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Further Mathematics Paper 2, Nov/Dec. 2008  
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 16

(a) A body, decelerating at 0.7ms-2 passes a certain point with velocity 32ms-1.  Find
            (i)  its velocity  after 8 seconds
            (ii)  the distance covered.

(b) A body P of mass 60kg, moving with velocity 5ms-1, collides with another body Q, of mass 50kg, moving with velocity 16ms-1 in the opposite direction.  After collision, P moves with velocity 4ms-1 in its  direction.  Calculate the:

  • velocity of Q immediately after the collision;
  • time it takes P to stop if it moves with a constant retardation of 0.25ms-2 after the collision.
_____________________________________________________________________________________________________
Observation

Candidates’ performance in the part (a) of the question was fair.  They were able to apply the appropriate equations of motion thus –
v = u+at,  u = 32 m/s, a = -0.7ms-2  and t = 8.
Thus v = 32 – 5.6  = 26.4 m/s
S = ut + ½ at2  =  32(8) - ½ (o.7) 64   =  233.6m

In the part (b), candidates’ performance was very poor.  They were not able to apply the momentum equation correctly.  They were expected to show that if v is the velocity after collision, then
60 x5 – 50 x 16 = 60 x 4 + 50v where v  =  -740   =  14.8 m/s in its direction. The time it
                                                                         50
will take P to come to  rest after collision if it is travelling with a constant retardation of 0.25m/s2 is given by v = u + at, where v = 0.  Therefore 0 = 14.8 - 0.25t.  Hence t =   14.8
                                                                                                      0.25
      = 59.2 seconds.

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