2023 HCI H2 PH Prelim P3 SS

2023 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 1 2023 H2 Physics Preliminary Examination Paper 3 Suggested Solutions 1 (a) The work done per unit mass , by an external force, in bringing a small test mass from infinity to that point, without any change in kinetic energy. B1 (b) (i) As the satellite goes to a higher orbit, gravitational potential energy, GMmU R=− will become less negative as R increases, hence the gravitational potential energy of the satellite increases. B1 (ii) The gravitational force between the Earth and the satellite provides for the centripetal force that keeps the satellite of mass m in its circular orbit. 2 2 GMm mv RR GMv R = = B1 A0 (iii) Since G and M are constant, it follows from part (ii) that 1v R Hence, ( ) 3 1old new old 3 new -1 6610 10   m 7780   m s6890 10   m 7620 m s Rvv R − ==   = M1 A0 (iv) 1. (Change in kinetic energy) = (final kinetic energy) – (initial kinetic energy) = ½ (120) 76202 – ½ (120) 77802 = -1.48 x 108 J M1 A1 2. Let KE be the kinetic energy, GPE the gravitational potential energy, and TE the total energy. Using part (ii), we have 211KE 22 GMmmv R== Furthermore, we have GPE =  GMm R− Hence, 11TE = KE + GPE = 22 GMm GMm GMm R R R + − =− Since TE = -KE (needs to be derived), (change in total energy) = (negative of change in kinetic energy) = +1.48 x 108 J M1 A1 Total: 8

2023 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 2 2 (a) (i) N: the number of molecules in the ideal gas m: the mass of an ideal gas molecule. V: the volume of the ideal gas or volume occupied by the ideal gas (molecules) Note: volume of the container (holding the gas) is accepted, but discouraged, since the question does not mention any container for the ideal gas. 2c : mean square speed of the ideal gas molecules. “Average of the squared velocities” is accepted (-1 for each mistake, max -2 marks.) molecules/ particles/ atoms are all accepted but make sure it is used consistently throughout. B2 (ii) 22 2 2 2 11 (1)33 1Substitute into (1): 3 3 1 3 2 3 2 13mean kinetic energy of a molecule 22 3Internal energy of the ideal gas, total KE of ideal gas = 2 k k Nmp c pV Nm cV pV NkT Nm c NkT m c kT E m c kT U N E NkT =  = −−−−− == = == == OR: 2 2 1 (1)3 (2) Internal energy of the ideal gas, 1 3 3total KE of ideal gas 2 2 2 k Nmpc V pV NkT U N E Nm c pV NkT = −−−−− = −−−−− = = = = = B1 B1 (b) (i) From Formula list, 2 2 2 3 2 13 22 3 40 3 kE kT m c kT mc T k uc k = = = = 27 2 23 (40 1.66 10 )(2380) 3(1.38 10 )T − − =  T = 9080 K B1 B1 OR: 2 2 2 2 2 3 2 13 22 13 ,where molar mass22 3 0.040 (0.040)(2380) 3 3(8.31) 9090 K k