A player caught a cricket ball of mass $$150 \text{ g}$$ moving at a speed of $$20 \text{ m/s}$$. If the catching process is completed in $$0.1 \text{ s}$$, the magnitude of force exerted by the ball on the hand of the player is:

A cricket ball of mass 150 g moving at 20 m/s is caught in 0.1 s. Find the force on the player's hand.

Recall Newton's second law in terms of impulse: $$F = \frac{\Delta p}{\Delta t}$$ where $$\Delta p$$ is the change in momentum and $$\Delta t$$ is the time interval.

The initial momentum is $$p_i = mv = 0.15 \times 20 = 3$$ kg m/s, and the final momentum is $$p_f = 0$$ (ball is brought to rest). Hence, $$|\Delta p| = |p_f - p_i| = 3 \text{ kg m/s}$$.

Substituting into the formula gives $$F = \frac{|\Delta p|}{\Delta t} = \frac{3}{0.1} = 30 \text{ N}$$.

By Newton's third law, the force exerted by the ball on the player's hand equals the force exerted by the hand on the ball.

The correct answer is Option (4): 30 N.