HGV 2026 Physics P2 MS

1 MARK SCHEME FOR 2026 4G3 PHYSICS PRELIM EXAM PAPER 2 SECTION A QN. SUGGESTED SOLUTION MARK(S) AWARDED 1a A to B: decreasing acceleration or accelerating at a decreasing rate B to C: constant velocity or zero acceleration 1 m 1 m 1b Area of straight-line triangle from A to B = 1 6.6 40 132 m2  = Since area under the curved line AB is greater than area of triangle (132 m), hence distance must be more than 130 m. 1 m for calculation 1 m for explaining based on area under v-t graph to distance 2a • One downward arrow (from centre) labelled “weight” or “W”. • One upward arrow labelled ‘thrust” or “motor force” 1 m 1 m 2b 672.8 10 10 2.8 10 N W mg W = =   =  1 m for calculation and correct ans 2c 66 67 77 2.8 10 1.4 3.92 10 N constant force = constant force = 3.92 10 N 2.8 10 3.192 10 N 3.2 10 N R R F ma FW = =   =  +  +  =    1 m for finding FR 1 m for correct answer 2d upward force by the exhaust gases on the rocket 1 m for correct answer 2e The overall mass/ weight of the rocket decreases as fuel is burned or air resistance decreases as the atmosphere becomes less dense Hence upward resultant force becomes larger and acceleration increases. 1 m for decrease in weight or air resistance 1 m for larger resultant force 3a ( )47 10 13 6110 J 6100 J W Fs= =   = 1 m for correct working 1 m for correct ans 3bi 5.6 230 11 14 168 J 14 000 J E IVt= =    1 m for correct working 1 m for correct ans 3bii There is energy transferred (lost) thermally to the motor or There is work done against friction in the cables/ air resistance as the boxes are lifted 1 m for valid reason 3c Using COE, GPE = KE + energy lost Since 15% of energy is lost, KE = 0.85 GPE 2 Energy converted to KE 0.85 6110 5193.5 J 1 5193.5 J2 5193.5(2) 14.866 m/s 15 m/s47 mv v =  = = = =  1 m for 0.85 or calculating energy loss 1 m for correctly applying COE formula 1 m for final answer

2 QN. SUGGESTED SOLUTION MARK(S) AWARDED 4a The sum of anticlockwise moments is equal to the sum of clockwise moments about the same pivot. 1 m for correct definition 4b 1 m for correct label and position 4c ( ) ( ) ( )2 Taking moments about X, total clockwise moments total anti-clockw ise moments 2.3 40 1.6 10 1.3 33.478 N 33 NB T F = =+ = 1m for correct working 1m for correct answer 4d T2 decreases. Smaller perpendicular distance from 40 N to pivot, hence total clockwise moments decrease, A smaller anti-clockwise moment is required to keep the rod in equilibrium and hence smaller T2. 1 m for decrease in T2 1 m for decrease in total clockwise moments 5a Because atmospheric pressure acting on the mercury in the dish is equal to the pressure exerted by a 76 cm column of mercury 1 m 5b P = hpg = 0.76 x 13600 x 10 = 1.0 x 105 Pa 1m for correct working 1m for correct answer 5ci Mercury level increases. 5cii When the container of air is heated, the molecules move at higher speeds. They collide with the walls of the container with a larger force and higher frequency. This results in higher force/ area or pressure 1 m for greater force and frequency 1 m for larger force/ area or pressure 6a identify r 180 45 30 90 15 (2.2)sin15 (1)sin 35 i i = − − − =  = = 1 m for correct working 1 m for correct answer 6b 1 sin 12.2 sin 27 n c c c = = = Since angle of incidence at boundary (60 ) is greater than critical angle (27), the light ray is totally internally reflected at X. 1 m for correct working of c 1 m for correct explanation of TIR 6c 8 8 3.0 102.2 1.4 10 m/s v v = = 1 m for correct answer W 1.3 m

