Candidates’ performance was also reported to be poor in this question. They were reported not to have applied the correct formulae in part (a) and were also unable to resolve the forces in the part (b) correctly.
In part (a), candidates were expected to recall that Force = mass x acceleration i.e. F = ma. Therefore
a = = = 0.1ms-2. Using the formula s = ut + ½ at2, where u = 0, a = 0.1 and t = 6, S = 0 + ½ (0.1)62 = 1.8m.
In part (b), candidates were expected to represent the given information in a diagram as shown:
T2 T1
30 ° 60°
-------------------
30N
Resolving the forces horizontally gave T2 Cos 30o = T1Cos60o. This implied that T2 = .... (equation 1). Resolving the forces vertically gave T1Sin60 + T2 Sin 30° = 30N........... (equation 2). From (`equation 1),T2 cos 30° = T1 cos 60° i.e. T2 = . This implied that T1 = T2. substituting T2 for T1 in equation 2 gave ( T2 × ) + = 30 i.e. 2T2 = 30. Therefore T2 = = 15.00N. T1 = × 15.00 = 25.98 N.
