TMJC MYE 2024 Section B solutions

CANDIDATE NAME ( ) CIVICS GROUP _________________________________________________________________________ H2 PHYSICS 9749 28 June 2024 2 hours 30 mins Candidates answer on the Question Paper. No Additional Materials are required. Section B: Structured Questions READ THESE INSTRUCTIONS FIRST Write your name and Civics Group in the spaces at the top of the 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. Answer all questions. The number of marks is given in brackets [ ] at the end of each question or part question. You are to spend 2 hours 30 mins on Section B. ______________________________________________________________________________ This document consists of 27 printed pages and 1 blank page. TAMPINES MERIDIAN JUNIOR COLLEGE JC2 MID-YEAR EXAMINATION For Examiner’s Use Percentage Section A / 15 Section B 1 / 10 2 / 10 3 / 9 4 / 16 5 / 10 6 / 10 7 / 15 8 / 10 9 / 10 Deductions Subtotal Section A&B / 115 / 100

2 Tampines Meridian Junior College 2024 JC2 Mid-Year Examination H2 Physics Data speed of light in free space c = permeability of free space o = permittivity of free space ε0 = = elementary charge e = the Planck constant h = unified atomic mass constant u = rest mass of electron me = rest mass of proton mp = molar gas constant R = 8.31 J K−1 mol−1 the Avogadro constant NA = the Boltzmann constant k = gravitational constant G = acceleration of free fall g = 813.00 10 m s − 714 10 H m −− 12 18.85 10 F m−− ( )( ) 911/ 36 10 F m −− 191.60 10 C− 346.63 10 J s− 271.66 10 kg− 319.11 10 kg− 271.67 10 kg− 23 16.02 10 mol − 23 11.38 10 J K−− 11 2 26.67 10 N m kg−− 29.81 m s−

3 Tampines Meridian Junior College 2024 JC2 Mid-Year Examination H2 Physics Formulae uniformly accelerated motion s = ut + at2 v2 = u2 + 2as work done on / by a gas W = pV hydrostatic pressure p = gh gravitational potential  = temperature T /K = T /°C + 273.15 pressure of an ideal gas p = mean translational kinetic energy of an ideal gas molecule E = kT displacement of particle in s.h.m. x = xo sin t velocity of particle in s.h.m. v = vo cos t =   electric current I = Anvq resistors in series R = … resistors in parallel = … electric potential V = alternating current / voltage X = xo sin t magnetic flux density due to a long straight wire B = magnetic flux density due to a flat circular coil B = magnetic flux density due to a long solenoid B = radioactive decay x = decay constant = 1 2 GM r− 21 3 Nm c V 3 2 22 oxx − 12RR++ 1 / R 121/ 1/RR++ 04 Q r 0 2 d   I 0 2 N r  I 0n I ( )0 expxt −  1 2 ln2 t

4 Tampines Meridian Junior College 2024 JC2 Mid-Year Examination H2 Physics Section B: Structured Questions Answer all the questions in the spaces provided. 1 (a) State Newton’s law of gravitation. …………………………………………………………………………….................... [1] Two point masses attract each other with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. B1 Comments: Average done. There are several key terms which students tend to miss out. proportional – hence the need for constant of proportionality G product of masses – not product of objects (which does not make sense) square of the distance – not just the distance. (b) A moon is in a circular orbit of radius r about a planet of mass M. Assume that the planet and the moon are point masses isolated in space. Show that the orbital period T of the moon is given by the expression 23 2 4 rT GM = where G is the gravitational constant. Explain your working. [2] Gravitational force provides centripetal force 2 2 2 2 2 23 2 4 GMm mrr GM rrT Tr GM    = =  = B1 B1 Comments: Average done. Some do not understand the meaning of “gravitational force provides for centripetal force”. Some wrongly write that “centripetal force provides for gravitational force”, or that “gravitational force is provided by centripetal force”. Some students got the formula for gravitational force wrong.

5 Tampines Meridian Junior College 2024 JC2 Mid-Year Examination H2 Physics (c) Fig. 1.1 shows the planet Mars and the orbits of its two moons, Phobos and Deimos. Deimos has an orbital radius of 23 500 km and takes 30.3 hours to orbit Mars, while Phobos takes 7.7 hours to orbit Mars. Assume the moons orbit Mars with circular orbits. Fig. 1.1 (not to scale) (i) Determine the orbital radius of Phobos. orbital radius = ……………………. m [2] For Deimos: ( ) 22 33 2 10 22 430.3 60 60 (23500 10 ) 41.19 10 1.30 10 GM GM     =   =   For Phobos: ( ) 22 3 2 83 47.7 3600 47.68 10 rGM rGM   = = Dividing the second equation by the first equation gives: 23 2 3 3 83 10 22 6 (7.7 60 60) [C1](30.3 60 60) (23500 10 ) 7.68 10 1.19 10 1.30 10 9.43 10 m r r r  =    = = C1 A1 Mars Phobos Deimos Alternatively: 23 23 23 PP DD Tr Tr Tr  =

6 Tampines Meridian Junior College 2024 JC2 Mid-Year Examination H2 Physics Comments: Poorly done. Students should recognise the proportional relationship between T2 and r3; using ratio to work this question would be the best method as this would minimise unnecessary careless mistakes. 1) Some tried to find the mass M, obtaining the wrong value as they did not do the necessary conversion to SI units (e.g. T, r). 2) Some dropped either the square of the period and/or the cube of the radius when they did their calculations. (ii) Hence, or otherwise, determine the orbital speed of Phobos. orbital speed = ………………… m s−1 [2] Thus the orbital speed of Phobos is: 62 2 9.43 10 7.7 3600 rv T  ==  12137 2100 m s −= C1 A1 Comments: Poorly done. Common mistakes include: 1) Some used the wrongly formulae v = r / T or v = 2π / T in their calculations. 2) Not using SI unit for the substitutions or wrong conversions 3) Small handful wrongly used the escape velocity formula. 4) Some correctly used GMv r= but substituted the wrong mass M. No e.c.f can be awarded if student used the wrong M which they have calculated in (i).

7 Tampines Meridian Junior College 2024 JC2 Mid-Year Examination H2 Physics (d) The mass of Phobos is 9.6 × 1015 kg and the mass of Mars is 6.4 × 1023 kg. Determine the total energy of Phobos in its circular orbit around Mars. total energy = ………………… J [2] Total energy of orbit = kinetic energy + gravitational potential energy. ( )( ) 2 11 23 15215 6 22 1Total energy 2 1 6.67 10 6.4 10 9.6 109.6 10 21372 9.43 10 2.15 10 J GMmmv r − = + −      =  −  =−  11 23 15 Note that it can be shown that (refer to lecture notes): E , E 22 Hence the following alternatives are accepted: a. Total energy 2 6.67 10 6.4 10 9.6 10 (2 KT GMm GMm rr GMm r − = =− =−     =− ( )( ) 22 6 2 215 22 2.2 10 J)9.43 10 1b. Total energy 2 1 9.6 10 2137 2.2 10 J2 KE mv =−  =− =− =−  =−  M1 A1 Comments: Poorly done 1) Many students calculated potential energy instead of total energy. 2) Students also wrote the formulae wrongly, such as using r2 instead of r 3) Others dropped the negative sign, wrongly thinking that total energy must be positive. 4) Careless mistakes made such as dropping the square in kinetic energy formula when they