2023 JPJC Prelim H2 Phy P3 Solutions

2023/JPJC/Prelim/9749/03 [Turn over Answers to 2023 JC2 Preliminary Examination Paper 3 (H2 Physics) Suggested Solutions: No. Solution Remarks 1(a)(i) horizontal component 1 67cos14 65.01 65.0 m s− = = ≈  vertical component 1 67sin14 16.21 16.2 m s− = = ≈  [1] for correct answer [1] for correct answer 1(a)(ii) horizontal displacement 65.01 4.9 318.5 319 m = × = ≈ [1] for correct substitution and answer 1(a)(iii) Using 2 yy y 1 2s ut at= + , Taking downward as positive, ( )( ) ( )( ) 2 y 116.21 4.9 9.81 4.92 38.34 38.3 m s = −+ = ≈ [1] for correct substitution [1] for correct answer 1(a)(iv) angle of the slope to the horizontal 1 38.34tan 318.5 6.864 6.86 θ − =  = ≈  [1] for correct substitution [1] for correct answer 1(b)(i) [1] for correct new path Note: new path under present path new path

2 2023/JPJC/Prelim/9749/03 No. Solution Remarks 1(b)(ii)1. The path of the ball at the start is determined by the direction of the force of the club acting on the ball or air resistance has acted only for a short period of time. [1] for correct answer 1(b)(ii)2. The horizontal velocity is much reduced by air resistance. [1] for correct answer 2(a)(i) Gravitational force on satellite provides for centripetal force to keep satellite in orbit. 2 2 2 2 23 2 63 11 24 2 44 10(9.0 ) (6.67 )(5.97 10 1 50 0 ) 80 ES S E E GM M Mr r GM r Tr rT GM s ω π ππ − = =  = = ×  × = × [1] correct substitution [1] correct answer 2(a)(ii) Work done = increase in total energy of satellite ,, 11 24 76 9 22 (6.67 )( 10 10 1 5.97 )(500) 1 1 2 1.0 0 10 9.0 1.1 J10 Tf Ti Es Es fi EE GM M GM M rr − = −  =− −−    = −+ × = ×× ×× [1] correct substitution [1] correct answer 2(b)(i) The gravitational force acts towards centre of Earth provides the centripetal force and thus, gravitational force must lie on the plane of orbit which passes through centre of Earth. Additionally, in order for the satellite to orbit such that it remains at a fixed relative position above the Earth, its axis of rotation must coincide with Earth’s axis of rotation and thus it has to orbit in the equatorial plane. [1] [1] [1] 2(b)(ii) Geostationary satellites move in orbits where they are at a fixed relative position above the Earth. This allows the satellites to constantly receive and transmit telecommunication signals to and from the same position on the Earth’s surface. [1] 3(a)(i) The internal energy of a system is the sum of a random distribution of kinetic and potential energies associated with the molecules of a system. [1]

3 2023/JPJC/Prelim/9749/03 [Turn over No. Solution Remarks 3(a)(ii) UQW∆=+ U∆ – increase in internal energy Q – heat supplied to the system W – work done on the system [1] equation [1] correct definition of symbols 3(a)(iii) There is work done by gas against the atmosphere during expansion for gas at constant pressure. Using UQW∆=+  BYQ UW= ∆+ , amount of heat supplied has to be larger for the same amount of increase in internal energy. Since ΔT depends on ΔU, therefore a larger amount of heat must be supplied to have the same amount of ΔT. [1] [1] [1] 3(b) Let θ be the equilibrium temperature. mw cw (90 − θ) = mice Lf + mwater at 0 °C cw (θ − 0) (0.5) (4200) (90 − θ) = (0.1) (3.34 × 105) + (0.1) (4200) (θ) θ = 61.7 °C [1] substitution [1] answer 4(a) Simple harmonic motion is defined as an oscillatory motion in which the acceleration of an object is directly proportional to the displacement of the object from its equilibrium position, and the acceleration is always directed towards that position. [1] 4(b)(i) Angular frequency 1 2 18.5 6.80 rad s0.400 k Tm πω − = = = = Amplitude x0 max 1.40 6.80 v ω= = = 0.206 m [1] value for ω [1] answer 4(b)(ii) Maximum tension of the spring occurs at the lowest position. At the lowest position, acceleration is maximum = 2(6.80) (0.206) = 9.53 m s−2 At lowest point, T – mg = ma T = mg + ma = (0.400)(9.81) + (0.400)(9.53) T = 7.74 N [1] value for max acceleration [1] sub [1] ans