3 QN. SUGGESTED SOLUTION MARK(S) AWARDED 7a The water around the ice cube gets cooled, contracts, becomes denser and sinks. The warmer, less dense water rises. This forms a convection current that cools down all the water. 1 m for density and motion of water 1 m for convection currents 7bi (100)(4.2)(40 34) 2520 J 2500 J Q mc Q = = − =  1 m for correct answer 7bii (5)(4.2)(34 0) 714 J 710 J Q mc Q = = − =  1 m for correct answer 7biii by water gained by ice 2520 714 (5)( ) 1806 = 360 J/g5 lost f f f Q Q ml l l =+ =+ = 1 m for correct working (ecf from bi and bii) 1 m for correct answer 8a When temperature rises, the resistance of the thermistor decreases and the voltage across the thermistor decreases. Since the sum of the p.d. across the thermistor and the 1.6 k  resistor is fixed at 6.0 V, voltage across the 1.6k  resistor increases and voltmeter reading increases. When temperature rises, the resistance of the thermistor decreases and the voltage across the thermistor decreases. The current in the circuit increases and since 1600V  resistor is proportional to current, voltmeter reading increases 1 m for decrease in thermistor’s resistance 1 m for voltage relationship in series circuit 8bi At 20 C, V = 2.2 V or 2.25 V 2.2 1600 0.0014 A I I = = 1 m for identifying voltage 1 m for correct answer 8bii At 50 C, V = 4.1 V 16004.1 6.01600 (1600 )(4.1) 6.0(1600) 4.1 6560 9600 741 740 T T T T R R R R = + += += =    1 m for reading from graoh 1 m for correct potential divider working 1 m for correct answer 9a Air particles near the tuning fork vibrate more vigorously and collide with their neighbours and cause them to vibrate parallel to the direction of wave travel. Sound wave is transmitted through a series of compressions and rarefactions. 1 m for vibration and collision 1 m for compressions and rarefactions 9bi 2.5 waves in 50 ms 1 m for correct working

4 QN. SUGGESTED SOLUTION MARK(S) AWARDED In 1 s, no of waves = 3 1 (2.5) 5050 10 − = f = 50 Hz 1 m for correct answer 9bii 340 (50) 6.8 m vf    = = = 1 m for correct working (ecf from 9bi) 1 m for correct answer 9c 1 m for correct amplitude 1 m for correct frequency (1.25 waves) 9d • EM waves are transverse waves while sound waves are longitudinal waves. • EM waves do not need medium to travel while sound waves require a medium for its transmission. • EM waves travel at speed of 3 x 108 m/s in air while sound waves travel at a speed of approximately 330 m/s in air. 1 m for any correct comparison Total 2 marks 10a Hydrogen-2 and hydrogen -3 both have the same number of protons (1) but different number of neutrons. 1 m for correct explanation 10b 32 32 0 15 16 1PS −→+ 1 m for correct answer 10ci technetium-99m 1 m 10cii It emits gamma radiation, which can penetrate the body to be detected externally. It has a short half-life (6 hours), which reduces the patient's long- term exposure to radiation. 1 m for penetrating ability 1 m for safety reason 10di No of protons decreases by 2 No of neutrons decreases by 2 or No of nucleons decreases by 4 1 m for proton 1 m for neutrons 10dii 12/4 = 3 half lives 1 1 1 1fraction remaining 2 2 2 8=   = No of atoms decayed = 1(1 )(7000) 61258−= 1 m for 3 half lives 1 m for correct answer Accept exact number 10diii Radon emits alpha particles, which are highly ionising and can damage cells/DNA or causing cell mutations, when inhaled. 1 m for effects when inhaled as effects are damaging inside the body SECTION B

5 QN. SUGGESTED SOLUTION MARK(S) AWARDED 11ai 1 m for magnetic field lines (lines are closer together nearer the wire) 1 m for direction (anticlockwise) 11aii Using Fleming’s Left Hand Rule, magnetic field (index finger) is pointing from N to S and current (middle finger) is flowing upwards, thumb points into the plane of the paper. 1 m for correct direction drawn on figure 1 m for correct explanation of LHR 11aiii The wire vibrates in and out of the plane of the paper 50