PLMGS 2026 Physics P2 MS

V2 - Solution for Pure Physics Prelim Paper 2 (2026) Solution Marks/Comments 1a a = (v-u)/t -10 = (0-25)/t t = 2.5s B1 b Max height = area under graph = ½ x 2.5 x 25 = 31.25 = 31 m M1 (o.e) A1 (2sf) ci B1 – Start from 25m/s End with 0 m/s earlier than t1 B1 - Decreasing deceleration shape Do not penalize if slope is near zero at horizontal axis. ii As ball B moves upward, its speed decreases. Hence, the (downward) air resistance acting against the ball decreases. Since resultant force = weight + air resistance, the resultant force decreases. Hence, the ball falls with decreasing deceleration. 1. Air resistance decreases 2. Fnet = W+ air resistance 3. Fnet decreases 4. Decreasing acceleration Hit any 2points award 1 mark Hit all 4 points full marks Allow ecf max 1 mark from ci B1 B1 7 marks B

2ai B1 – Weight, downwards (Do not mark down on CG position) 2aii B1 – both tension and weight Arrow of weight is longer. 2aiii B1 – straight line from 0m/s to position A B1 – decreasing acceleration profile from A to B bi The resultant force is opposite to direction of motion/velocity. B1 bii Fnet = ma Take upward as positive, T – W = ma T –(60 x10) = 60 (3.0) T = 180 + 600 = 780 N M1 (calculate weight 600N correctly) A1 7 marks 3a It is a point where entire weight of the object appears to act. B1 bi m = W/g = 3.0/10 = 0.30 kg B1 (2 sf) bii Acw moment = Cw moment F x 1.10 = 44.0 x (1.70-1.10) + 3.0 x (4.00 – 1.10) = 44.0 x (0.60) + 3.0 x (2.90) F = 31.9090 = 32 N M1 A1 (accept 2sf or 3sf) weight Tension

biii Total upward force = Total downward force Fupward = 44.0 + 3.0 + 32 = 79 N B1 A1 (accept 2 or 3 sf) 6 marks 4a Diff in gpe = mg(h1 -h2) = 330 (4.0 – 1.1) = 957 = 960 J M1 A1 (2 sf) b W.D = change in GPE – KE at Y 52 x d = 957 – 540 d = 8.019 = 8.0 m M1 A1 (2sf) c Energy in the gravitational potential store of the child is transferred to the kinetic store of the child and internal store.(of the child & surrounding) B1 B1 d ½ mv2 = 540 ½ (330/10) v2 = 540 v = 5.7 m/s M1 A1 ( accept 2 to 3 sf) 8 marks 5a Weight of oil = weight of water (Vρg)oil = (Vρg)water 0.272 x 0.80 10000 x ρ x 10 = 0.245 x 0.80 10000 x 1000 x 10 ρ = 900.735 = 901 kg/m3 M1 A1 Note: No need to convert V to SI unit is ok. Do not penalise as they will cancel off and still give same answer. b Pressure at B = (hρg)water + Patm = 0.245 x 1000 x 10 +101000 = 103450 = 103000 Pa M1 A1 c Height should be equal/ the same. Pressure is independent of volume/cross sectional area Alternative: Since P = hρg. Density & gravitational field strength are constant, Pressure will be dependent only on the height of the liquid column. B1 B1 6 marks

