questi on Answers Marks 1(a) B1: t = 0 - 10 s B1: t = 10 - 22 s 1(b) a = (0 - 15)/10 = -1.5 m/s2 deceleration = 1.5 m/s2 A1 1(c) Distance travelled = area under the graph = 0.5 × 10 × 15 + 0.5 × 12 × 24 = 219 m Average speed = total distance / total time = 219 / 22 = 10 m/s (2 s.f.) C1: showing evidence distance = area under the graph A1 2(a) • Moving air molecules collide with inner wall of container and exerts a force on it. • Since pressure is the force exerted per unit area, the air molecules exert pressure on the surface of wall. B1 B1 (22, -24)

3 2(b) • Using Newton’s third law, downwards force of water on the rocket is equal to the upwards force of rocket on the water. • Using Newton’s second law, this upwards force is larger than sum of weight and air resistance. Positive net force is upwards. B1 B1 2(c) • The weight of the rocket decreases since water is expelled. • As rocket accelerates upwards, the air resistance increases. • Pressure (of air) decreases • Less water expelled per second / water expelled at slower speed B2: Any 2 answe rs 3(a) Product of force and perpendicular distance between pivot and line of action of force. B1 3(b) Moment = F × d = 80 × 0.90 = 72 Nm A1 3(c) Pivot: worker’s shoulder ACW moment of weight of bucket = F × d = 87 × 1.3 = 113.1 Nm CW moment of weight of ladder = 72 Nm ACW > CW. Therefore the moment due to force L must be CW so ladder can be balanced. ACW = CW 113.1 = 72 + force at L × 0.54 force at L = 76 N Direction: downwards C1 A1 B1 4(a) Energy cannot be created or destroyed. It can only be transferred from one store to another such that the total energy in isolated system remains constant. B1 4(b) Decrease in Ep = Increase in Ek mgh = 0.5mv2 10 × 0.45 = 0.5 × v2 v = 3.0 m/s or 3 m/s C1 A1 4(c) • No, she is incorrect. Not enough (kinetic) energy (at C) to travel beyond D / to top of circle or more Ep needed originally to travel beyond D / to top of circle or Ep gained cannot be greater than original Ep (transferred to Ek) • Car stops at D and goes backwards OR maximum height is D/0.45m/level of A B1 B1 5(a) • Conduction of heat from the heater to the man via air molecules. • Infra-red heat/radiation from the heater to the man B1 B1 5(b) Shiny surface reflects infra-red heat/radiation to man OR poor absorber of infra-red heat/radiation B1

4 5(c) • Sweater is made of a material that is an insulator of heat. • Sweater traps air too and air is an insulator of heat, reducing heat transfer from warm man to cooler surroundings. B1 B1 6(a) Angle of incidence in the optically denser medium whereby the angle of refraction in the optically less dense medium is 90O. 6(b), 6(c) B1 B1 6(d) • At side XZ, the angle of refraction (r) of the ray increases since lower refractive index = constant sin i / higher sin r • At side XY, the angle of incidence of the ray decreases. Since angle of incidence is smaller than the larger critical angle, refraction of light into the air takes place at side XZ. B1 B1 7(a)(i) Smallest frequency: infra-red wave Intermediate frequency: visible light Largest frequency: ultraviolet wave B1: correct waves in any sequen ce B1 7(a)(ii) Damaging effect: (skin) cancer / cataracts Property: ultraviolet radiation is ionizing (reject high energy) B1 B1 7(b) 500 nm = 500 × 10-9 m v = fλ f = (3.0 × 108) / (500 × 10-9) = 6.0 × 1014 Hz (ecf) C1 A1 8(a) Work done by the battery to push 1 C of charge around the complete circuit is 12 J. 8(b) Resistance of 2 parallel resistors = (1/30 + 1/20)-1 = 12 Ω Total resistance = 12 + 28 + 2 = 42 Ω A1 A1 8(c) V = (28 / Rtotal ) × emf = (28/42) × 12 = 8.0 V C1 A1 i r

