2023 HCI Prelim P2 AS

2022 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 1 2023 Preliminary Examination H1 Physics Paper 2 Suggested Solution Qn 1 Answer Marks (a)(i) Actual dimensions of 330 ml soft drink can radius r = 3.2 cm, height h= 12.2 cm (the volume of the can is greater than the 330 ml of soft drink contained) Acceptable range 3 5 cm 2 5 cm. r .   , 14 cm 10 cm h Volume of a can, 2V r h  Acceptable range for volume of can,    3550 196 cmV Accept if student quotes 330 cm3 A1 (a) (ii) Consider cans as being blocks of dimensions 2r x 2r x h. These blocks are stacked in the crate ……………………………. ………………………………. (independent of can dimensions) M1 M1 A1 3 1No of blocks in 1 m , 2 2N r r h     2 Empty space per block, 2 2 4 empty block canV V V r r h r r h r h                  2 2 3 Total empty space in crate 1 44 4 14 0 21 m emptyN V r hr h .          

2022 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 2 (b)(i) M1 A1 (b)(ii) time for car to move 2 km = t Using 21 2s ut at  1 9 20002 21.1 s t t t     time for jet to move 2 km = t’ time difference = 1.2 s [A1] Alternatively, time for car to move 2 km = t   2 2 1 2 12000 9 2 21 1 s s ut at t t .     [M1] time for jet to move 2 km = t’ s1 distance travelled by jet in first 10 s s2 distance travelled by jet from 10 s to finish           2 1 2 2 1 2 2 10 10 5 10 250 m2 150 10 15 10 2 2000 1250 50 10 15 10 20002 22 3 s s s t t s s t t t .                         time difference = 1.2 s [A1] M1 M1 A1 Max Marks 9  100 1000 0 93600 3.1 s v u at t t               2 1 110 50 50 50 15 10 10 20002 2 1250 100 15 150 10 20002 15 200 3000 0 22.3 s t t t t t t t                      t t

2022 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 3 Qn 2 Answer Marks (a) Weight component of the block normal to the slope = 1.6 (9.81) cos 30 = 13.59 N In the normal direction, net force = 0 Normal contact force by the scale on the block = weight component of the block normal to the slope = 13.59 N By Newton’s 3rd Law of Motion, Normal contact force by the block on the scale (i.e. this force determine the reading on the scale) is equal and opposite to the normal contact force by the scale on the block. Thus, the reading on the scale = 13.59 / 9.81 = 1.39 or 1.4 kg. B1 B1 B1 A1 (b)(i) Consider the block: Let the velocity of the block immediately after collision be v. By conservation of energy, Loss in KE = Gain in GPE 2 -1 1 2 2 0.5 2.2 m s mv mgh v gh g     B1 (b)(ii) By N2L:     1.6 0.5 9.81 0 0.2 17.7 N m v uF t     B1 A1 (b)(iii) By conservation of linear momentum, mballuball = mblockvblock + mballvball uball = mblockvblock / mball + vball =   -11.6 2.2 19.1 42 m s0.058   B1 A1 1.6 kg before after 58 g u 2.2 m s-1 1.6 kg 58 g 19.1 m s-1

2022 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 4 (c) Relative speed of approach = 42 m s-1   -1 0.5 9.81 19.1 21 m s relative speed of approach Relative speed of separation     Thus, the collision is inelastic. OR Initial KE = ½ (0.058)(42)2 = 51 J Final KE = ½ (0.058)(19.1)2 + ½ (1.6)(0.5 x 9.81) = 14.5 J Since total KE s not conserved, the collision is inelastic. B1 A1 Max Marks 11 Qn 3 Answer Marks (a) Mass/kg Length/m 0.100 0.140 0.200 0.180 0.300 0.220 0.400 0.260 0.500 0.300 1 mark for each 2 correct values A2 (b) Correct plotting of all data – 1 mark 0.10 0.15 0.20 0.25 0.30 0.00 0.10 0.20 0.30 0.40 0.50 length / m mass / kg

2022 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 5 BFL drawn – 1 mark (c) By Hooke’s law,   10.100 10 25 N m0.140 0.100 F kx mg kx mgk x        M1 A1 (d)(i) From Fig. 3.2, Length = 0.240 m Extension = 0.240 – 0.100 = 0.140 m A1 (d)(ii)    221 1EPE 25 0.140 0.245 J2 2kx   M1 A1 (e)(i) Work done to support spring = Loss in EPE + Gain in GPE = -0.245 + mgh = -0.245 + (0.350)(10)(0.140) = 0.245 J M1 A1 (e)(ii) By conservation of energy,    2 3 loss in GPE gain in EPE 1 2 2 2 350 10 10 25 0.28 m mgx kx mgx k        M1 A1 (e)(iii) Maximum acceleration occurs when the support is just removed i.e. the mass is free falling. Hence maximum acceleration is 10 m s2. A1 (f) Line with gentler gradient Same y-intercept at length = 0.10 m Labelled line S (deduct one mark for each missing marking point) A2 Max Marks 16

2022 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 6 Qn 4 Answer Marks (a) e.m.f. of a source is the energy converted per unit electric charge to electrical energy from other forms in driving charge round a complete circuit Potential difference between two points in a circuit is the energy converted per unit electric charge from electrical energy to other forms of energy when charge moved between the two points. B1 B1 (b) (i) In Fig. 5.2, the ratio of V/I decreases as current or voltage increases . Resistance of thermistor is given by ratio V/I. As current or voltage increases, the thermistor’s resistance decreases with increasing temperature. B1 B1 (b)(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 As the thermistor and filament bulb are in parallel, the current through filament bulb and thermistor should add up to 8.0 A. The p.d. across the filament bulb = p.d. across the thermistor = e.m.f. of the battery. A1 A1 A1 (c)(i) Correct equation ( E - IR = V ) [1] Correct line sketched (line in blue) [1] (c)(ii) 1. the current through the filament bulb = 4.0 A 2. the potential difference across the filament bulb = 4.0 V, 3. the potential difference across the resistor = 10.0 V. A1 A1 A1 [Total : 12 marks ] 0.00 2.00 4.00 6.00 8.00 10.00 0 2 4 6 8 10 12 14 16 filament bulb fixed resistor

2022 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 7

2022 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 8 Qn 5 Answer Marks (a) B3 (b)(i) It was observed that most of the α-particles (more than 99%) will pass straight through or emerge scattered over a small angle. As -particles are charged (and some α-particles are observed to be scattered / deflected), it shows the existence of a charged nucleus within an atom. Only a very small fraction of -particles is observed to be backscattered (i.e. suffer deflections of more than 90o). This shows that the size of the nucleus is so small that the probability of an -particle coming close enough to a nucleus to be deflected over a large angle is very low. B1 A1 B1 A1 (c)(i) The -particle consists of 2 protons and 2 neutrons. A1 (c)(ii) Loss of mass, Δm = (4.00260 + 9.01212) – (1.00867 + 12.00000) = 0.00605 u = 1.0043 x 10-29 kg Energy equivalence of the loss mass, E = (Δm)c2 = (1.0043 x 10-29)(3.0 x 108)2 = 9.04 x 10-13 J B1 M1 A1 (c)(iii) Since energy is released after the reaction