2019 RI Prelims Paper 3 Soln

© Raffles Institution [Turn over 2019 Preliminary Examinations H2 Physics Paper 3 Solutions 1 (a) The moment of a force about a point is defined as the product of the force and the perpendicular distance from the point to the line of action of the force. B1 B1 (b) (i) B1 (ii) The weight of the load produces an anti-clockwise moment about P. The counterweight produces an addi tional clockwise moment that can balance the moment due to heavier loads. B1 B1 (iii) Taking moments about P, W  (20  3.2) + 5.0  104  (7.6  3.2) = 3.5  105  3.2 + 8.4  104  (3.2 + 3.2) W = 8.56  104 N M1 A1 Marker’s Comments: 1 (b) (iii) The normal contact force acting on the crane at P is unknown. Students who apply the principle of moments about any other position needs to include this normal contact force in their equation to obtain the correct answer. 2 (a) The gravitational potential at a point is the work done per unit mass by an external force in bringing a small test mass from infinity to that point. *B1 is not awarded if student wrote: - energy instead of work done - ob ject instead of mass B1 B1 (b) *There are a few ways to get x, the easiest way is to equate the gravitational field strengths due to Earth and Moon because at Q, the resultant gravitational field strength is zero.  Earth Moon 22 8 8 22 8 24 8 3.8 10 3.8 10 7.4 103.8 10 6.0 10 3.4 10 m EM M E gg GM GM x x Mx xM xx x          OR M1 M1 P

2 © Raffles Institution       6 8 11 22 11 24 6 8 8 1.3 10 3.8 10 6.67 10 7.4 10 6.67 10 6.0 10 1.3 10 3.8 10 3.4 10 m QM E MEGM GM xx xx x                      M1 M1 (c) (i) The resultant force due to Earth and Moon on the satellite provides the centripetal force.    net 2 2 8 1.22 21.22 520 27 24 60 60 3.2 10 m CFF mr d d       Or (this method is too time consuming to solve)     Earth Moon 22 8 24 22 11 22 8 8 1.22 3.8 10 6.0 10 7.4 101.22 6.67 10 520 3.8 10 3.2 10 m net EM FF F GM m GM m d d d d d           M1 M1 M1 M1 (ii)         11 24 11 22 88 8 6- 1 6.67 10 6.0 10 6.67 10 7.4 10 3.2 10 3.8 10 3.2 10 1.33 10 J kg total E S EM EM GM GM dd                   *1 mark if student only found the potential due to the Earth or the Moon. C1 M1 A1 (c) The minimum energy required should be the difference in potential energy between Q and (c)(ii). Or the energy required should be just enough to reach Q. B1 Marker’s Comments: 2 (b) As this is a show question, you just need to show the expression followed by the proper substitution. After that, you can use your calculator to get the final answer. You do not need to show the mathematical steps on how you arrive at your final answer. Do note that for show questions, you need to show all substitution values instead of leaving them in notations e.g. for mass of Earth, you need to show 7.4 × 10 22 instead of leaving it as ME.

3 © Raffles Institution [Turn over You are not advised to prove by substi tuting the value that you are supposed to show and conclude that the left hand side of the equation is equal to the right hand side. (c) (i) Similar comments as 2(b). (ii) There are several students who did not know that gravitational potential is ALWAYS negative and left out the negative sign in their calculations. The total gravitational potential due to Earth and Moon at the location of the satellite is the additional of the gravitational potential due to Earth alone and the potential due to Moon alone. Man y students found the difference instead. 3 (a) (i) A stationary wave is formed when two progressive waves of the same frequency (wavelength), speed and amplitude, moving along the same line towards each other, meet and superpose. B1 B1 (ii) 1. shape of wave M1 (antinode at S 1 and S2) correct label M1 2. (i) At M, the path difference betw een the two microwaves reaching M is the separation between S 1 and S 2, which is equal to 0.12/0.040 = 3 wavelengths. Hence the waves will produce a constructive interference and an intensity maximum is formed at M. *in fact, M is at the position of 3rd order maximum. M1 A1 (ii) Either (2× 4) + 4 (due to two 0th orders and two 3rd orders) = 12 OR (5 × 2) + 2 = 12 *5 Antinodes correspond to 5 antinodal lines, and 2 more along S 1S2 line B1 S1 S2 1.2 m A A A A A A A N N N N N N

4 © Raffles Institution (b) (i) (ii) 3 9 4 sin for very small angles sin tan 2.3 10 2 550 10tan 1.2 5.74 10 m d dd d                B1 Marker’s comments: 3 (a) (i) Conditions required to form a stationary wave were generally not well stated. Quite a few students gave the conditions to form an observable interference pattern instead. (ii) 1. Very very few students scored 2/2 for this part. Common mistakes were:  Majority drew nodes at S 1 and S2 instead of antinodes.  Quite a few students did not draw the stationary wave correctly (that is, one solid curve, and one dotted curve of opposite phase).  Forgot to label A and N for antinodes and nodes.  Labelled A and N wrongly  Wrong number of half-wavel ength segments between the two sources. 2. (i) Generally well done for this part. Some students took 0.18 m as the path difference, and deduced a minimum instead (no ecf for this.) (ii) Very few students got the correct answer of 12 maxima. 3 (b) (i) Very few well-drawn curves. Should include the two secondary maxima as well, with equal spacing between the centre and the 1 st minimum, and between the 1 st and 2 nd minimum. The intensities of the secondary maxima should be much lower than the central maximum. (Do refer to your lecture notes for the correct drawing.) (ii) Not many showed clear working using correct equations. On the whole, this question was not well scored. intensity O distance along screen Correct shape, B1 First and second secondary maxima shown; same distance between central max and first min, and between first min and second min B1

5 © Raffles Institution [Turn over 4 (a) The First Law of Thermodynamics states that the increase in the internal energy of a system is equal to the sum of the heat supplied to the system and the work done on the system, and the internal energy of a system depends only on its state. B1 B1 (b) (i) Work done by gas = p  V = 1.00  105  (4.00  2.00)  103 = 200 J M1 A1 (ii) change work done on gas / J heat supplied to gas / J Increase in internal energy / J A to B 310 0 310 B1 B to C 0  610  610 B1 C to A  200 500 300 B1 Marker’s Comments: 4 (a) This part was not very well-answered. If “change in internal energy” is used, then “the change in internal energy is equal to the sum of the heat transfer to/from the system and the work done on/by the system”. So its easier to just say “increase in internal energy”. Some students forgot to include the 2 nd part “the internal energy of a system depends only on its state”. (b) (i) Some students forgot to write “work done = pV” (ii) This part was generally alright. 5 (a) The resistance of a resistor is defined as the ratio of the potential difference across it to the current flowing through it. B1 (b) (i) Since the value of R is comparable to the resistance of the voltmeter, only a fraction