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 RR R E kT m c kT m c RT m mc T R c R = = == = == =

2023 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 3 (mass of the argon atom is approximately the same as mass of the argon nucleus, it does not matter whether you take into account the mass of electrons) (ii) Argon-40 is highly likely that the argon formed from the nuclear decays will be present in the Moon’s atmosphere, since the temperature on the surface of the Moon is much lower than temperature calculated in (b)(i), the majority of argon-40 atoms would have speeds lower than the escape speed from Moon, and are unable to escape from the surface of the Moon. B1 B1 Total 8

2023 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 4 3 (a) -1 max 0 0 2 2 (2.5)(0.034) 0.534 m sv x fx   = = = = Amplitude and maximum velocity values correctly indicated Point A is labelled correctly A1 A1 (b) 2 2 2 2 0 11 1.2(2 (2.5)) (0.034)22 TE m x == 0.171 J= Or 22 max 11 1.2(0.534) 0.171 J22 TE mv= = = M1 A1 (c) (i) Any point in second quadrant. (e.c.f if point A is wrongly labelled) A1 (ii) P K TE E E+= When 2 T PK EEE== 22 022 0 1 1 2 22 0.034 0.024 m 22 mx md xd   = = = = d = - 0.024 m (no answer mark for + 0.024 m) (e.c.f if point B is wrongly labelled as + 0.024 m) M1 A1 displacement / m velocity / m s-1 0.53 - 0.53 - 0.034 0.034 A B d = - 0.024

2023 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 5 (d) ( ) ( ) 0 0 1cos cos 22 x x − =− → = rad or 0.785 rad4  = M1 A1 Total: 9 4 (a) Two images are just resolved or distinguishable when the central maximum of one diffraction pattern coincides with the first minimum of the diffraction pattern of the other. B1 B1 (b) Applying Rayleigh criterion, 9 6 2 600 10 3.00 10 rad20.0 10b  − − − = = =   16 16 18.14 10 tan(3. m0 .0 41 40) 210d − =   = M1 A1 (c) (i) 1 9 1 3 9 1 3 2F 3 or second 4 order maxima, sin 2 sin 2(400 10 )when 400 nm, sin 10 500 2(700 10 )when 700 nm, sin 10 500 Angular spread 48. 0 2 .578 4.427 2 .82 590 3.578 4 o o ooo d d       − − − − − − − =  =  = = =  = = =   = − = M1 M1 A1 (ii) Advantage: 𝜃 is larger, so for the same Δ𝜃, the fractional uncertainty / percentage uncertainty in 𝜽 is smaller. B1 Total: 8

2023 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 6 5 (a) (i) Resistance of thermistor is given by the ratio of p.d. across thermistor, V to current through thermistor, I. In Fig. 5.2, the ratio of V/I decreases as current or voltage increases. As current or voltage increases, the thermistor will heat up. Thus, the thermistor’s resistance decreases with increasing temperature indicating that it is a ntc thermistor. Alternatively, Resistance of thermistor is given by the ratio of p.d. across thermistor, V to current through thermistor, I. In Fig. 5.2, the gradient of the straight line that joins the origin to a point on the I-V graph of thermistor increases, the reciprocal of this gradient gives the ratio of V/I, i.e. V/I decreases as current or voltage increases. As current or voltage increases, the thermistor will heat up. Thus, the thermistor’s resistance decreases with increasing temperature indicating that it is a ntc thermistor. B1 B1 Or B1 B1 (ii) From the graph, when current through battery is 8.0 A, 1. the current through the filament bulb = 5.5 A 2. the current through the thermistor = 2.5 A 3. the e.m.f. of the battery = 8.0 V A1 A1 A1 (b) (i) Correct sketch of a straight line passing through the origin with gradient = 1/2.5 = 0.4. Straight line pass through (2.5, 1.0) and (10, 4). A1 (ii) the current through the filament bulb = 4.0 A A1 (iii) the potential difference across the resistor = 4.0 x 2.5 = 10 V A1 Total marks 8

2023 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 7 6 (a) (i) , 3500170 297.5 V 2000 SS S peak PP VN VVN = → = =  , 297.5 210 V 2 S rmsV == M1 A1 (ii) , , , ,, 210 1.62 A130 mean power output (210)(1.62) 340 W S rms S rms S rms S rms S rmsoutput V I R I P V I = → = = = = = = Power input = Power output (ideal transformer) peak power input = 2 outputP , , , (2 340)2 340 4.00 A 170 P peak P peak P peakV I I =   = = M1 M1 A1 (b) As the number of turns in the secondary coil is halved while input voltage and number of turns in the primary coil remain the same, the induced voltage in the secondary coil is halved. The power output is proportional to the square of the voltage. The resistance of the heater is unchanged, so the new mean power dissipated is 1 4 P . B1 A1 Total: 7

2023 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 8 7 (a) When the energy level of atom