6a The more energetic/faster particles close to the surface of the water have enough energy to overcome intermolecular /attractive forces and escape. The (average) kinetic energy of the remaining particles is now lower, leading to a decrease in temperature. B1 B1 b In one second, Q = mLf = 0.020 1000 x 3.34 x 105 = 6.68 J % efficiency = 6.68 45 × 100% = 14.844 = 15% M1 (for convert mass to per kg) M1 (allow error forward in Q=6.68) A1 (ans in 2sf) 5 marks 7a Amplitude of wave X: 10 cm (accept 10.0 cm as ½ smallest div) Wavelength of wave Y: 0.4m (accept 0.40m as ½ smallest div) B1 b Wave Y. It has a larger amplitude. B1 c Agree. Since both waves have the same speed and wavelength, By v=fλ, they have the same frequency and hence same pitch. B1 B1 4 marks 8a R = ½ x 18 = 9.0 Ω A1 bi I=V/R = 6.2 / 18 = 0.34 A B1 ii V = 0V V = 6.2V B1 B1 c The voltmeter reading decreases. Total resistance decreases as the (fixed) resistor is parallel to the resistance wire. Hence, (total) current increases. Since resistance across CB is fixed, by V=IR, voltage across CB increases. Thus voltage of AC/voltmeter reading decreases. B1 (award if student can conclude current increase) B1

OR The voltmeter reading decreases. Total resistance decreases as the (fixed) resistor is parallel to the resistance wire. By potential divider rule, voltage across AC decreases. B1 B1 d Maximum voltage of lamp in Fig. 8.2 is less than 6.2V. By P = V2/R, the maximum power is less and thus less bright. B1 7 marks 9a P = VI Current of fan, I = P / V = 300 / 240 = 1.25 A current of heater = 10.4 A P = V/I I = 2500 / 240 = 10.4 A Total current = 1.25 + 10.41 = 11.66 A Since total current exceeds 11A, the fuse will melt and is not suitable. M1 (correct current of either fan or heater) B1 ( correct explanation, allow 1 mark based on deduction from calculation) b The fuse is wrongly connected to the neutral wire. A fuse should be connected to the live wire. The earth wire is wrongly connected to the heater coil . The earth wire should be connected to the metal casing of the heater. (must mention the metal case) B1 B1 ci AB downwards, CD upwards B1 cii When a current flows through AB, it generates a magnetic field. This magnetic field interacts with the magnetic field of the magnet to produce a force. B1 ciii To reverse the direction of current every half a revolution so as to ensure continuous rotation. B1 civ By Lenz’s law, the induced e.m.f must be in a direction that produce a magnetic effect which opposes the changes in the magnetic flux linkages. Since induced e.m.f is opposite polarity to supply e.m.f, the net e.m.f decreases and current decreases. B1 8 marks

10ai The force acting on the blade (or force exerted by wind) is larger than resistive force acting against the blade when the wind is larger than the cut-in speed. (accept overcome resistive force) B1 aii The turbine (blades) might get damaged. OR The current produced might be too large/excessive for safe use/ system to handle. B1 b At wind speed of 12 m/s, power is at rated = 650 W P = VI I = P/V = 650/ 48 = 13.5 = 14 A M1 A1 (2sf but do not deduct for sf) ci average energy generated per day 2000 365= = 5.48 kWh Also accept if in Joules, 5.48 kWh = 5480x 60 x 60 = 19.7 x 106 J B1 (accept 2 or 3 sf) cii energy consumed by the home per day ( ) ( ) ( )0.6 6 0.02 10 0.75 3=  +  +  6.05= kWh Since the average energy generated is less than the energy consumed per day, the suggestion is not suitable. B1 B1 di The air particles are decelerated/slow down upon hitting on the blades, so by Newton’s second law, there must be a (net) force by blade on the air particles. By Newton’s third law, there must also be a reaction-pair force by air particles on the blades, which turns the blades. B1 B1 dii.1 Mass of air which hits the blades per second = 0.25 × 9000 60 = 37.5 kg/s B1 (accept 2 or 3 SF)

dii.2 Amount of kinetic energy transferred to the blades per second = ½ mv2 - ½ mu2 = ½ (37.5)(42-12) = 281.25W Power of the turbine = 281 W Allow full ecf from dii.1 for mass. M1 (diff in KE) A1 (accept 2 or 3sf) 12 marks Section B 11ai As the plane flies to the right, wing PQ cuts the magnetic flux/field lines. By Faraday’s law of electromagnetic induction, an induced e.m.f is produced. OR As the plane flies to the right, there is a rate of change of magnetic flux linkages formed with wing PQ. By Faraday’s