CWSS 2026 Physics Prelim P2 MS

COMMONWEALTH SECONDARY SCHOOL SECONDARY FOUR EXPRESS PHYSICS 6091 PRELIMINARY EXAMINATIONS 2026 MARK SCHEME Section A (70 marks) Deduct a maximum of one mark for error in sf. Q Answers Marks / Remarks Accept Reject 1 (a) • For an object in equilibrium, • the sum of the clockwise moments about a pivot / point is equal to the sum of anticlockwise moments • about the same pivot / point. [2] for all three points. [1] for any two. (b) (i) The whole / entire weight of the metal sheet appears / seems / is considered to act through Point B. [1] If B is not referenced in answer. (b) (ii) perpendicular distance indicated correctly [1]

Page 2 of 20 Q Answers Marks / Remarks Accept Reject Moment = force x perpendicular distance (accept symbol form) Or: 0.50 N x 0.026 m Or: 0.50 N x 0.025 m Or: 0.50 N x 0.024 m Or: 0.50 N x 2.6 cm Or: 0.50 N x 2.5 cm Or: 0.50 N x 2.4 cm [1] for working if distance is indicated correctly No ecf from first part. 0.013 N m Or: 1.3 N cm Or: 0.012 N m Or: 1.2 N cm [1] (iii) The final position of B is vertically below pivot A. [1] B is below A. The weight of the sheet has no moment as its perpendicular distance from pivot A is zero. [1] Perpendicular distance is zero without mention of weight Total mark for Q 1 [8] 2 (a) (i) ½mv2 Or: [½ x 970 kg x (27 m/s)2 Or [½ x 970 kg x (15 m/s)2 Or: [½ x 970 kg x (27 m/s)2] – [½ x 970 kg x (15 m/s)2] [1] ½ (970 kg) (27 m/s – 15 m/s)2 2.4 x 105 J or 240 000 J or 240 kJ [1] (a) (ii) Work done = force x distance Or: Force = energy transferred out / distance Or: 2.4 x 105 J / 40 m [1] error carried forward from (i)

Page 3 of 20 Q Answers Marks / Remarks Accept Reject 6100 N [1] J/m (b) (i) geothermal [1] (ii) can be replenished naturally Or: • “will not run out” • “infinite” • “will not be depleted” [1] “reused” “recycled” ‘regrown’ ‘reproduced’ ‘Unlimited’ ‘Cannot be easily diminished’ Total mark for Q 2 [6] 3 (a) Specific heat capacity, c, of a material is the change in amount of its internal energy per unit mass Or: amount of internal energy transferred per unit mass [1] 1 kg’ in place of ‘per unit mass’ for each unit change in its temperature. [1] (b) Q = 𝑚𝑐𝜃 = 20 kg x 4200 J/(kg oC) x (100 - 20) oC = 6 720 000 J = 6.7 x 106 J or 6 700 000 J or 6.7 MJ or 6700 kJ [1] (c) Specific latent heat of vaporisation is the amount of energy transferred per unit mass of a substance [1] to change the state of a substance between the liquid and gaseous states at constant temperature. [1] (d) L = 𝑚𝑙 = 0.50 x 20 kg x 2.26 x 106 J/kg [1]

Page 4 of 20 Q Answers Marks / Remarks Accept Reject = 22.6 x 106 J = 2.3 x 107 J or 23 000 000 J or 23 MJ or 23 000 kJ (e) Power = energy transferred / time Or: Time = energy transferred / power Or: (6.7 x 106 J + 2.3 x 107 J) / 2.5 x 103 W Or: (6.7 x 106 J + 2.3 x 107 J) / 2.5 kW Or: answer in (b) / 2.5 x 103 W Or: answer in (d) / 2.5 x 103 W [1] error carried forward from (b) and /or (d) 12 000 s or 1.2 X 104 s or 12 ks [1] (f) The calculation in (e) did not consider the energy that was transferred to the internal store of the air (above the water) / metal tank / pressure cap / tubing / parts of the stripper / surroundings. [1] Total mark for Q 3 [9] 4 (a) (i) 4 x 5.0 cm = 20.0 cm Or: 0.200 m Or: 20 cm [1] (a) (ii) v = f Or: 10 Hz x 20.0 cm Or: 10 Hz x 0.200 m [1] error carried forward from (i) 200 cm/s or 2.00 m/s or 2.0 m/s [1] J/m (iii) 1.0 cm or 0.010 m [1]

