Prelim H2P3 MS

weight of book support force on book by table 1 Anderson Serangoon Junior College 2021 H2 Physics Prelim Mark Scheme Paper 3 (80 marks) 1a Taking right as positive, s = ut 4.40 = 5.40 ×t t = 0.8148 s Taking downwards as positive, s = ut + ½at2 h = ½ ×9.81 ×0.8148 2 = 3.2565 ≈ 3.26 m C1 A1 1bi downward pointing arrow labelled weight upward pointing arrow labelled air resistance no credit if magnitude of air resistance exceeds that of weight B1 1bii Air resistance increases (as velocity increases) Weight (or mass) is constant, so resultant force decreases Hence, acceleration decreases B1 B1 B1 1biii At terminal velocity, air resistance equals weight With larger mass, weight is larger. (Greater air resistance), so greater terminal velocity B1 B1 1biv 1. gravitational potential energy to kinetic energy and thermal/internal energy B1 1biv 2. gravitational potential energy to thermal/internal energy B1 2ai Either The support force on book by table and the weight of book are both acting on the book. (N3L states that forces act on different bodies.) They are different types of forces, gravitational and electromagnetic/contact forces. (N3L states that forces must be of the same type.) (So, they are not a pair of action-reaction forces.) Or Reaction force of contact force by table on book is contact force that book exerts on table. Reaction force of weight of book is gravitational force that book exerts on Earth. B1 B1 B1 B1 9749/03/ASRJC/2021PRELIM [Turn Over weight air resistance

2 (So, weight of book and contact force by table on book are not a pair of action-reaction forces.) Credit 1 mark only if student did not mention type of force. 2aii Since book is resting on table, there is no net force on the book, so the two forces are equal and opposite. B1 2b Suppose two colliding bodies A and B (where A and B is an isolated system), By Newton’s third law, force A exert on B, FAB is equal in magnitude and opposite in direction to force B exert on A, FBA. FAB = – FBA Duration of collision is the same for A and B. By Newton’s second law, net force on A, FBA is equal to rate of change of momentum of A. Net force on B, FAB is equal to rate of change of momentum of B. Hence, total (rate of) change of momentum is 0. B1 B1 B1 2c By Conservation of Linear Momentum Sum of initial momentum = Sum of final momentum (28)(88) + (17)(53) = (28)(67)+ (17)v2 v2 = 87.6 m s1 Loss in kinetic energy = total initial kinetic energy – total final kinetic energy = ½ (28)(88)2 + ½ (17)(53)2 – (½ (28)(67)2 + ½ (17)(87.6)2) = 4200 J C1 C1 A1 2d Correct shape – smooth curves for both lines. Line for steel objects has larger peak and smaller duration than line for rubber, with approximately equal area under the two lines. B1 B1 3a base units: kg m s–2 × m = kg m2 s–2 A1 3bi distance of COG from P (= GP) = 17 cos 45° – 4.0 = 8.02 cm (or using Pythagoras Theorem: ) moment = 0.15 × 8.02 × 10–2 C1 A1 9749/03/ASRJC/2021PRELIM force time0

3 = 1.2 × 10–2 N m 3bii (line of action of) weight acts through pivot/P or distance between (line of action of) weight and pivot/P is zero (so) weight does not have a moment about pivot/P M1 A1 3ci upthrust = 6.20 – 5.60 = 0.60 N C1 A1 3cii Δp = ΔF / A = 0.60 / 1.2 × 10–3 = 500 Pa C1 A1 3ciii upthrust increases when density increases and since upthrust + force on spring = weight of cylinder so extension decreases M1 A1 4a The heat input is used to break intermolecular bonds between water molecules/ increasing the potential energy of molecules, and do work against the atmosphere as it expands when it changes phase. The average kinetic energy of molecules remains unchanged, and hence no change in temperature. B1 B1 B1 4b As light passes two slits instead of one, the total power that passes through the slits is doubled. (Intensity of central bright fringe increases by four times due to constructive interference as waves from both slits arrive in phase.) At dark fringes, destructive interference as waves from both slits arrive with a phase difference of 180o, so the intensity of dark fringes becomes zero. The total power / average intensity delivered onto the screen is hence doubled, so that energy is conserved. B1 B1 B1 5a weight provides the centripetal force (or acceleration of free fall is centripetal acceleration) 9.81 = 0.130 × ω2 ω = 8.687 = 8.7 rad s−1 B1 M1 A0 5b force in cord weight = centripetal force ‒ T – W = mrω2 force constant k = 5.0/0.018 (L – 0.013) × 5.0/0.018 – 5.0 = 5.0/9.81 × L × 8.72 L = 0.172 m = 17.2 cm C1 C1 A1 6ai When light intensity is maximum, resistance of LDR = 1200  C1 9749/03/ASRJC/2021PRELIM [Turn Over

