CJC.H2.PRELIM.2023.P3.Suggested Solutions

1 CANDIDATE NAME CLASS 2T PHYSICS 9749/3 Paper 3: Longer Structured Questions September 2023 2 hours 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, graphs or rough working. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Section A Answer all questions. Section B Answer one question only. Circle on the cover page the question number attempted in Section B. You are advised to spend one and a half hours on Section A and half an hour on Section B. The number of marks is given in brackets [ ] at the end of each question or part question. Suggested Solutions This document consists of 33 printed pages and 0 blank page. [Turn over FOR EXAMINER’S USE SECTION A Q1 / 8 Q2 / 8 Q3 / 8 Q4 / 14 Q5 / 8 Q6 / 11 Q7 / 3 SECTION B Q8 / 20 Q9 / 20 TOTAL /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 [Turn over

4 Section A Answer all questions in this section in the spaces provided. 1 A ball of mass 10 g is dropped from a height and falls through air. The variation with time t of the speed of the ball v is shown in Fig. 1.1 Fig 1.1 (a) (i) Use Fig 1.1. to determine the acceleration of the ball at time t = 0. Show your construction on Fig 1.1. acceleration = ……………………. m s -2 [3] Solution:

5 acceleration = gradient of the v – t graph at t = 0 = 40 0 40 − − = 10 m s-2 1 1 1 (ii) By reference to your answer in (a)(i), suggest the difference, if any, between your answer and the acceleration of free fall. …………………………………………………………………………………………………... …………………………………………………………………………………………………... …………………………………………………………………………………………………... ……………………………………...……………………………………………………….. [2] Solution: The value obtained in (a )(i) is close to the value of the acceleration of the free fall. At time t = 0, the velocity of the ball is (close to) 0, so no (or very little) drag force acting on it. The net force = weight, acceleration = g 1 1 (b) Calculate the maximum resistive force acting on the ball. force = …..…………………… N [1] Solution: Maximum resistive force acts when the ball is at terminal velocity (a = 0) At this point, Fnet = mg – Fresistive = 0 Fresistive = mg = 0.01 x 9.81 = 0.0981 N 1

6 (c) On Fig 1.1, draw another curve to show the variation with time t of the speed of the ball v if the ball was dropped in a more viscous medium. [2] [Total: 8] Solution: 1 mark – rate of increase of v is lower than the original graph. Note: mg – kv = ma, k is larger and hence acc is smaller 1 mark – object reaches terminal velocity at a lower velocity and at an earlier time. Note: When mg = kv, terminal vel is reached or when a = 0, Hence acc = 0 is reached earlier when k is larger. 2 A binary star consists of two stars A and B. The two stars may be considered to be isolated in space. The centres of the two stars are separated by a constant distance, as illustrated in Fig. 2.1. Fig. 2.1 Star A of mass M A has a larger mass than star B of mass MB such that MA = 4MB. The stars are in circular orbits about each other such that the centre of their orbits is at a fixed point O. The radius of orbit of star A and star B are Ar and Br respectively. The period of each orbit is T. star A mass MA star B mass MB O

7 (a) Explain why the two stars must always be directly opposite as they move in the circular orbit. ……………………………………………………………………………………………….. ……………………………………………………………………………………………….. ……………………………………………………………………………………………….. ……………………………………………………………………………………………. [2] Solution: The centripetal force acting on each star is provided by the gravitational force between the two stars. The direction of the gravitational force acts along the line joining their centres of mass (which is the diameter of the circular orbit). 1 1 (b) Show that 4B A r r = . Explain your working. [2] Solution: The gravitational force on each star provides the centripetal force required for its respective circular motion about O. Since the stars have the same period, they have the same angular velocity ω. For star A, ( ) 2A AA2 ........... (1)B AB GM M Mr rr ω= + For star B, ( ) 2A BB2 ........... (2)B AB GM M Mr rr ω= + ------------------------------ Equating (1) and (2) OR Since the stars have the same period, they have the same angular velocity. Therefore the magnitude of centripetal force acting on each star is equal. 22 AA BB ........... (3)Mr Mrωω= 1 1

8 -------------------------------- ( ) BA B AB B B A 4 4 shown rM M rM M r r = = = (c) If the period T is 104 days and the separation of the centres of the stars is 111.1 10× m, (i) calculate the angular velocity of star A, and angular velocity = …………………………………. rad s-1 [1] L1 Solution: angular velocity 7122 7.0 10 rad s104 24 60 60T ππω −−= = = ×××× 1 (ii) determine the mass of each star. mass M A of star A = …………………………………. kg mass MB of star B = …………………………………. kg [3] L2 Solution: 11 AB 11 AA 10 1 0 AB 1.1 10 4 1.1 10 2.2 10 m and 8.8 10 m rr rr rr += × += × = ×= × 1

9 The gravitational force on star A provides the centripetal force for its circular motion about O. ( ) 2A AA2 AB BGM M Mr rr ω= + ( ) ( ) 22 AA B2B AB2 AB rr rGM rM Grr ωω += ⇒= +