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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
General Comments
Weakness/Remedies
Strength
Question 8

Question 8

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                     5 kg                                                                                                         
                       6kg
                      
The diagram shows a light inextensible string that passes over a smooth pulley and carries masses of 6 kg and 5 kg at its ends.  When the system is released from rest:

  1. find the acceleration of each mass.
  2. calculate, correct to two decimal places, the distance moved in 2 seconds.

                                                      [ Take g = 10ms-2]      

_____________________________________________________________________________________________________
Observation

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.

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