2013 NYJC H2 Physics P2 Answer

© NYJC 2013 JC2/Prelim/H2/9646/02 [Turn over NANYANG JUNIOR COLLEGE Science Department JC 2 PRELIMINARY EXAMINATION Higher 2 Candidate Name Class Tutor Name PHYSICS 9646/02 Paper 2 Structured Questions 24 September 2013 1 hour 45 minutes Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your name, class and tutor name on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 2 3 4 5 6 7 8 Total This document consists of 20 printed pages Nanyang Junior College

2 © NYJC 2013 JC2/Prelim/H2/9646/02 Data Formulae uniformly accelerated motion, s = ut + ½at2 v2 = u2 + 2as work done on/by a gas, W = pΔV hydrostatic pressure, p = Ρgh gravitational potential,  = /Gm r displacement of particle in s.h.m. x = xo sin ωt velocity of particle in s.h.m. v = vo cos ωt =  22 xxo   mean kinetic energy of a molecule of an ideal gas E = 3 2 kT resistors in series, R = R1 + R2 + … resistors in parallel, 1/R = 1/R1 + 1/R2 + … electric potential, V = Q / 4πεor alternating current/voltage, x = xo sin ωt transmission coefficient, T α exp(-2kd) where k =   2 2 8 m U E h   radioactive decay, x = xo exp (-λt) decay constant λ = 21 693.0 t speed of light in free space, c = 3.00 x 108 m s-1 permeability of free space, μo = 4π x 10-7 H m-1 permittivity of free space, εo = 8.85 x 10-12 Fm-1 (1 / (36 π)) x 10-9 Fm-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 J K-1 gravitational constant, G = 6.67 x 10-11 N m2 kg-2 acceleration of free fall, g = 9.81 m s-2

3 © NYJC 2013 JC2/Prelim/H2/9646/02 [Turn over For Examiner’s Use 1 (a) State the relation between force and momentum. …………………………………………………………………………………………….. [1] (b) A rigid bar of mass 450 g is held horizontally by two supports A and B, as shown in Fig. 1.1. Fig. 1.1 The support A is 45 cm from the centre of gravity C of the bar and the support B is 25 cm from C. A ball of mass 140 g falls vertically onto the bar such that it hits the bar at a distance of 50 cm from C, as shown in Fig. 1.1. The variation with time of the velocity of the ball before, during and after hitting the bar is shown in Fig. 1.2. Fig. 1.2 Force is proportional (or equal) to the rate of change of momentum [B1].

4 © NYJC 2013 JC2/Prelim/H2/9646/02 For Examiner’s Use For the time that the ball is in contact with the bar, use Fig. 1.2 to determine (i) the magnitude of the change in momentum of the ball, change in momentum = ……………….. kg m s-1 [2] (ii) the magnitude of the force exerted by the ball on the bar. force by ball = ……………….. N [2] (c) Hence, calculate the magnitude of the force exerted on the bar by support A for the time that the ball is in contact with the bar. force by support A = ……………….. N [2] ∆p = 140 x 10-3 [5.4 – (– 4.0)] [M1] = 1.32 kg m s-1 [A1] ∆ [B1] ( )( ) ( )( ) N [A1] Ex mi ’ C mm : Most candidates failed to realise that the change in momentum of the ball is only due to the resultant force on the ball, which is not solely the Fbar on ball. Taking moments about B, ( ) ( )( ) ( ) [M1] [A1] Ex mi ’ C mm : Some candidates failed to calculate the weight of the bar, using only the mass of the bar instead.

5 © NYJC 2013 JC2/Prelim/H2/9646/02 [Turn over For Examiner’s Use 2 An unpowered artificial satellite of mass m has been placed in a stable orbit around the Sun in the same direction as that of the Earth. It is at a distance of 0.99 R from the Sun, where R is the orbital radius of the Earth as shown in Fig. 2.1. Fig. 2.1 (a) Ignore the very small force the satellite act s on the Earth. S how that the period of the Earth round the Sun TE is given by where MS is the mass of the Sun. Ex mi ’ C mm : Most candidates did the proof correctly. But the presentations show that many students learned the solution by heart from somewhere. The standard answer given started like this: Fg= Fc mr2 = GMsME/R2. I suppose what they mean is the gravitational force provides the centripetal force. I would like to emphasise that there is no centripetal force. It is better to say the gravitational force accelerates the Earth towards the Sun. The , y w ’ 2 nd law, F = ma GMsME/R2 = mr2 R 0.99R Sun Earth satellite 2 3/24 E S TR GM  2 2 2 2 2 3/2 The gravitational force of the Sun on the Earth accelerates it towards the centre of the circular motion. F = ma 2() 2() 4 SE E E S E E S GM M MR TR GM R TR TR GM       

6 © NYJC 2013 JC2/Prelim/H2/9646/02 For Examiner’s Use Bear in mind that, it is a force that produces the acceleration, not an acceleration produces a force! [2] (b) Show that the resultant force on the satellite is given by 20.99 SGM m R , given that the mass of the Sun is 3.33 x 105 times the mass of Earth. 5 2 2 2 2 2 2 2 ()3.33 101.02 10000(0.99 ) (0.01 ) 1.02 0.03 0.99 S SS E S S S MGmGM m GM mGM m xF R R R R GM m GM m GM m R R R         Ex mi ’ Comments: The steady and patient candidates got this one correct. Only a few messed up the working and still put in the final answer, tried to pass off fake products as genuine. A few of them put in F = 0, and then equate the force on the satellite by the Sun and that by the Earth. That means they do not know the net force cannot not be zero because the satellite is doing a circular motion. [2] (c) Hence determine the period of the satellite round the Sun in terms of the period of the Earth TE. Ex mi ’ Comments: More than half of the candidates got this one right. The rest used R as the radius of the circular motion instead of 0.99R for the satellite. A few were not aware that the resultant force on the satellite was already proved in part (b). Period of satellite = ………………………… [2] 2 2 2 2 2 2 20.99 (0.99 )( ) 2() 4 S S E S GM m mR TR GM R TR T R TGM      

7 © NYJC 2013 JC2/Prelim/H2/9646/02 [Turn over For Examiner’s Use (d) ‘Si c h i i g i g d h S i i , i i i q i i i m.’ Comment on the statement. ………………………………………………………………………………………………… …………………………………………………………………………………………….. [1] Ex mi ’ Comments: 1. A few students thought that it is in equilibrium though they have proved that the net force in (c) which is not zero. 2. M y d c d ‘ d d w ’ f m m wh : ‘ h i still a gravitational f c ’ f m h S , h k y w d ‘ f c cc i i z ’ did not occur. 3. Few said there is other celestial body around. 4. Many said it is in translational equilibrium and not in rotational equilibrium because it is in circular motion. They obviously not knowing that an object is in rotational equilibrium as far as it is not ro