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 th