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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 18

(a)       A bullet of mass 0.084kg is fired horizontally into a stationary block of mass 20kg on a smooth                      horizontal floor. If they both move with a velocity of 0.24ms-1 after impact, calculate, correct            to two decimal places, the initial velocity of the bullet.

(b)       The distances, s, travelled by a body at anytime t seconds is given by s = - + 5t. Calculate,             correct to three significant figures, the distance between the points when the body is             momentarily at rest.

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Observation

Although it was reported that majority of the candidates who attempted this question performed badly in it, it was also reported that more candidates attempted this question when compared with question 17 and their performance in this question was also better than what it was in question 17.

In part (a), candidates were expected to apply the principle of conservation of momentum which was m1u1   + m2u2 = m1v1 + m2v2, where m are the masses, u = initial velocity and v = final velocity.  Here,        u1 = ?, u2 = 0, m1 = 0.084 kg, m2 = 20 kg, v1 = v2 = 0.24ms-1.  Therefore, 0.084(u1) + 20(0) = 0.084 (0.24) + 20(0.24).  Solving this equation gave u1 = 57.38 ms-1.

In part b, candidates were expected to recall that the body will be momentarily at rest when the velocity = 0.  The velocity was obtained by differentiating s with respect to t  i.e.   = 2t2 – 7t + 5.  When  = 0, it implied that 2t2 – 7t + 5 = 0.  Solving this quadratic equation gave t = 1 and  seconds.
When t = , s =   -    + 5( = 1.0517 m.  When t = 1, S =   -   + 5  =  2.1667 m.  Therefore, the required distance was 2.1667 – 1.0517 = 1.12 m, correct to 3 significant figures.