WAEC 2019 · Paper 2 · Q7

(a)
A ball of mass $0.1\text{ kg}$ approaching a tennis player with a velocity of $10\text{ m s}^{-1}$ is hit in the opposite direction with a velocity of $15\text{ m s}^{-1}$ . If the time of impact between the racket and the ball is $0.01$ second, calculate the magnitude of the force with which the ball was hit.

Worked solution (try it first)

Take the direction the ball is hit as positive: $u = -10\text{ m s}^{-1}$ and $v = 15\text{ m s}^{-1}$ .
Change in momentum $= m(v - u)$
$= 0.1(15 - (-10))$
$= 2.5\text{ N s}$ .
$F = \dfrac{\text{change in momentum}}{t}$
$= \dfrac{2.5}{0.01}$
$= 250\text{ N}$ .