2021 RI Prelims H2 Phy Paper 3 Soln

© Raffles Institution [Turn over 2021 Raffles Institution Preliminary Examinations – H2 Physics Paper 3 – Suggested Solutions 1 (a) vut a  Sub into the 2nd equation    22 22 1 22 2 2 vu vuvusu v aa as v u vu a s      B1 Students are required to simplify the equation and not leave it in its raw form (eg   1 22 vu vuvusu v aa    and expect to get the full mark. (b) Displacement required 0.233.0 2.1 1.015 m2  Minimum speed of release required means that velocity at 1.015 m is zero.   22 2 1 2 0 2 9.81 1.015 4.46 m s vu a s u u     B1 M1 A1 OR, using conservation of energy Loss in kinetic energy = Gain in GPE     2 2 2 1 1 02 1 2 2 9.81 1.015 4.46 m s yf i yf i y y y mu mgh mgh ug h hg s u u        The displacement of the ball has to be taken with reference to a fixed point on the ball. Students are required to either calculate the di splacement of the ball from the centre of the ball as it is released (2.1 m from the ground) to the centre of the ball as it enters the net (3.0 m from the ground + radius of the ball); or from the base of the ball (2.1 m from the ground minus the radius of the ball) to the hoop (3.0 m from the ground). A few students assumed height to be displacement. (c) (i) Horizontally  1 cos 6.7 1.3 cos50 6.7 8.02 m s1.3cos50 xxsu t u t u u       M1

2 © Raffles Institution (ii) Vertically       22 2 11 sin50 9.81 1.322 18.0sin50 1.3 9.81 1.32 0.3226 m yy ysu t a tu t          height = 2.5  0.3226 = 2.18 m M1 A1 A1 Students should realise that displacement is a vector that takes into account the direction of velocity. Thus, there is no need to calculate the displacement to the max height and then the displacement again from the max height before adding everything together to find the final displacement. Again, some students forgot to include the negative sign for acceleration (assuming up as positive) and ended up with 8.65 m (this is approximately 3 storeys high) and yet did not think that this number was plausible. Some students also assumed that the ball travelled at constant speed. If this is so, the ball will not return to the ground. 2 (a) Gravitational potential at infinity is zero. Since gravitational force is attractive in nat ure, to bring a mass from infinity to a point in the gravitational field, the direction of the external force is opposite to the direction of displacement of the mass. This results in negative work done by the external force B1 B1 It is important that students include a statement that the work done by the external force is negative, rather than ending the explanation as external force is opposite in direction to the displacement of the test mass from infinity to the point in the gravitational field. A common mistake amon