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?

1 answer

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.