Further Mathematics Paper 2, Nov/Dec. 2012
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
Weakness/Remedies
Strength
Question 18

(a)        Two bodies in motion reach a point at the same time with velocities of 8 ms-1 and 22 ms-1 and accelerations of 20 ms-2 and 18 ms-2 respectively.  In what time will the bodies be 15 m apart?
(b)        Two balls of masses 25g and 15g moving in opposite directions with speeds of 8 ms-1 and 3 ms-1 respectively, collide.  After collision, the 25g ball continues in its original direction with a speed of 5 ms-1. Calculate the change in momentum of the 15 g ball due to the collision.

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Observation

The Chief Examiner also reported that this question was poorly attempted by majority of
the candidates who attempted it.  Candidates were reported not to apply the distance
formula correctly and their responses indicated poor knowledge of conservation of linear
momentum.

In part(a), candidates were expected to respond as follows:
Distance covered by first body = S1 = u1t + ½ a1t2  = 8t + 10t2
Distance covered by second body = S2 = u2t + ½a2t2 = 22t + 9t2
Since the distance between them is 15 m, it implied that S1 – S2 = 15m                                                                              i.e. 10t2 + 8t -  which gave t2 – 14t = 15.  Solving this quadratic             equation gave t = 15 seconds.

In part (b), candidates’ expected response was:

Total momentum before impact = 25 x 8 + 15(-3)  = 155 gms-1
Total momentum after impact   = 25 x 5 + 15v = (125 + 15v) gms-1
Since momentum is conserved, then 155 = 125 + 15v.  Solving this equation gave
v = 2 ms-1.  The change in momentum of the 15 g ball was given by m(v-u)                                      = 15 = 15 x 5 = 75 gms-1.