2021 HCI H3 Physics Prelim 2021 Solutions Marker sComments

9814 H3 Physics Prelim 2021 Suggested Solutions and Markers’ Comments 1 (a) 0 0 0 0 0 (3 ) CC AA CA C PV PV TT P V PV TT   TC = 3T0 A1 Comments: Quite well-done. Majority of candidates managed to get the correct expression for temperature at C. However, the approaches varied. Most used PV=nRT instead of the suggested method. It is expected that candidates compare temperatures at A and C, but there were a few compared temperatures at B and C, and gave the relationship without clear working - the mark was awarded, given the benefit of doubt. (b) Work done by gas during process A to B = 0 Work done by gas during process B to C Work done by gas during process C to A = - Po (VA - VC)  = - Po (3Vo - Vo) = - 2PoVo Net work done by gas = 3nRTo ln 3 - 2PoVo = 3nRTo ln 3 - 2nRTo since PoVo = nRTo = nRTo (3 ln 3 - 2) B1 B1 B1 A0 Comments: Quite well done. Most students could show the correct expression for net work done by gas. We preferred that candidates explicitly state that the net work done is the sum of work done by gas during each process AB, BC, CA before showing the math. More candidates had difficulty determining the work done by gas during process BC, as it required integration. There were signs that some candidates actually “reverse-engineered” the proof. (c) A to B: By First Law of Thermodynamics, as w = 0, U = q Since temperature increases, U and hence q are positive i.e. heat is added to the gas. B to C: as temperature is constant, U = 0, q = - w B1 3 3 ln 3 ln3 o o o o V B V V B V o PdV nRT dVV nRT V nRT      

2 © Hwa Chong Institution 9814 H3 Physics / Prelim 2021 Since gas is expanding, work done on the gas, w is negative. To keep U = 0, q must be positive i.e. heat must be added to the gas. C to A: Temperature decreases, U is negative, gas is compressed, hence w is positive. Hence q must be negative i.e. heat is lost. B1 B1 Comments: Quite well done. Most candidates were able to clearly explain whether heat is supplied during each process, analysing work done by gas using the change in volume, the change in internal energy using change in temperature or PV, and applying First Law of Thermodynamics. Candidates are reminded to explain/define all notations. Marks are deducted for incomplete/wrong explanations and undefined notations. 2 (a) By Newton’s 3rd law, the tension force (F) acting on both springs is the same. 1 1 2 2F k x k x The total extension/compression of the springs is 12e x x It can be found that 2 1 12 kxe kk  Since the system performs simple harmonic motion, F ma , where a is the acceleration of the oscillation.   2 2 11 2 122 1 1 2 1 2 2 2 2 k x m e m e T m k kkk e m e Tk k T k k            B1 B1 B1 B1 Comments: There were a handful of remarkable, detailed solutions seen. Thank you very much! Many only stated that the relationship for the effective spring constant connected in series and did not really show how it was derived using Newton’s 2nd Law and applying simple harmonic motion. A handful of students did not mention at all about any forces involved in the spring- mass system in their workings. They just did a “reverse-engineering”. (b) As the objects move down the ramp, by PCOE, the gravitational potential energy will be converted to rotational kinetic energy (because they are rolling without slipping) and translational kinetic energy of the objects. The greater the moment of inertia, the greater the rotational kinetic energy. This results in less energy to be converted to translational kinetic energy. The hollow cylinder has the greatest moment of inertia ( 2MR ). It has the greatest rotational kinetic energy and the least translational kinetic energy. Hence it moves down the ramp the slowest. B1 B1 B1

3 © Hwa Chong Institution 9814 H3 Physics / Prelim 2021 Likewise, the solid sphere has the least moment of inertia ( 22 5 MR ). It has the least rotational kinetic energy and the greatest translational kinetic energy. Hence, it moves down the ramp the fastest. Therefore, the solid sphere (C) would reach the bottom first. (The moment of inertia of solid cylinder is 2 2 MR ) (Energy loss due to the rolling friction is negligible.) Alternatively,   22 2 1100 22 2 1/ i i f f CM CM CM CM KE GPE KE GPE Mgh I mv ghv I MR             The greater the moment of inertia, the less the centre of mass velocity and the translational kinetic energy of the object. Hence, the hollow cylinder moves down the slowest and the solid sphere moves down the fastest. Comments: Many students mentioned that the GPE is converted to both rotational KE and translational KE as the objects roll down and figured out the role played by the moment of inertia in their rotations. 3 (a) Kepler’s Third Law is 2 23 4 GM Ta   1 solar mass =     322 3 22 1 AU44 1 year solarMa GGT  Leaving everything in solar-system units, we get     33 6 22 919 3.68 10 14.53 aM T    solar masses M1 A1 Comments: Most could recall the Kepler’s Third law which was also printed in the formulae page. A significant number of students did not cube or square the respective variables in the equations. Some were troubled in converting their answer to solar masses. (b)(i) Since M is very large, we may assume it to be stationary, while m is in orbit around M. The total energy Etotal is the sum of the kinetic energy of m and the gravitational potential energy of the system, 21 2 total GMmE mv r A1

4 © Hwa Chong Institution 9814 H3 Physics / Prelim 2021 Comments: Surprisingly, a significant number of students just expressed the total energy for a circular orbit in H2 physics. The question did not specify that the orbit was circular, hence a general expression for the total energy was expected. (b)(ii) It is given that,  eff total rU E KE . Decomposing the velocity v in its radial and tangential components vr and vt, we get 2211 22 total r t GMmE mv mv r   Using that the angular momentum L = r x p = rmv sin θ, where v sin θ = vt is the perpendicular component of the motion, we get 2 2 2 1 2 2 total r L GMmE mv rmr   Taking   21 2 rrKE mv , we find that 2 22 eff L GMmU rmr A1 A1 A1 Comments: “Reversed-engineering” students had a hard time to find the effective potential. Some just threw in L into their equations without properly using the concept of angular momentum. (c)(i) Determining the escape velocity vesc by applying the principle of conservation of energy, total energy at altitude of 160 000 m = total energy at infinity 21 02 GMmmv r      11 22 1 56 2 6.67 10 7.36 102 2273 m s 1.6 10 1.74 10 esc GMv r         Since the speed of the Apollo 11 was larger than the escape speed from an altitude of 160 000 m, the spacecraft would not stay within the Moon’s gravitational well. B1 A1 Comments: Most of the students recognized that the Apollo 11’s total energy must be zero or greater than zero and showed properly that its speed is much larger than the escape speed from the Moon’s gravitational well. (c)(ii) By the principle of conservation of angular momentum, A A P Pmv r mv r 56(1670)(1.1 10 1.74 10 ) AAvr    93.09 10 A A r v  (