HCI 2022 Prelim P3 Guide

2022 C2 Prelim H2 Physics Paper 3 Suggested Solutions Setter: Wei Hong Marker: Question Answer Marks 1(a)(i) The gravitational potential at a point in a gravitational field is the work done per unit mass by an external agent in bringing a small point mass from infinity to that point in the field at constant speed. A2 1(a)(ii) Gravitational force is attractive in nature and the potential is set to be zero at infinity. To move a mass from infinity to a point in the field (of the source mass), the force exerted on the mass by the external agent will be in opposite direction to the displacement of the mass. Thus negative work is done by the external force (agent). A1 A1 1(b)(i) During the path AC, the gravitational force is directed towards A and is weakening. Gravitational force is zero at C. A1 A1 1(b)(ii) 1m for KEB < KEA 1m for minimum point at C A2 Max Marks 8 Setter: Soo Yen Marker: Question Answer Marks 2(a) Arrows drawn Processes are labelled A is a curve (steeper than C at common point) B is a vertical line C is a curve A3 0 Kinetic energy x x0

2(b) Isovolumetric/ isochoric A1 2(c) Net heat transfer out of cylinder A1 2(d) zero A1 2(e) Process B and C offers heat transfer out of the cylinder to the bath. This results in the melting of ice in the bath. Net heat transfer out = (100)(334) = 33400 J Applying 1st Law of Thermodynamics to process ABC, ΔU = Q + W For one cycle, there is no internal change in energy. Hence net work done on the gas = + 33400J M1 A1 Max Marks 8 Setter: BiaoJin Marker: Question Answer Marks 3(a) Since the acceleration of the object can be written in the form ay = , where  is a constant, the acceleration of the object is directly proportional to its displacement from its equilibrium point. The negative sign in the equation kay m=− shows that this acceleration is always directed in the opposite direction to the displacement. B1 B1 3b(i) Reading off directly from the given equation, k M = . Substituting in values from the table and converting to SI units, 2 3 0 25 10 12 9099 12 9150 10 -1. . . radsk m − = = = =  A1 3b(ii) From (b)(i), we have 12 9099 -1. rads = Using 00vy = , we have 0 0 0 31 0 024012 9099 . .m. vy = = = A1

3(b)(iii) 1m for drawing an ellipse 1m for indicating correctly the intercepts 1m for correctly indicating both points A and B. B1 B1 B1 3(b)(iv) Marking points: 1m for correct shape of KE graph, with maximum at equilibrium position and 0 KE at lowest position. 1m for correct shape of EPE graph, with positive EPE at equilibrium position and EPE at lowest position greater than the value of KE at the equilibrium position. B2 Max Marks 9 Setter: Lih Juinn Marker: Question Answer Marks 4(a) The e.m.f. of accumulator is the sum of p.d.s across the external resistor and the resistance wire. e.m.f. = 8.00 V. A1 4(b) Null deflection means the no current flows through the unknown cell. Thus, terminal p.d across unknown cell = p.d. across AB terminal p.d. across unknown cell = 4.00 V e.m.f. of unknown cell = terminal p.d across unknown cell = 4.00 V M1 A1 4(c)(i) p.d. across AC = (e.m.f. of accumulator) [ (72.0/120.0) R1 / (72.0/120.0) R1 + R1 ] = 3.00 V M1 A1 4(c)(ii) p.d. across R2 = (p.d. across AC) = 3.00 V A1 4(c)(iii) 4.00 X 12.0 / ( r + 12.0 ) = 3.00 r = 4.0  M1 A1 Max Marks 8

