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 answ