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 Further Mathematics Paper 2, Nov/Dec. 2013
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
Weakness/Remedies
Strength
Question 17

(a)        A lorry of mass 1200 kg was kept in motion by a force of 120N. If the lorry was initially at              rest, calculate the distance covered in 6 seconds.

(b)        An object weighing 30N is kept in equilibrium by two strings inclined at 30o and 60° to the              horizontal. Find the tensions in the strings.

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Observation

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.