Observation

The Chief Examiner reported that candidates performed well in this question. Majority of them were reported to have applied the formulae correctly and showed good understanding of the topic. In part (a), candidates were expected to obtain the effective force first and then use it to calculate the acceleration. The effective force, F = tension in the rope – weight of the bucket = 150 – 80 = 70 N. Candidates would then apply the formula F = ma, where F = effective force, m = mass and a = acceleration. Here, F = 70 N and m = 8 kg.Therefore, a =  =  = 8.75 ms-2. In part (b)(i), candidates were expected to calculate the acceleration of the mass first. This was obtained using the formula v2 = u2 + 2as, where v = final velocity, u = initial velocity, a = acceleration and s = distance covered. Here, v = 25 ms-1, u = 15 ms-1 and s = 50 m. Therefore,a =  =  =  = 4 ms-2. The required force, F, would now be calculated using the formula F = ma. Here, m = 12 kg, a = 4 ms-2. Therefore, F = 4 × 12 = 48 N. In part (b)(ii), the appropriate formula to use was v2 = u2 + 2as. By making s the subject of the formula we have s = . Here, v = 35 ms-1, u = 15 ms-1 and a = 4 ms-2. Therefore, s =  =  = 125 m. This meant that the distance covered when the speed was 35 ms-1 was 125 m.