Q.A cricket ball of mass $0.15\ \text{kg}$ moving at $12\ \text{m/s}$ towards a batsman is hit back along the same line at $20\ \text{m/s}$ in the opposite direction. If the bat remains in contact with the ball for $0.01\ \text{s}$, find the impulse imparted to the ball and the average force exerted by the bat on it.

Given: mass $m = 0.15\ \text{kg}$, contact time $\Delta t = 0.01\ \text{s}$. Taking the ball's original direction of motion (toward the batsman) as positive, the initial velocity is $u = +12\ \text{m/s}$; since the ball is hit straight back along the same line, its final velocity is $v = -20\ \text{m/s}$.

Change in momentum: $\Delta p = m(v - u) = 0.15\,[(-20) - (12)] = 0.15 \times (-32) = -4.8\ \text{kg}\,\text{m/s}$ The magnitude of the impulse delivered to the ball is therefore $4.8\ \text{kg}\,\text{m/s}$, directed away from the bowler (opposite to the ball's original motion) -- exactly matching the direction of $\Delta p$.

Average force: by the impulse-momentum theorem, $J = F_{\text{avg}}\,\Delta t = \Delta p$, so $F_{\text{avg}} = \frac{|\Delta p|}{\Delta t} = \frac{4.8\ \text{kg}\,\text{m/s}}{0.01\ \text{s}} = 480\ \text{N}$

Assign a sign convention (ball's original direction = positive), compute $\Delta p = m(v-u)$ using the reversed final velocity as negative, then divide by the contact time to get the average force.

• Treating the ball's speed as unchanged (just $12\ \text{m/s}$ or $20\ \text{m/s}$) instead of correctly subtracting a negative final velocity from a positive initial one, which understates the true change in momentum.
• Forgetting that reversing direction, not just changing speed, is what makes the momentum change so large.