Setter: Koon Loon Marker: Question Answer Marks 5(a)(i) Square loop rotate about axis CD. A1 5(a)(ii) Square loop will rotate clockwise when viewed from C. The current in sides ab and bc of square loop interact with the components of the magnetic flux density that is perpendicular to the respective sides and hence experience a force. By Fleming’s Left Hand Rule, the forces acting on sides ab and bc acts in the direction out of the paper. Similarly, the forces acting on sides cd and da can be determined to act in the direction into the paper. Thus, there is a couple resulting in clockwise torque (viewed from C) about the axis CD. A1 B1 B1 B1 5(b)(i) When a and c are 2.00 m apart, the area enclosed by the loop consists of four triangular sections, each having hypotenuse of 3.00 m, height of 1.00 m, and base of 22 00.100.3 − = 2.83 m The decrease in the enclosed area is 34.3)]83.2)(00.1([42)00.3( 21 =−=−= fAiAA m2 The average induced e.m.f. is 0 100 3 34 3 340 100 A ( . )( . )B.t t .  = = = = V B1 B1 B1 A1 5(b)(ii) The induced current in the loop, 3 34 0 33410 0 . .R. = = =I A A1 Max Marks 10 Setter: Jit Ning Marker: Question Answer Marks 6(a) Frequency = 50 Hz Peak Voltage = 340 V Root-mean-square voltage = 240 V A1 A1 A1 6(b)(i) Voltage = 240/20 = 12 V A1 6(b)(i) Power consumed by 6.0 resistor, = = = 22 12.0 24.0 6.0 VPW R Power generated by primary circuit, ==24.0 25.263 0.95 PW Current in the primary circuit, 25.263 0.10526 0.105 A 240 PI V = = =  A1 A1 A1

Max Marks 7 Setter: Soo Yen Marker: Question Answer Marks 7(a) Photoelectric Effect A1 7(b) ℎ𝑐 𝜆 = 𝜙 + 𝐾 (6.63 × 10−34 )(3 × 108 ) (500 × 10−9 ) = (1.6 × 10−19 )(1.0 ) + 𝐾 𝐾 = 2.38 × 10−19 𝐽 = 1.49 𝑒𝑉 M1 A1 7(c) 𝑝 = √2𝑚𝐾 𝑝 = √2(9.11 × 10−31 )(2.38 × 10−19 ) = 6.58 × 10−25 𝑁𝑠 M1 A1 7(d) 𝑃 = 𝑁(ℎ𝑐 𝜆 ) 𝑡 25 × 10−6 = ( 𝑁 𝑡 ) (6.63 × 10−34 )(3 × 108 ) (500 × 10−9 ) Number of incoming photons per unit time, 𝑁 𝑡 = 6.285 × 1013 Number of electron ejected per unit time = (0.2) ( 𝑁 𝑡 ) = 1.257 × 1013 Electron current, 𝐼 = (1.6 × 10−19)(1.257 × 1013) = 2.01 × 10−6 A M1 M1 A1 7(e) The metal must able to at least eject electrons from the least energetic photons. 𝜙 = ℎ𝑐 𝜆 𝜙 = (6.63 × 10−34 )(3 × 108 ) (700 × 10−9 ) = 2.84 × 10−19 = 1.78 𝑒𝑉 M1 A1 Max Marks 10 Section B Setter: Caleb Marker: Question Answer Marks 8(a)(i) When two or more waves overlap (meet) The resultant displacement at any point and instance is the vector sum of the displacements caused by the individual waves at that point at that instance. B1 A1

8(a)(ii) The two waves must be of the same type (i.e both waves must be electromagnetic/sound waves) The two waves must be coherent. The two waves must have similar amplitudes. If the two waves are transverse waves, they must either be unpolarised or polarized in the same plane. (1 mark for each correct condition. Maximum of 3 marks) B3 8(b)(i) 180° (The two sources are in antiphase.) (As the waves that arrive at line MM have a path difference of 0, since a minima is detected, the sources must be in antiphase). A1 8(b)(ii)1. The lines above are possible EE lines. (Drawn line should be at least 2 wavelengths long). A1 8(b)(ii)2. The lines above are possible FF lines. (Drawn line should be at least 2 wavelengths long.) A1 8(b)(iii) Line FF: Antinodal line B1 8(b)(iv) A stationary wave/ standing wave. (As the interference pattern is produced by two coherent sources of waves meeting along a line). A1 8(b)(v) Number of wavefronts between S1 and S2 is 6. Distance between S1 and S2 = 6 λ = 3.0 mm λ = 3.0 / 6 = 0.5 mm (Shown) M1

8(b)(vi) As a stationary wave is formed, the distance between each minima = ½ λ = 0.25 mm Number of half-wavelengths in 3.0 mm = 3.0/0.25 = 12 Number of minimas detected (not inclusive of the sources) = 11 (Draw out a diagram if you are uncertain. If the sources are counted, there will be a total of 13 minimas) M1 M1 A1 8(b)(vii) (As the distance of the sources to line CC is much gre