2021 JPJC Prelim H2 Physics P3 Solutions

[Turn over 2021/JPJC/Prelim/9749/03 Answers to 2021 JC2 Preliminary Examination Paper 3 (H2 Physics) No. Solution Remarks 1(a)(i) Upthrust is the net upward force acting on a body submerged in a fluid due to the difference in pressure at the top and bottom of the immersed portion of the body. [1] for correct answer 1(a)(ii) Upthrust sea 23 4 5 4 (7.5 10 )(1.03 10 )(9.81)5 606.3 606 N U V g         [1] for correct equation and numerical substitution [1] for correct answer 1(a)(iii) 606.3 (8.0)(9.81) 527.8 530 N (shown) U T mg T U mg      [1] for correct method and numerical substitution 1(b)(i) [1] for correct T, U and D 1(b)(ii) Resolving forces horizontally, sin35 --- (1)TD  Resolving forces vertically, cos35 --- (2) cos35 T W U UWT   Substitute (2) into (1), [1] for correct equations by resolution of forces sea bed W T D direction of current 35° sea surface U

2 2021/JPJC/Prelim/9749/03 No. Solution Remarks  sea 32 ( )tan35 tan35 (1.03 10 )(7.5 10 )(9.81) (8.0)(9.81) tan35 475.7 476 N D U W Vg mg          [1] for correct numerical substitution [1] for correct answer 2(a)(i) Work done per unit mass in bringing a small test mass from infinity to that point. [1] for correct answer 2(a)(ii) Potential at infinity is taken to be zero. Due to the attractive nature of the gravitational force, work done by an external agent to bring any mass from infinity to that point is always negative. Hence the potential at any point must always be negative. [1] [1] 2(b) [1] for correct shape [1] for at least 3 correct plots (i.e. r, 2r and 4r) 2(c)(i) Gravitational force provide the centripetal force for the spacecraft to move in circular orbit. 2 2 11 24 31 (6.67 10 )(6.0 10 ) 4(6400000) 3953 4.0 10 m s (shown) GMm mv RR GMv R        [1] for statement [1] for correct equation and numerical substitution

3 2021/JPJC/Prelim/9749/03 [Turn over No. Solution Remarks 2(c)(ii) 1. 11 24 11 24 66 10 (6.67 10 )(6.0 10 )(8600) (6.67 10 )(6.0 10 )(860 0) 3(6.4 10 ) 4(6.4 10 ) 4.48 10 J fi fi U U U GMm GMm RR                                       [1] for correct equation and numerical substitution [1] for correct answer 2(c)(ii) 2. 11 24 6 31 22 22 10 (6.67 10 )(6.0 10 ) 3(6.4 10 ) 4565 4.57 10 m s 11KE 22 1 (8600)(4565 3953 )2 2.24 10 J f fi GMv R mv mv              Since total energy = kinetic energy + gravitational potential energy and there is more loss in GPE than gain in KE, the total energy decrease. [1] for correct quantitative method [1] for correct explanation and conclusion 3(a) The first law of thermodynamics states that the increase in internal energy of a system is the sum of the heat supplied to the system and the work done on the system. [1] correct definition 3(b)(i) For adiabatic process, 0 500 J U Q W Q UW       [1] correct answer 3(b)(ii) Since internal energy decreases, temperature of gas decreases. [1] correct answer 3(c)(i)        11 For constant volume process, 0 3 2 3 2.5 8.31 500 3002 6232.5 6230 J 6232.5molar heat capacity 2.5 200 12.465 12.5 J mo l K U Q W W Q U nR T             [1] for correct equation and substitution [1] for correct Q [1] for correct molar heat capacity