5 8(d) radius = 0.00050 m R = ρl / A ρ = 2.0 × (π × 0.000502) / 1.0 = 1.6 × 10-6 Ωm (ecf) C1 A1 9(a) uranium-235 has (3) fewer neutrons (in the nucleus) B1 9(b) the nucleus splits (and releases more neutrons) B1 9(c) Alpha particle = 𝐻𝐻𝐻𝐻2 4 Beta particle = 𝐻𝐻−1 0 P roton number of protactinium = 92 – (2 – 1) = 91 C1 A1 9(d)(i) Half of initial count rate of 59 counts/s = 59 / 2 = 29.5 counts/s Half life = 64 s OR Half of initial count rate of 60 counts/s = 60 / 2 = 30 counts/s Half life = 64 s A1 9(d)(ii) 59 counts/s -> 29.5 counts/s -> 14.75 counts/s -> 7.4 counts/s OR 59 × 1 23 = 7.4 counts/s OR 60 counts/s -> 30 counts/s -> 15 counts/s -> 7.5 counts/s OR 60 × 1 23 = 7.5 counts/s A1 10(a) (i) from B to A (ii) top of page B1 B1 10(b) Using Fleming’s left hand rule, point the index finger in the direction of the magnetic field and middle finger in the direction of the current. The thumb gives the direction of the force. OR Based on the direction of the magnetic field due to the current in the wire and direction of the magnetic field due to the magnets, we are able to determine that the force will act along the stronger magnetic field towards the weaker magnetic field. B1 B1 OR B1 B1

6 10(c) The split-ring commuter ensures that the current direction changes at every half turn. After half a turn, force acts towards bottom of page because current now flows from A to B. The coil continues to rotate in a clockwise direction (about the axis). B1 B1 10(d) Disagree. (no marks for disagreeing without explanation) By moving the 2 solenoids further apart will bring the external magnetic field further away from the coil ABCD. This causes the external magnetic field to become weaker. A weaker magnetic field produces a weaker force on the coil and hence a weaker turning effect and thus slower rotation of the coil. B1 B1 10(e) Insert bar magnet into a solenoid with alternating current (a.c.) – with diagram Slowly withdraw magnet in the east west direction. B1 B1 11 (a) I = P / V = 6000 / 230 = 26 A MCB is suitable as the operating current is slightly lower than the MCB current rating. B1 B1 11(b)(i) Energy consumption in kWh = power in kW x time in hour = 6000/1000 x 4 + 2800/1000 x 24 + 1500/1000 x 2 = 94.2 kWh or 94 kWh A1 11(b)(ii) Cost = 94.2 x 0.327 x 30 days = $924 OR Cost = 94 x 0.327 x 30 days = $922 A1 (ecf) 11(c) Water conductivity or damp condition could cause short circuit OR electrocution B1 11(d)(i) Rcold = resistance per unit length x length = 0.012 x 3.0 = 0.036 Ω B1 11(d)(ii) 𝑹𝑹𝒉𝒉𝒉𝒉𝒉𝒉 = 𝑹𝑹𝒄𝒄𝒉𝒉𝒄𝒄𝒄𝒄 [𝟏𝟏 + 𝜶𝜶 (𝑻𝑻𝒉𝒉𝒉𝒉𝒉𝒉 − 𝑻𝑻𝒄𝒄𝒉𝒉𝒄𝒄𝒄𝒄)] = 0.036 (1 + 0.00393 (62 – 30)) = 0.041 Ω B1

7 11(d)(iii) Higher resistance leads to higher temperature as Rhot is higher than Rcold. Power loss calculation: At 62°C: P = (6.8)² × 0.041 = 1.9 W At 30°C: P = (6.8)² × 0.036 = 1.7 W Higher resistance leads to higher power loss. This power loss manifests as heat, causing further temperature rise. B1 B1 B1 12(a) 1.0 × 105 N of force is acting on an unit