2021 J2 H2 Physics Prelim P2 MS Sharing

Anglo-Chinese Junior College 2021 H2 Preliminary Exam Paper 2 MS H2 (9749) Physics JC2 2021 Page 1 of 9 Annotations used in marking BOD - Benefit of doubt ECF - Error carried forward POT - Powers of ten error TE - Transfer error CE - Calculation error XP - Wrong physics ENG - Generally bad english, phrasing and expression PP - Poor presentation of answers Note: For POT and TE, we can award the M mark, not the A mark. Qn Suggested MS 1 (a)(i) 1 1 1 2 2 2 1 0.1 2 0.2 2 1.8 126 5.1 v T m v T m                         l l 0.055 5.5%   OR Find max and min value Take average and correct answer (a)(ii)      3 1 1.8 1.26 5.1 10 21.08805 m s v        1 0.055 0.055 21.00805 1 m s v v v         121 1 m sv  (b)(i) Average value of 20.7 m s–1 falls within the range due to experimental error. (b)(ii) Random error Repeat the experiment more times so that the average value will tend towards the true value.

Anglo-Chinese Junior College 2021 H2 Preliminary Exam Paper 2 MS H2 (9749) Physics JC2 2021 Page 2 of 9 2 (a)   24 1.5 x x sv t  116 m s (b)   tan28 16 tan28 8.51 y x y v v v     18.5 m s (c)(i) Straight line with negative gradient Gradient = 9.81 m s–2 , t = 0.87 s  cut at 0.84 s < t < 0.90 s Stop at t = 1.5 s, magnitude of vy must be less than initial (c)(ii) maximum height = area under the v-t graph   1 0.866 8.52 3.7 m   (c)(iii)         2 2 1k.e. 2 g.p.e 1 162 9.81 3.7 xmv mgh  k.e. 3.5g.p.e  (iv) Steeper initial gradient, which decreases over time. Area above and below must be the same. Tangent to curve where it cuts the time-axis is parallel to original graph and must be the correct shape. 8.5 0.87

Anglo-Chinese Junior College 2021 H2 Preliminary Exam Paper 2 MS H2 (9749) Physics JC2 2021 Page 3 of 9 3 (a) Upthrust is a force on a partially or fully submerged object , acting in the opposite direction to weight Equal in magnitude to the weight of the fluid displaced (b)      Upthrust 1000 0.40 0.50 9.81 1962 N waterVg      weight upthrust 950 9.81 1962 7357.5 N BET    7360 N (shown) (c)(i) Beam is in equilibrium. Taking moments about A, Total clockwise moment = total anticlockwise moment    3.003.00cos58 cos58 1.24sin322 BE beam CDT m g T      where 1.24 sin 32° = perpendicular distance from CD to O       7360 3.00cos58 80.0 9.81 1.50cos58 1.24sin32 18 700 N CDT      (c)(ii) Net force on bar must be zero in all directions for translational equilibrium Possible justification includes  F must have an upward component to cancel out the net downward force from TCD, TBE, and the weight of the beam.  F must have a rightward component to cancel out the leftward force from TCD. (c)(iii) Beam is in equilibrium. Taking moments about D, TBE provides clockwise moment. Hence, F must provide anticlockwise moment, so the direction of F is below AB.

Anglo-Chinese Junior College 2021 H2 Preliminary Exam Paper 2 MS H2 (9749) Physics JC2 2021 Page 4 of 9 4 (a)(i) When the voltmeter reading is zero, the p.d. across AB is equal to the p.d. across the 100  resistor.   AC 2 2 100 100 160 100 90 10100 160 34 6 10 m 35 cm x l x .         (ii) eff 30 0 234 12 82 R I . . .    eff AC AC 1 1 1 160 1 1 1 12 82 100 160 R R R .R     AC 13 49 13 5 R. .   (iii) If the battery has internal resistance, the terminal potential difference across the battery will be smaller. With the same current, the effective resistance of the external circuit is smaller. Hence, the resistance of the resistance wire AC will be a smaller value.

Anglo-Chinese Junior College 2021 H2 Preliminary Exam Paper 2 MS H2 (9749) Physics JC2 2021 Page 5 of 9 (b)(i) For half wave rectification, 2 0 rms 0 22 2 V T V T V   2 rms ave 2 0 2 1 2 12 1 2 13 5 2 7 W VP R V R . .                 (ii) Sinusoidal waveform with an amplitude of three divisions since the peak voltage is 12 V. 1 1 50 0 020 s 20 ms T f .     Period of each cycle is 4 divisions and three cycles are drawn.

Anglo-Chinese Junior College 2021 H2 Preliminary Exam Paper 2 MS H2 (9749) Physics JC2 2021 Page 6 of 9 5 (a)(i) Out of the page (a)(ii)1. BE = 35 μT (a)(ii)2. As the d increases (and 1/d tends to zero), the magnetic field due to the wire decreases to 0. Hence the y-intercept is only due to the magnetic field due to the Earth. (a)(ii)3. From graph, (140, 80) So this implies B = BW + BE at this position. So BW = 80 – 35 = 45 μT Apply 0 2 w IB d   IW = 6 7 112 ( )(45 10 )(4 10 )140    = 1.6 A (a)(iii) Arrows must be labelled in the opposite direction and of the same magnitude. Within 2nd and 4th quadrant. (b) Direct current in electromagnet sets up an external magnetic field on the disc By Faraday’s law, since disc is rotating, there is a rate of change in magnetic flux linkage as the disc enters the region of the external magnetic field. This results in an induced emf and a corresponding induced eddy currents in the disc. By Lenz’s law, these eddy currents will results in a force that oppose the change that is causing it. Hence the velocity of the disc will decrease. wire A wire B d FA FB

Anglo-Chinese Junior College 2021 H2 Preliminary Exam Paper 2 MS H2 (9749) Physics JC2 2021 Page 7 of 9 6 (a)(i) A photon has an energy which is dependent on its frequency (E = hf), which is lower than the work function, hence not sufficient emit the electrons. For waves, the energy is dependent on the amplitude/intensity of the wave. (a)(ii) A single photon is absorbed by a single electron and the interaction is almost instantaneous. For waves, it will take time for the photon to absorb and accumulate sufficient energy to be ejected. (b)(i)1.      34 8 9 6.63 10 3.0 10 680 10 hcE        192.9 10 JE  (b)(i)2.   I                    2 23 3 19 2 1.5 103.2 10 2 2.9 10 N hc P t A d N t 16 11.9 10 sN t  (b)(ii)     34 9 28 1 cos30 6.63 10 cos30 680 10 8.4 10 kg m s i hp                   28 28 on one photon 8.4 10 8.4 10 fippF tt     27 on surface by one photon 1.7 10f