To calculate the average force the player exerted on the ball, we can use the impulse-momentum theorem, which states that the impulse (force times time) equals the change in momentum of the object.

Given Data:

• Mass of the ball, $m = 0.25 \, \text{kg}$
• Initial speed of the ball, $v_i = 20 \, \text{m/s}$ (in the original direction)
• Final speed of the ball, $v_f = -25 \, \text{m/s}$ (opposite direction, hence negative)
• Contact time, $\Delta t = 5 \, \text{ms} = 5 \times 10^{-3} \, \text{s}$

Step 1: Calculate the Change in Momentum

The change in momentum $\Delta p$ is given by: $\Delta p = m \cdot (v_f - v_i)$ Substitute the values: $\Delta p = 0.25 \times (-25 - 20)$ $\Delta p = 0.25 \times (-45)$ $\Delta p = -11.25 \, \text{kg} \cdot \text{m/s}$

Step 2: Calculate the Average Force

Using the impulse-momentum theorem: $F_{\text{avg}} = \frac{\Delta p}{\Delta t}$ Substitute the values: $F_{\text{avg}} = \frac{-11.25}{5 \times 10^{-3}}$ $F_{\text{avg}} = -2250 \, \text{N}$

Answer:

The average force the player exerted on the ball is 2250 N in the opposite direction of the ball's initial motion.

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Would you like further explanation or have any questions?

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Here are 5 related questions to deepen understanding:

1. How would the force change if the contact time was doubled?
2. What happens to the momentum of the ball after it’s kicked?
3. How would the answer differ if the ball's mass were 0.5 kg?
4. Can we calculate the average acceleration of the ball during contact?
5. How would we solve this if we were only given the impulse applied?

Tip: When objects change direction, remember to account for direction as positive or negative to avoid errors in calculations.