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WAEC 2012 · Paper 2 · Q18✱

  1. (a)

    Two bodies in motion reach a point at the same time with velocities of 88 m s−1^{-1} and 2222 m s−1^{-1} and accelerations of 2020 m s−2^{-2} and 1818 m s−2^{-2} respectively. After what time will the first body be 15 m ahead of the second?

  2. (b)

    Two balls of masses 25 g and 15 g moving in opposite directions with speeds of 88 m s−1^{-1} and 33 m s−1^{-1} respectively, collide. After collision, the 25 g ball continues in its original direction with a speed of 55 m s−1^{-1}. Calculate the change in momentum of the 15 g ball due to the collision.

Worked solution (try it first)

(a)

  1. Distance of the first body after tt s: s1=8t+12(20)t2=8t+10t2s_1 = 8t + \frac12(20)t^2 = 8t + 10t^2.
  2. Distance of the second body: s2=22t+12(18)t2=22t+9t2s_2 = 22t + \frac12(18)t^2 = 22t + 9t^2.
  3. The first body is 15 m ahead: s1−s2=15s_1 - s_2 = 15, so t2−14t=15t^2 - 14t = 15.
  4. Rearrange and factorise: t2−14t−15=0t^2 - 14t - 15 = 0, i.e. (t−15)(t+1)=0(t - 15)(t + 1) = 0.
  5. Time is positive, so t=15t = 15 s.

(b)

  1. Take the 25 g ball's direction as positive.
  2. Momentum before: 25×8+15×(−3)=15525 \times 8 + 15 \times (-3) = 155 g m s−1^{-1}.
  3. Momentum after: 25×5+15v=125+15v25 \times 5 + 15v = 125 + 15v.
  4. Momentum is conserved: 125+15v=155125 + 15v = 155, so v=2v = 2 m s−1^{-1}.
  5. Change in momentum of the 15 g ball: 15(2−(−3))=7515(2 - (-3)) = 75 g m s−1^{-1} (0.0750.075 kg m s−1^{-1}).

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