Page 5 of 20 Q Answers Marks / Remarks Accept Reject (b) A – down or downward D – down or downward G – up or upward All three correct: [2] Any two correct: [1] Total mark for Q 4 [6] 5 (a) Image accurately drawn (+/- 1mm) and labeled as I. [1] A ray accurately drawn with reflected ray extended to image with broken lines. [1] two rays instead of one ray • no image is indicated • missing arrows on rays, • having arrows on dotted lines, • arrows in wrong direction.

Page 6 of 20 Q Answers Marks / Remarks Accept Reject (b) (i) Refractive index is the ratio of the speed of light in vacuum to the speed of light in the medium [1] ‘air’ instead of ‘vacuum’ (ii) Angle = 35 o [1] (iii) n = sin i / sin r Or: sin (65)/ sin (30) [1] sin i /sin r appears alone without being equated to n. 1.8 [1] (c) (i) • missing arrows on ray • arrows in wrong direction 1 ray through optical centre [1] 1 ray through focal point [1] Object labelled as O and drawn with an upward arrow. [1] (ii) Projector [1] Total mark for Q 5 [10]

Page 7 of 20 Q Answers Marks / Remarks Accept Reject 6 (a) (i) The current flowing through the lamp is not directly proportional to the pd across it. Or: the graph is not a straight line that passes-through origin Or: the graph is a curve. Or: the resistance of the lamp increases as I / V / temperature increases. Or: I and V do not increase proportionally to each other Or: Current does not increase proportionally as its potential difference increases. [1] Aww for not stating ‘pass through origin’ (ii) R = V / I Or: 4.50 V / 0.070 A Or: 4.50 V / 70 mA [1] 64 Ω or 64.3 Ω [1] (b) The resistance of the filament lamp decreases. [1] As temperature decreases, the resistance of the thermistor increases. [1] The p.d. across thermistor increases. Since the emf is shared between the filament lamp and the thermistor, the p.d. across the filament lamp decreases. Or: Current through the lamp decreases. [1] Total mark for Q 6 [6]

Page 8 of 20 Q Answers Marks / Remarks Accept Reject 7 (a) Conduction, convection and radiation [1] (b) (i) Metal is a good conductor of heat Or: metal increases the rate of transfer of energy through conduction. [1] (ii) Black is a good (or better) radiator / emitter of infra-red radiation Or: black increases the rate of transfer of energy through radiation. [1] Ignore ‘absorber’ ‘radiant heat’ in place of ‘radiation’ ‘heat’ or ‘thermal energy’ in place of ‘radiation’ (iii) To increase the surface area with air. Or: increases the rate of transfer of energy to air [1] (c) Rate of energy transferred to the heat sink from the component = the rate of energy transferred from the heat sink to the surroundings. Or: Amount of energy transferred to the heat sink from the component = amount of energy transferred from the heat sink to [1] Total mark for Q 7 [5]

Page 9 of 20 Q Answers Marks / Remarks Accept Reject 8 (a) The guitar string is magnetised by induction in the presence of the cylindrical magnet. [1] misconception or contradiction. The coil is magnetized by the magnet. The magnetic field of the string linking the coil changes as the string vibrates. Or: The magnetic flux in the coil changes as the string vibrates. [1] Do not award this mark if the first mark is not awarded Not clear how the changing field comes about. The magnetic field due to the magnet changes. By Faraday’s law, an e.m.f. is induced in the coil. Since the circuit is a closed circuit, an induced current therefore can flow hence generating an electrical signal between its terminals. [1] Do not award this mark if the first mark is not awarded (b) Nylon is a non-magnetic material Or: Nylon cannot be magnetised by induction in the presence of the cylindrical magnet. [1] The coil does not experience a changing magnetic flux as the string vibrates and thus no e.m.f. will be induced in t