AES 2026 Physics Prelim P1 - P2 MS

Page 1 of 6 Secondary 4 Express Preliminary Examination 2026 Physics 6091 Answer Keys and Marking Scheme PAPER 1 – 40 M Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 D B A D B A C C D D Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 C C B D C B B D A B Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 C C B C A B C C C A Q31 Q32 Q33 Q34 Q35 Q36 Q37 Q38 Q39 Q40 C C D B B A B C B A PAPER 2 SECTION A – 70 M 1a 40 s B1 1b From 48 s to 54 s, the weight is greater than the lift (or upward force) of the airplane. The resultant force hence airplane moves upward but at constant deceleration, eventually coming to a momentary rest. From 54 to 60 s, the airplane moves downward at constant acceleration. [3 points – 2 marks, 2 points – 1 mark] B2 1c 2 2.0 - (-4.0)a = 66-60 = 1.0 m/s C1 A1 2a F-45 = 90(1.5) F= 180 N C1 A1 2b 180(40) 60 = E 100 E= 12 000 J C1 A1 2c The perpendicular distance between the line of action of weight through G and the rear wheel decreases. This causes the anticlockwise moment about the rear wheel to decrease. B1 B1

Page 2 of 6 3a The internal kinetic energy of water molecules increases. B1 3b Convection B1 3c Particles in the hotplate gain kinetic energy and vibrate more vigorously. They collide into adjacent glass particles and transfer energy to them. B1 B1 3d 2 4.8P = π(0.040) = 950 Pa (2 sf only) M1 A1 4a Ultrasound is a longitudinal wave where the particles vibrate in parallel to the direction of ultrasound wave motion. B1 4b -31500(0.032 10 )d = 2 = 0.024 m  C1 A1 4c Ultrasound works by reflection of sound, while X rays work by penetration and absorption. B1 5ai A region where an electric charge experiences an electric force. B1 5aii At least 3 lines to show attraction between charges Point from + to – B1 B1 5bi During rubbing, electrons from the hair are transferred to the comb. With more protons than electrons, the hair becomes positively charged. B1 5bii Negative charges in the foil are repelled away, inducing positive charges on the side nearer to the comb. The attractive force between the opposite charges is greater than the weight of the foil, the foil moves up towards the comb. B1 B1 6a The work done by a source to drive a unit charge around a complete circuit. B1 6b V = 2.0 1.5 = 3.0 V  B1 6c 9.0 – 3.0 = 6.0 V [ ecf from 6b] 6.0R = 2.0+2.0 = 1.5  B1 A1

Page 3 of 6 6d The new wire has higher resistance, hence, the effective resistance of the circuit increases. This leads to a smaller current and potential difference in the light bulbs. Light bulbs are less bright. B1 B1 7a When the fuse blows and/or switch is opened, the heating coils are completely disconnected from the live wire X at high potential. This prevents electric shock if a person touches it. B1 B1 7b 2000I = 230 = 8.7 A Total current = 8.7 + 8.7 = 17.4 A Suitable as the 20 A fuse rating is just slightly higher than the current in the circuit. M1 M1 A1 7c Earth wire The shower system does not have a metallic casing / is non-metal. B1 B1 8a The current flowing in the copper wire produces magnetic fields around it, which interacts with the magnetic field of the u shaped magnet. This results in a downward force which causes the wire to move down. B1 B1 8b Arrow drawn from right to left in the wire between the poles B1 8c Increase the emf (reject add more batteries) / Increase current / Attached iron around wire / Change to a stronger magnet B1 9a As the wheel spins, it causes a continuous change to the magnetic fields between the poles of the electromagnet. (By Faraday’s law,) this causes an induced emf in the wheel, which in turns induces a current in the wheel. B1 B1 9b (By Lenz’s law,) the induced current creates a magnetic field that opposes the rotation of the wheel. B1 9c Iron is a soft magnetic material/easy to magnetise, the strength of the electromagnet increases. B1 9d The energy in the kinetic store of the metal wheel is transferred mechanically to the internal stores of the metal wheel and the surroundings. B1 B1

Page 4 of 6 10a 1 neutron from the nucleus of iodine becomes a proton. An electron is emitted during the decay. B1 B1 10b 131 131 0 0 5453 -1 0I Xe + e + /energy → Correct reactant and products Correct nucleon and proton numbers [accept if gamma/energy is not included in the equation] B1 B1 10ci Building materials / Waste products from nuclear power station / Radon gas / Rocks / Food that contains potassium B1 10cii 5300 - 220 = 5080 counts/min 3 half-lives → 5080 ÷ 2 ÷ 2 ÷ 2 = 635 counts/min C1 A1 10d Gamma rays. It has high penetrating ability so it can pass through human tissue easily. B1 11a B1 11b1 As light intensity increases, the resistance of the LDR decreases. By the potential divider formula, the voltage across the LDR decreases. B1 B1 11b2 As the sum of voltage across LDR and fixed resistor is 6.0 V, the voltage across the fixed resistor increases. B1 11c 6.0I = 800+1000 = 0.0033 A C1 A1 11d 4000Vout = 6.0800+4000 = 5.0 V  The output voltage is 5.0 V, which is more than 3.0 V. Therefore, the LED would be turned on. M1 A1 B1 11e From 0 V to about 1.5 V, the resistance is infinitely high. From 1.5 V to 3.9 V, the resistance decreases as the voltage increases. B1 B1

Page 5 of 6 11f LED is more energy efficient than filament. Or LED can be adjusted to operate at desired surrounding brightness. B1 PAPER 2 SECTION B – 10 M 12a 650 °C B1 12b 0 s to 600 s: Solid particles gain energy and vibrate more vigorously about their fixed positions. 600 s to 100 s: Particles gain sufficient energy to overcome the intermolecular forces of attraction, thus beginning to slide past each other. B1 B1 12c Q = 0.300(880)(650) = 170 000 (2sf) or 172 000 J (3sf) C1 A1 12d 171600P = 600 = 290 W (2sf) or 286 W (3sf) [ecf from 12c] C1 A1 12e f f 286(400) 0.300L L 380000 J/kg (2sf) or 381000 J/kg (3sf) = = [ecf from 12d] C1 A1 12f Heat capacity of the ceramic container / heater are not considered. OR Some energy is transferred into the internal store of the surroundings due to absence of lid. B1 13a The ratio of the speed of light in vacuum to the speed of light in the medium. B1 13b sin 45°2.42 = sin r r = 17.0° C1 A1 13c 8 88 3.0×102.42 = v v = 1.2 ×10 m/s (2sf) or 1.24 ×10 m/s ( 3sf) C1 A1

Page 6 of 6 13d Show that angle of incidence at AB = 28.0° Since light is travelling from optically denser diamond to optically less dense air, and angle of incidence is greater than critical angle, the light ray will be total internally reflected at AB. M1 B1 13e Higher refractive index will lead to a smaller critical angle. More light rays will strike at an angle greater than critical angle. Therefore, more total internal reflection. B1 B1 13f Refractive index of sweat and grease is greater than that of air. OR The critical angle of light in diamond into sweat and grease increases, leading to fewer total internal reflection. B1