XMS 2026 Physics P2 MS

Marking Scheme Physics (6091) Preliminary Examination 2026 Accept 2 sf Paper 2 – Section A Qn Marking Scheme Marks Marker’s Report 1a W = mg m = 2400/10 m = 240 kg d = m/v d = 240/(64) d = 3.75 d = 3.8 kg/m3 [1] [1] 1b 1.0 g/cm3 = 1000 kg/m3 Compares density using same unit Density of the cube is lower than density of water, therefore it would float on water. [1] [1] ecf 1c Mass of X = 240 – 3.8 = 236.2 kg Mass of Y = 20 kg Effective density = total mass/total volume = 256.2 / 64 = 4.0 kg/m3 [1] [1] [1] Total [7] Mark scheme will use these abbreviations / + R A Ig ref ecf AW AVP ORA OWTTE underline ( ) alternatives statements on both sides of the + are needed for that mark reject accept (for answers correctly cued by the question) ignore as irrelevant with reference to error carried forward alternative wording (where responses vary more than usual) alternative valid point or reverse argument or words to that effect actual word given must be used by candidate (grammatical variants excepted) the word / phrase in brackets is not required but sets the context

Qn Marking Scheme Marks Marker’s Report 2 T1 and T2 drawn to scale with 113 accurately draw between them. Parallogram or tip-to-tail diagram completed correctly Direction of resultant force or T3 drawn correctly/arrows drawn correctly T3 = 7.9 N (7.1 N to 8.7 N) [1] [1] [1] [1] Total [4] Qn Marking Scheme Marks Marker’s Report 3a When the pan is empty, only the 2.0 N weight and the weight of the steelyard acts on the steelyard. At R, 2.0 N weight produces a CWM and that implies that the weight of the steelyard must produce an ACWM for the steelyard to be in equilibrium. Thus, the weight of the steelyard must act between P to Q, to the left of pivot. [1] [1] 3b Moment due to weight of steelyard = moment due to 2.0 N weight = 2.0 x 0.05 m = 0.10 Nm direction = anticlockwise [1] [1] Allow 1 sf 3c Max CWM = 0.65 x 2.0 = 1.3 Nm ACWM due to weight of steelyard = 0.10 Nm ACWM left for weight of other objects = 1.3 – 0.1 = 1.2 Nm Maximum weight of other objects = 1.2 / 0.15 = 8.0 N [1] [1] ecf Total [6]

Qn Marking Scheme Marks Marker’s Report 4a Freq. = 4/8 = 0.50 Hz Wavelength = 0.40 m V = f  V = 0.50 (0.4) = 0.20 m/s [1] [1] [1] 4b 1.5 second later > wave moved 0.2 x 1.5 = 0.30 m to the right. d1 = -2.0 cm d2 = 0.60 m from t = 4.0 s to t = 5.5 s, p moves upwards until d1 is 2.0 cm then moves downwards (until d1 = -2.0 cm). [1] [1] [1] ecf Total 6 Qn Marking Scheme Marks Marker’s Report 5a I = (6.0 – 3.5)/100 I = 0.025 A R = 3.5 V/ 0.025 A R = 140  [1] [1] Accept alternative method. 5b For 25 C to 40 C the resistance of the thermistor decreases, by potential divider rule it would cause more p.d. to be across R and hence the bulb. P = V 2/R, the power and thus the brightness of the bulb would increase. [1] [1] Total [4] Qn Marking Scheme Marks Marker’s Report 6a So that if they melt and break the circuit, the components of the circuit would be at low potential/ not Live. [1] 6b Current draw of 1 lamp = 60/220 = 0.27 A Max total lamps = 3.1A / 0.27 A = 11.48… lamps More lamps = 11 – 3 = 8 lamps [1] [1] accept whole number 6c The earth wire is to provide an alternate path of low resistance when there is fault in the circuit making the casing live. By doing so it would increase the current in the circuit and causing the fuse to melt and break the circuit. [1] [1] Total [5]

