This question was reported to be poorly answered by majority of the candidates who attempted it. Candidates were expected to convert the masses to weight. Thus, the 6 kg mass gave a weight of 6 x 10 = 60N while the weight of the 5 kg mass was 50N. Since the 6 kg mass is heavier, it is likely to pull the 5 kg mass towards itself. Therefore, if T is the tension in the string, the effective force moving downwards was 60 – T = 6a ....(eqn. 1). Similarly, the effective force pulling the 5 kg mass upwards was T – 50 = 5a ....(eqn.2). Solving these two equations simultaneously gave 10 = 11a, hence a = or 0.91 ms-2.
Candidates were expected to calculate the distance by applying the formula s = ut + ½ at2 where s = distance, u = initial velocity = 0 since the body was initially at rest, t = 2 secs and a = or 0.91 ms-2. Therefore s = 0(2) + ½ 22 = or 1.82m.
