College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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**Physics Problem: Impulse and Force Exerted on a Volleyball**

A 0.266-kg volleyball approaches a player horizontally with a speed of 16.1 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of 22.4 m/s.

**(a)** What impulse is delivered to the ball by the player? (Take the direction of final velocity to be the positive direction. Indicate the direction with the sign of your answer.)

\[ \text{Impulse} = \ \underline{\hspace{1cm}} \ \text{kg} \cdot \text{m/s} \]

**(b)** If the player's fist is in contact with the ball for 0.0600 s, find the magnitude of the average force exerted on the player's fist.

\[ \text{Average Force} = \ \underline{\hspace{1cm}} \ \text{N} \]

### Explanation:
- **Impulse (a)**: Impulse is defined as the change in momentum of an object when a force is applied over a period of time. It is given by the equation:

\[ \text{Impulse} (J) = m \cdot (v_f - v_i) \]

Where:
  - \( m \) is the mass of the ball (0.266 kg),
  - \( v_f \) is the final velocity (-22.4 m/s, taking the opposite direction as positive),
  - \( v_i \) is the initial velocity (16.1 m/s).

- **Average Force (b)**: The average force exerted can be calculated using the impulse-momentum theorem:

\[ \text{Average Force} (F_{avg}) = \frac{\text{Impulse}}{\Delta t} \]

Where:
  - \(\Delta t\) is the time interval (0.0600 s).
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Transcribed Image Text:**Physics Problem: Impulse and Force Exerted on a Volleyball** A 0.266-kg volleyball approaches a player horizontally with a speed of 16.1 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of 22.4 m/s. **(a)** What impulse is delivered to the ball by the player? (Take the direction of final velocity to be the positive direction. Indicate the direction with the sign of your answer.) \[ \text{Impulse} = \ \underline{\hspace{1cm}} \ \text{kg} \cdot \text{m/s} \] **(b)** If the player's fist is in contact with the ball for 0.0600 s, find the magnitude of the average force exerted on the player's fist. \[ \text{Average Force} = \ \underline{\hspace{1cm}} \ \text{N} \] ### Explanation: - **Impulse (a)**: Impulse is defined as the change in momentum of an object when a force is applied over a period of time. It is given by the equation: \[ \text{Impulse} (J) = m \cdot (v_f - v_i) \] Where: - \( m \) is the mass of the ball (0.266 kg), - \( v_f \) is the final velocity (-22.4 m/s, taking the opposite direction as positive), - \( v_i \) is the initial velocity (16.1 m/s). - **Average Force (b)**: The average force exerted can be calculated using the impulse-momentum theorem: \[ \text{Average Force} (F_{avg}) = \frac{\text{Impulse}}{\Delta t} \] Where: - \(\Delta t\) is the time interval (0.0600 s).
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