Qn Marking Scheme Marks Marker’s Report 7a [1] 7b Shape direction of field [1] [1] ecf 7c The compass needle would follow the magnetic field lines it experiences… … and point towards the left. [1] [1] ecf Total 5 Qn Marking Scheme Marks Marker’s Report 8ai Using flemming’s right hand rule, with the thumb point upwards in the direction of the force, and the index finger pointing to the right in the direction of the magnetic field, current represented by the third finger would be… …in the BA direction/into paper. [1] [1] 8aii When the wire is held stationary, there will be no rate of change of magnetic field lines linking the wire. Thus, no induced emf would be formed and the needle would not deflect. [1] [1 8bi Sine graph starting from (0,0) Max V is 2 V and 4 s for 1 complete cycle. [1] [1] [-1] if not labelled properly, and not min 2 cycles 8bii Period halved Max V doubled [1] [1] [-1] if not labelled properly, And not min 2 cycles 7

Qn Marking Scheme Marks Marker’s Report 9a It is an electron. [1] 9bi Pa91 233 [1] 9bii Xe54 137 [1] 9biii Sr38 94 [1] 9c 100% > 50% > 25% >12.5% > 6.25 > 3.125 > 1.5625 > 0.78125 7 half-lives = 7 × 75 = 525 = 530 s (2sf) [1] [1] Total [6] Qn Marking Scheme Marks Marker’s Report 10a t = 1.5 s to t = 1.6 s: negative linear line from 15 m/s to 12 m/s. t = 1.6 s to t = 2.8 s: positive linear line from 12 m/s to 0 m/s. [1] [1] 10b acceleration is the rate of change velocity. [1] 10c a = (-12 – 15) / (1.6 – 1.5) a = -270 m/s2 [1] [1] 10d As the ball is falling without air resistance, the only force/resultant force acting on it is its own weight. Since F = ma and W = mg, thus a = g. [1] [1] 10e Initial height = ½ (1.5)(15) = 11.3 m Maximum rebound = 1/2 (1.2)(12) = 7.2 m 11.3 > 7.2 Therefore, initial height > maximum rebound [1] [1] [1] ecf no ecf Total: [10] -

Qn Marking Scheme Marks Marker’s Report 11a Energy cannot be created nor destroyed it can only be transferred from one store to another. The total energy in an isolated system is constant. [1] 11bi P = 4(120 000) P = 480 000 J/s [1] [1] 11bii Power needed to be absorbed by coolant = 480 000 J/s 480 000 = (flow rate) c (change in temp) 480 000 = 16 c (25-10) c = 2 000 J/(kg C) [1] [1] ecf 11c copper, painted black copper, painted black [1] [1] 11d It is to increase the contact area of the pipes with each other… …to increase the rate of transfer via conduction (and electromagnetic radiation). [1] [1] 11e Any acceptable answer e.g. Increase the number of turns/contact area of the pipes within the heat exchanger. Cover the heat exchanger with an insulator to reduce the heat transferred to the surrounding. [1] Total 10

Paper 2 – Section B Qn Marking Scheme Marks Marker’s Report 12a Egp = mgh Egp = 80(10)(5) Egp = 4 000 J [1] 12b Ek = ½ mv2 = ½ 80 (5.02 – 4.02) = ½ 80 (25 – 16) = 360 J [1] [1] 12c (At A, it has energy in the chemical potential store and kinetic store). Energy is transferred mechanically from the chemical potential store to the kinetic store (increasing it) and gravitational potential store. [1] [1] 12di Ek at C = ½ mv2 = ½ 80 5.02 = 1000 Ek at D = 4000 + 1000 = 5000 J WD = F x d 5000 = 500 x d d = 10 m [1] [1] [1] 12 dii Any two: The energy is transferred to the internal store of the cyclist and bicycle. The energy is transferred out of the system to the surroundings by heating. The energy is transferred (mechanically) out of the system to the ground. The energy is transferred out of the system by the propagation of (sound) waves. [2] Total: 10

Qn Marking Scheme Marks Marker’s Report 13a n1 sin 1 = n2 sin 2 1 sin 90 = n2 sin 70 n2 = 1/sin 70 = 1.06 = 1.1 (2 sf) [1] [1] 13b 1 sin X = 1.06 sin 40 X = 42.9 X = 43 [1] [1] Ecf Note truncation error: 1 sin X = 1.1 sin 40 X = 45 13ci n =c/v 1.06 = 3.0 ×108 /v v = 2.8 x 108 m/s [