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WAEC 2008 · Paper 2 · Q16✱

  1. (a)

    A body, decelerating at 0.7 m s−20.7\text{ m s}^{-2}, passes a certain point with velocity 32 m s−132\text{ m s}^{-1}. Find: (i) its velocity after 8 seconds; (ii) the distance covered in that time.

    Separate values with commas, e.g. 3, −2

  2. (b)

    A body PP of mass 60 kg60\text{ kg}, moving with velocity 5 m s−15\text{ m s}^{-1}, collides with another body QQ of mass 50 kg50\text{ kg} moving with velocity 16 m s−116\text{ m s}^{-1} in the opposite direction. After the collision, PP moves with velocity 4 m s−14\text{ m s}^{-1} in its original direction. Calculate the: (i) velocity of QQ immediately after the collision; (ii) time it takes PP to stop if it moves with a constant retardation of 0.25 m s−20.25\text{ m s}^{-2} after the collision.

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

  1. Use v=u+atv = u + at with u=32u = 32, a=−0.7a = -0.7, t=8t = 8: v=32−5.6=26.4 m s−1v = 32 - 5.6 = 26.4\text{ m s}^{-1}.

(ii)

  1. Use s=ut+12at2s = ut + \frac12at^2: s=32(8)−12(0.7)(64)s = 32(8) - \frac12(0.7)(64)
    =256−22.4= 256 - 22.4
    =233.6 m= 233.6\text{ m}.

(b)(i)

  1. Take PP's original direction as positive.
  2. Momentum before: 60(5)+50(−16)=300−800=−50060(5) + 50(-16) = 300 - 800 = -500.
  3. Momentum after: 60(4)+50v=240+50v60(4) + 50v = 240 + 50v.
  4. Momentum is conserved: 240+50v=−500240 + 50v = -500, so 50v=−74050v = -740 and v=−14.8v = -14.8.
  5. So QQ moves at 14.8 m s−114.8\text{ m s}^{-1}, still in its original direction.

(ii)

  1. After the collision PP has u=4u = 4 and a=−0.25a = -0.25.
  2. It stops when v=0v = 0: 0=4−0.25t0 = 4 - 0.25t.
  3. So t=40.25=16t = \frac{4}{0.25} = 16 seconds.

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