Question

A 0.16 kg cricket ball is thrown horizontally at 30. m/s and after it is struck by a cricket paddle, it travels at -30. m/s. If the contact time of the paddle and ball was 0.0007 seconds, what was the average force the paddle exerted on the ball?

Answer 1

To calculate the average force exerted by the paddle on the cricket ball, we can use the impulse-momentum theorem, which states:

\[
\text{Impulse} = \Delta p = F_{\text{avg}} \Delta t
\]

Where:
- \(\Delta p\) is the change in momentum,
- \(F_{\text{avg}}\) is the average force,
- \(\Delta t\) is the contact time.

First, we need to calculate the initial and final momentum of the ball:

1. **Initial momentum (\(p_i\))**:
- Mass of the ball, \(m = 0.16 \, \text{kg}\)
- Initial velocity, \(v_i = 30 \, \text{m/s}\)

\[
p_i = m \cdot v_i = 0.16 \, \text{kg} \cdot 30 \, \text{m/s} = 4.8 \, \text{kg m/s}
\]

2. **Final momentum (\(p_f\))**:
- Final velocity, \(v_f = -30 \, \text{m/s}\)

\[
p_f = m \cdot v_f = 0.16 \, \text{kg} \cdot (-30) \, \text{m/s} = -4.8 \, \text{kg m/s}
\]

3. **Change in momentum (\(\Delta p\))**:
\[
\Delta p = p_f - p_i = -4.8 \, \text{kg m/s} - 4.8 \, \text{kg m/s} = -9.6 \, \text{kg m/s}
\]

4. **Now, use the change in momentum to find the average force**. Given the contact time \(\Delta t = 0.0007 \, \text{s}\):

\[
F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{-9.6 \, \text{kg m/s}}{0.0007 \, \text{s}} \approx -13714.29 \, \text{N}
\]

Thus, the average force exerted by the paddle on the ball is approximately:

\[
\boxed{-13714 \, \text{N}}
\]

The negative sign indicates that the force exerted by the paddle is in the opposite direction of the ball's initial motion.