A cricket ball has mass of 0.16 kg and it hits the bat with a speed of 25m/s after being in contact with the bat for 0.0013s the ball rebounces with a speed of 22m/s in the exact opposite direction.

To solve this problem, we need to calculate the change in momentum of the cricket ball and the average force exerted on the ball by the bat.

1. Break Down the Problem

   1. Calculate the initial momentum of the ball.
   2. Calculate the final momentum of the ball.
   3. Determine the change in momentum.
   4. Calculate the average force exerted on the ball.

2. Relevant Concepts

   1. Momentum: $p = mv$, where $m$ is mass and $v$ is velocity.
   2. Change in Momentum: $\Delta p = p_{\text{final}} - p_{\text{initial}}$.
   3. Impulse-Momentum Theorem: $\Delta p = F_{\text{avg}} \cdot \Delta t$, where $F_{\text{avg}}$ is the average force and $\Delta t$ is the time duration.

3. Analysis and Detail

   1. Initial Momentum: $p_{\text{initial}} = m \cdot v_{\text{initial}} = 0.16 \, \text{kg} \times 25 \, \text{m/s} = 4 \, \text{kg} \cdot \text{m/s}$
   2. Final Momentum: $p_{\text{final}} = m \cdot v_{\text{final}} = 0.16 \, \text{kg} \times (-22) \, \text{m/s} = -3.52 \, \text{kg} \cdot \text{m/s}$ (Note: The velocity is negative because the ball rebounds in the opposite direction.)
   3. Change in Momentum: $\Delta p = p_{\text{final}} - p_{\text{initial}} = -3.52 \, \text{kg} \cdot \text{m/s} - 4 \, \text{kg} \cdot \text{m/s} = -7.52 \, \text{kg} \cdot \text{m/s}$
   4. Average Force: $F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{-7.52 \, \text{kg} \cdot \text{m/s}}{0.0013 \, \text{s}} = -5784.62 \, \text{N}$

4. Verify and Summarize

   • The calculations show that the change in momentum is $-7.52 \, \text{kg} \cdot \text{m/s}$.
   • The negative sign in the average force indicates the direction of the force is opposite to the initial direction of the ball.

Final Answer

The average force exerted on the ball by the bat is approximately $-5784.62 \, \text{N}$.