Question

(a) Explain why a cricket player moves his hands backward while catching a fast cricket ball.
​(b) A 150 g ball, travelling at 30 m/s, strikes the palm of a player's hand and is stopped in 0.05 second. Find the force exerted by the ball on the hand.

Answer 1

(a) A player moves his hands backward to reduce the effect of the force of fast cricket ball on his hands. A fast-moving ball has a high momentum, and when the player catches the ball, the momentum of the ball changes to zero in a very short span of time. That's why it hits his hands. Hence, to reduce the effect of the force, the player pulls his hands back so that he can increase the time in which the momentum of the ball falls to zero. By increasing the time, rate of change of momentum decreases which impacts less force on the hands of the player.

(b) Mass of the ball = 150 g = 0.15 kg

Initial velocity of the ball = 30 m/s

Final velocity of the ball = 0 m/s

Time taken by the ball to stop = 0.05 s

Force exerted by the ball on the hand of the player = Rate of change of momentum of the ball = $\frac{\mathrm{Initial}\mathrm{momentum}-\mathrm{Final}\mathrm{momentum}}{\mathrm{Time}\mathrm{taken}}$

Force $=\frac{0.15\mathrm{kg}\times 30\mathrm{m}/\mathrm{s}-0.15\mathrm{kg}\times 0}{0.05\mathrm{s}}=\frac{4.5\mathrm{kg}\mathrm{m}/\mathrm{s}}{0.05\mathrm{s}}=90\mathrm{N}$

Hence, the force exerted by the ball on the hand of the player is 90 N.