4 2023/JPJC/Prelim/9749/03 No. Solution Remarks 4(c)(i) [1] shape [1] 4 numerical values Allow e.c.f. 4(c)(ii) [1] for same amplitude and smaller maximum speed 5(a)(i) Since copper has one conduction electron per atom, the number density of charge carriers is equal to the number of atoms per unit volume. 3 3 53 massnumber of mole molar mass number of mole per unit volume density molar mass 8.9 10 63.5 10 1.40157 10 m − − = = ×= × = × [1] v / m s−1 x / m 0 +0.206 −0.206 +1.40 −1.40 0 v / m s−1 x / m 0 +0.206 −0.206 +1.40 −1.40 Z 0

5 2023/JPJC/Prelim/9749/03 [Turn over No. Solution Remarks 5 23 28 28 3 number of atoms per unit volume number of mole per unit volume 1.40157 10 6.02 10 8.44 10 8.4 10 m AN − = × = ×× × = × ≈× 28 3 Hence number density of charge carriers 8.4 10 m −= × [1] answer before rounding off 5(a)(ii) 23 3 28 19 7 71 Using 1.8 1017 10 8.4 10 1.60 102 4.97 10 5.0 10 m s Anvq v v π − −− − −− =  ×× =× × × ×× ×  = × ≈× I [1] for correct substitution [1] for correct answer 5(b) Current is the same in all segments of the wire, including the thinner segment. For the thinner segment, the cross-sectional area is smaller. The charge carriers hence move with a higher drift velocity (number density of charge carriers does not change). [1] same current and smaller cross- sectional area [1] higher drift velocity 6(a) Faraday’s law of electromagnetic induction states that th e induced e.m.f. in a coil is directly proportional to the rate of change of magnetic flux linkage through the coil. [1] 6(b)(i) 0 7 3 400 4 10 3. 81.6 1.19 10 T Bn µ π − − = = ×× ×  = × I ( ) 2 3 4 0.04080 1.19 10 2 1.2 10 Wb NBA π− − Φ= = × × ××  = × [1] value of B [1] sub [1] ans 6(b)(ii) meanE t ∆Φ= ∆ 4 4 2 1.2 10 0.30 8.0 10 V − − ××= = × [1] sub [1] ans

6 2023/JPJC/Prelim/9749/03 No. Solution Remarks 6(b)(iii) [1] E = 0 V from t = 0 to 0.5 s & 0.8 s to 1.4 s [1] E = 8.0 × 10 −4 V from t = 0.5 s to 0.8 s [1] E = −2.0 × 10 −4 V from t =1.4 s to 2.0 s 6(b)(iv) The magnitude of the maximum e.m.f. induced in the coil increases, The iron core in the solenoid increases the magnetic flux density in the coil, resulting in increase in the magnetic flux linkage in the coil. Hence by Faraday’s law, the magnitude of the maximum e.m.f. induced in the coil increases, [1] state [1] explain 7(a)(i) 2r.m.s. current 22 1.4 A o rms ΙΙ = = = [1] substitution [1] answer 7(a)(ii) r.m.s. voltage 2 240 V 2 o rms VV = = Mean power supplied rms rmsPV Ι= 2 240 22 240 W P   =     = OR Peak power Po = Vo Ιo = (240)(2) 480 W 480 240 W 22 oPP = = = [1] substitution [1] answer 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 8.0 × 10−4 E / V −2.0 × 10−4 t / s

7 2023/JPJC/Prelim/9749/03 [Turn over No. Solution Remarks 7(b)(i) 22 2 2 11 10 4.8 V 240 500 VN V VVN= → = →= 12 1 2 2 21 2 500 100 A 2 10 VN VN ΙΙ ΙΙ== →= →= Assumption: Transformer is 100% efficient / ideal with zero ohmic, magnetic leakage, copper and core losses. The output power is equal to the input power of the transformer. [1] answer [1] answer [1] reasonable answer 7(b)(ii) Welding Induction furnace Smelting scrap metal [1] reasonable answer 8(a)(i) When the vibrator