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.) kg • 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. N
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.) kg • 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. N
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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).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F674187d8-06e8-4f69-af48-239b1a5ca06a%2Fb9524815-675b-4448-9385-fa1cd08a7ca9%2Fsiil5d.jpeg&w=3840&q=75)
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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