Although it was reported that majority of the candidates who attempted this question performed badly in it, it was also reported that more candidates attempted this question when compared with question 17 and their performance in this question was also better than what it was in question 17.
In part (a), candidates were expected to apply the principle of conservation of momentum which was m1u1 + m2u2 = m1v1 + m2v2, where m are the masses, u = initial velocity and v = final velocity. Here, u1 = ?, u2 = 0, m1 = 0.084 kg, m2 = 20 kg, v1 = v2 = 0.24ms-1. Therefore, 0.084(u1) + 20(0) = 0.084 (0.24) + 20(0.24). Solving this equation gave u1 = 57.38 ms-1.
In part b, candidates were expected to recall that the body will be momentarily at rest when the velocity = 0. The velocity was obtained by differentiating s with respect to t i.e. = 2t2 – 7t + 5. When = 0, it implied that 2t2 – 7t + 5 = 0. Solving this quadratic equation gave t = 1 and seconds.
When t = , s = - + 5( = 1.0517 m. When t = 1, S = - + 5 = 2.1667 m. Therefore, the required distance was 2.1667 – 1.0517 = 1.12 m, correct to 3 significant figures.
