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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).](https://content.bartleby.com/qna-images/question/674187d8-06e8-4f69-af48-239b1a5ca06a/b9524815-675b-4448-9385-fa1cd08a7ca9/siil5d_thumbnail.jpeg)
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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