4 Total resistance = = 400  A1 6aii For minimum p.d. across R2, R1 = 400  total parallel resistance (R2 + LDR) is lower than R2 (minimum) p.d. across R2 in Fig. 6.1 is lower than that in Fig. 6.2 M1 M1 A1 6bi At balance length, no current in E1 or r, so E1 = VXY (Balance length XY = 100.0  37.5 = 62.5 cm) (Using potential divider principle,) VXZ = = (2.0) = 0.80 V VXY = LXY LXZ (VXZ) = (0.80) = 0.50 V Therefore E1 = VXY = 0.50 V M1 M1 M1 A0 6bii C1 C1 A1 7ai A1 7aii A1 7b the r.m.s. voltages are different, so no same power dissipated. (Explanation: the r.m.s. voltage for Fig 7.1 is but for Fig. 7.2 it is Vo) B1 7ci output power, P = Vr.m.s. × I r.m.s. = 120  0.64 = 76.8 W efficiency = (76.8/80)  100% C1 9749/03/ASRJC/2021PRELIM

5 = 0.96 or 96% A1 7cii Any one from: • heat losses due to resistance of windings / coils • heat losses in magnetising and demagnetising core / hysteresis losses in core • heat losses due to eddy currents in (iron) core • loss of flux in the (iron) core B1 8ai By Fleming’s Left Hand Rule, the force acting on the wire is out of the plane of paper. By Newton’s Third Law, the force on the magnet is into the plane of paper. B1 B1 8aii Force decreases by half / to half of its original value. OR from a maximum value at θ = 0° to a half its maximum value at θ = 60°. M1 A1 8bi Magnetic force = Bqv = 0.45 (1.60  10–19)(5.0) = 3.6  10-19 N M1 A0 8bii By Fleming’s Right Hand Rule, there is rate of cutting of flux which induced current to flow from B to A through the wire. OR By Fleming’s Left Hand Rule, magnetic force acting on electron is directed towards B. Electrons accumulate at end B leaving excess positive charges at end A. End A has a higher potential. M1 A1 8biii1 . Potential difference across the two ends of wire produces an electric field. Electrons in the wire experience an electric force of equal magnitude but directed oppositely to the magnetic force. Hence, equilibrium is achieved. B1 B1 8biii2 . Electron in equilibrium, FB = Bqv = qE, so E = FB / q E = 3.6  10–19 / 1.60  10–19 = 2.25 = 2.3 N C–1 C1 A1 8biii3 . E = V/d don’t accept using e.m.f. = BLv 2.25 = V / 0.20 V = 0.45 V A1 8ci The frame experiences an increase in flux linkage. By Faraday’s law, an emf is induced across XY. By Lenz’s Law, a current is induced in the frame and flows clockwise (X → Y→ Z → W), resulting in a magnetic force on XY to the left / against its motion. To maintain constant speed, there should be no net force. Hence, an external force needs to be applied to the right. OR M1 M1 A1 9749/03/ASRJC/2021PRELIM [Turn Over

6 The frame experiences an increase in flux linkage. By Faraday’s law, an emf is induced across XY. A current is induced in the frame, if no external force is applied, kinetic energy of the frame will be transformed to thermal ene