4 2021/JPJC/Prelim/9749/03 No. Solution Remarks 3(c)(ii) For the constant pressure process, in addition to the increase in internal energy, work is done by the gas hence more heat is supplied for the same change in temperature. This will result in a larger molar heat capacity. [1] for work done [1] for more heat supplied 4(a) bottom top 4 bottom bottom 1401.40 10 0.028 532 V VVE d V V    [1] for correct substitution [1] answer 4(b)(i)      electron electron electron 19 4 31 15 15 2 1.60 10 1.4 10 9.11 10 2.45884 10 2.5 10 m s 2 s.f. F qE m a qE qEa m            [1] correct substitution 4(b)(ii) Consider horizontal motion, 2 7 9 1 2 0.120 5.0 10 2.4 10 s x x xs u t a t t t     [1] correct substitution [1] answer 4(c) Consider vertical motion,    2 215 9 3 1 2 1 2.5 10 2.4 102 7.2 10 m y y ys u t a t        Since 0.72 cm is less than 1.4 cm (distance away from the bottom plate), the electron does not hit the plates, thus emerging from between the plates at the opposite end. Or To hit the plate, the minimum vertical displacement from straight- through direction = 1.4 cm   2 2 15 2 9 1 2 11.4 10 2.5 102 3.3 10 s y y ys u t a t t t        [1] correct substitution [1] answer [1] correct conclusion [1] correct substitution [1] answer

5 2021/JPJC/Prelim/9749/03 [Turn over No. Solution Remarks Since 3.3 ns is longer than the 2.4 ns, the electron does not hit the plates, thus emerging from between the plates at the opposite end. [1] correct conclusion 5(a)(i) Circular path downwards [1] for both sketch and description 5(a)(ii) Parabolic path upwards [1] for sketch [1] for description 5(b) Magnetic force on the electron provides for its centripetal force 2 31 8 19 2 (9.11 10 )(1.8 10 ) (1.60 10 )(0.012) 8.5 10 T mvBqv r mvB qr         When electron emerges undeviated, the magnetic and electric forces acting on the electron are equal in magnitude but opposite in direction. magnetic force = electric force 82 71 (1.8 10 )(8.5 10 ) 1.5 10 V m Bqv qE Ev B E vB          [1] correct B [1] substitution [1] answer B electron E electron

6 2021/JPJC/Prelim/9749/03 No. Solution Remarks 5(c) The upwards magnetic force is smaller than the downwards electric force. Hence the path curves downwards. (do not accept circular path) [1] for explanation [1] for description 6(a) The root-mean-square value of an alternating current is the value of steady direct current which would produce the same mean power as the alternating current in a given resistance. [1] answer 6(b)(i) Period, T = 20 ms Frequency, 3 11 50 Hz 20 10 f T     [1] answer 6(b)(ii) Peak value = 3.0 A [1] answer 6(b)(iii) Root-mean-square value peak value 2 3.0 2 2.1 A    [1] answer 6(c) Peak power, 2 00 2 3.0 5.0 45 W PR   I [1] correct graph [1] correct value of peak power 0 10 20 30 40 45 power / W time / ms

7 2021/JPJC/Prelim/9749/03 [Turn over No. Solution Remarks 6(d) 300 3 6000 0.15 A ps ps s s N N        [1] substitution [1] answer 7(a)(i) A photon is a quantum of electromagnetic radiation with energy given by hf, where h is the Planck’s constant and f is the frequency of the wave. [1] for correct answer 7(a)(ii) The energy required by the hydrogen atom for it to transit from the ground state to its first excited state is 10.2 eV. [1] for correct answer 7(b)(i) E3  E1:   31 19 19 19 9.0 10 24.6 10 15.6 10 J E E E            19 34 15 15 15.6 10 6.63 10 2.353 10 2.35 10 Hz E hf Ef h        [1] for correct expression and numerical substitution [1] for correct answer 7(b)(ii)     31 32 21 21 31 32 31 32 31 32 8 15 15 7 3.00 10 2.35 10 1.09 10 2.381 10 238 nm E E E E E E hc h f f hc h f f                   [1] for correct method and numerical substitution [1] for correct