CJC.H2.PRELIM.2023.P2.Suggested Solutions

CANDIDATE NAME CLASS 2T PHYSICS 9749/2 Paper 2: Structured Questions August 2023 2 hours Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your name and class in the spaces at the top of this page. Write in dark blue or black pen on both sides of the paper. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Answer all questions. The number of marks is given in brackets [ ] at the end of each question or part question. Suggested Solutions This document consists of 27 printed pages and 1 blank page. [Turn over FOR EXAMINER’S USE Q1 / 10 Q2 / 10 Q3 / 8 Q4 / 10 Q5 / 14 Q6 / 9 Q7 / 19 PAPER 2 / 80 Catholic Junior College JC2 Preliminary Examinations Higher 2

2 DATA speed of light in free space c = 3.00 x 108 m s-1 permeability of free space µ0 = 4π x 10-7 H m-1 permittivity of free space ε0 = 8.85 x 10-12 F m-1 (1/(36π)) x 10-9 F m-1 elementary charge e = 1.60 x 10-19 C the Planck constant h = 6.63 x 10-34 J s unified atomic mass constant u = 1.66 x 10-27 kg rest mass of electron me = 9.11 x 10-31 kg rest mass of proton mP = 1.67 x 10-27 kg molar gas constant R = 8.31 J K-1 mol-1 the Avogadro constant NA = 6.02 x 1023 mol-1 the Boltzmann constant k = 1.38 x 10-23 mol-1 gravitational constant G = 6.67 x 10-11 N m2 kg-2 acceleration of free fall g = 9.81 m s-2

3 FORMULAE uniformly accelerated motion s = u t + ½ a t2 v2 = u2 + 2as work done on / by a gas W = p ∆V hydrostatic pressure p = ρgh gravitational potential φ = - Gm r temperature T / K = T / ˚C + 273.15 pressure of an ideal gas p = 1 3 Nm V 〈c2〉 mean translational kinetic energy of an ideal gas molecule E = 3 2 kT displacement of particle in s.h.m. x = x0 sin ωt velocity of particle in s.h.m. v = v0 cos ωt = 22 0 x x− ±ω electric current I = Anvq resistors in series R = R1 + R2 + ... resistors in parallel 1/R = 1/R1 + 1/R2 + ... electric potential V = Q 4πεor alternating current / voltage x = x0 sin ωt magnetic flux density due to a long straight wire B = μoI 2πd magnetic flux density due to a flat circular coil B = μoNI 2r magnetic flux density due to a long solenoid B = μonI radioactive decay x = x0 exp(-λt) decay constant λ = 1 2 ln2 t

4 Answer all questions from this paper. 1 (a) A swimmer is swimming at a constant speed in a pool. Drag forces due to the water oppose the motion of the swimmer. Explain why the swimmer travels at constant speed. …………………………………………………………………………………………………. …………………………………………………………………………………………………. …………………………………………………………………………………………………. …...…………………………………………………………………………………..……..[2] Solution: The swimmer experiences a forward force from the water (thrust) which is equal and opposite to the drag force so that the net force is zero. According to Newton’s first law of motion, since there is no net force the swimmer will travel at a constant speed. 1 1 (b) The power output P of a swimmer used to overcome the drag forces travelling when at speed v is given by 3 D 1P= C ρAv2 where CD is the drag coefficient, ρ is the density of water and A is the frontal area of the swimmer. In an experiment to measure the C D for freestyle, the data for a particular swimmer is collected. The data is shown in Table 1.1. Table 1.1 quantity magnitude uncertainty P/W 294 ±2 ρ/kg m-3 1000 ±1 A/m2 0.20 ±0.01 v/m s-1 1.4 ±0.1 Determine the C D of the swimmer, with its actual uncertainty. Give your answer to an appropriate number of significant figures.

5 CD = ………………. ± ……..…… [4] Solution: D 33 2P 2(294)C= =ρAv 1000(0.20)(1.4) DC = 1.07 ρ ρ ∆ ∆∆∆ ∆= +++ = + + + ∆ = 2 1 0.01 0.133 294 1000 0.20 1.4 0.272 D D D D C P Av C P Av C C ∆= × ≈0.272 1.07 0.3DC = ±1.1 0.3DC 1 1 1 1 (c) (i) Derive, from the definition of power, an expression of the drag force F experienced by the swimmer in terms of the velocity v. [2] Solution: Since = = =W FsP F vtt ρ = = 31 2 DC AvPF vv ρ= 21 2 DF C Av 1 1 (ii) Hence or otherwise, calculate the work done by the swimmer to overcome the drag force when he swims a distance of 50 m at a constant speed of 1.4 m s-1.

6 work done = ………………. J [2] Solution: 22 D 11W=Fs = C ρAv s = (1.07)(1000)(0.20)(1.4) (50 )22 W = 10500 J 1 1 2 (a) Fig. 2.1 shows a mass initially travelling at right angles to the Earth’s uniform gravitational field. Fig. 2.1 Describe the subsequent motion of the mass. ………………………………………………………………………..………………………………… ………………………………………………………………………….…………………………… [1] Solution: Mass will follow a parabolic path towards the ground 1 (b) Fig. 2.2 shows an electron initially travelling parallel to a uniform electric field. Fig. 2.2 Describe the subsequent motion of the electron.

7 ………………………………………………………………………..………………………………… ………………………………………………………………………..………………………………… ………………………………………………………………………….……………………………[2] Solution: Electron will slow down to a stop and then accelerate to the left 1 1 (c) Fig. 2.3 shows a long molecule placed in a uniform electric field. Fig. 2.3 The ends of the molecule have equal but opposite charges. Describe and explain the initial motion of the molecule in the electric field. ……………………………………………………………………………………………………..…… ……………………………………………………………………………………………………..…… ……………………………………………………………………………………………………..…… …………………………………………………………………………………………………….. [2] Solution: Forces acting on ends of the molecule are equal and opposite or they form a couple This causes the molecule to rotate clockwise about its centre of mass 1 1 (d) Fig. 2.4 shows a sphere of weight 1.6 × 10-2 N with an electric charge of +2.0 μC. It is released from rest, in vacuum, between two parallel, vertical metal plates. The separation of the plates is 0.10 m. One plate has a potential of +80 V and the other plate has a potential of -80 V. 0.10 m

8 Fig. 2.4 (i) Determine the electric force experienced by the sphere. force = ……………… N [2] Solution: FE = qV/d = 2 × 10-6 × 160 / 0.10 = 3.2 × 10-3 N 1 1 (ii) On Fig. 2.4, sketch the path taken by the sphere after it is released. [1] Solution: Straight line pointing downward slightly to right (less than 20 degrees from vertical) 1 (e) The variations with separation of the gravitational potential energy UG and of the electric potential energy UE between two protons are shown in Fig. 2.5. +2.0 μC +2.0 μC 0.10 m +80 V -80 V +80 V -80 V

9 Fig. 2.5 Explain why the gravitational potential energy and the electric potential energy have opposite signs. ……………………………………………………………………………………………………..…… ……………………………………………………………………………………………………..…… ……………………………………………………………………………………………………..…… …………………………………………………………………………………………………….. [2] Solution: The gravitational for