1. A basketball (m = 625 g) bounces up a horizontal floor. The speeds of the ball before and after the bounce are 4.00 m/s and 2.50 m/s respectively. a) Draw a diagram showing the velocities of the ball before and after the bounce. b) Write down the algebraic expression for the change in momentum of the basketball (assume the upward direction to be positive). c) What is the magnitude and direction of the change in momentum of the basketball?

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### Physics Problem: Basketball Bounce

1. **A basketball (m = 625 g) bounces up a horizontal floor. The speeds of the ball before and after the bounce are 4.00 m/s and 2.50 m/s respectively.**

#### a) Draw a diagram showing the velocities of the ball before and after the bounce.

*Explanation of the Diagram:* 

Before the bounce:
- The basketball is moving downward (negative direction) with a speed of 4.00 m/s.

After the bounce:
- The basketball is moving upward (positive direction) with a speed of 2.50 m/s.

[Diagram not shown here, but should visually depict a downward arrow labeled 4.00 m/s before hitting the floor, and an upward arrow labeled 2.50 m/s after the bounce.]

#### b) Write down the algebraic expression for the change in momentum of the basketball (assume the upward direction to be positive).

**Change in momentum (Δp):**
- Momentum before the bounce: \( p_{\text{before}} = m \cdot v_{\text{before}} = 625 \, \text{g} \cdot (-4.00 \, \text{m/s}) \)
- Momentum after the bounce: \( p_{\text{after}} = m \cdot v_{\text{after}} = 625 \, \text{g} \cdot 2.50 \, \text{m/s} \)
- Change in momentum: \( \Delta p = p_{\text{after}} - p_{\text{before}} \)

#### c) What is the magnitude and direction of the change in momentum of the basketball?

**Calculation:**
Convert mass to kg: \( m = 625 \, \text{g} = 0.625 \, \text{kg} \)
- \( p_{\text{before}} = 0.625 \, \text{kg} \cdot (-4.00 \, \text{m/s}) = -2.50 \, \text{kg} \cdot \text{m/s} \)
- \( p_{\text{after}} = 0.625 \, \text{kg} \cdot 2.50 \, \text{m/s} = 1.5625 \, \text{kg} \cdot \text{m/s} \
Transcribed Image Text:### Physics Problem: Basketball Bounce 1. **A basketball (m = 625 g) bounces up a horizontal floor. The speeds of the ball before and after the bounce are 4.00 m/s and 2.50 m/s respectively.** #### a) Draw a diagram showing the velocities of the ball before and after the bounce. *Explanation of the Diagram:* Before the bounce: - The basketball is moving downward (negative direction) with a speed of 4.00 m/s. After the bounce: - The basketball is moving upward (positive direction) with a speed of 2.50 m/s. [Diagram not shown here, but should visually depict a downward arrow labeled 4.00 m/s before hitting the floor, and an upward arrow labeled 2.50 m/s after the bounce.] #### b) Write down the algebraic expression for the change in momentum of the basketball (assume the upward direction to be positive). **Change in momentum (Δp):** - Momentum before the bounce: \( p_{\text{before}} = m \cdot v_{\text{before}} = 625 \, \text{g} \cdot (-4.00 \, \text{m/s}) \) - Momentum after the bounce: \( p_{\text{after}} = m \cdot v_{\text{after}} = 625 \, \text{g} \cdot 2.50 \, \text{m/s} \) - Change in momentum: \( \Delta p = p_{\text{after}} - p_{\text{before}} \) #### c) What is the magnitude and direction of the change in momentum of the basketball? **Calculation:** Convert mass to kg: \( m = 625 \, \text{g} = 0.625 \, \text{kg} \) - \( p_{\text{before}} = 0.625 \, \text{kg} \cdot (-4.00 \, \text{m/s}) = -2.50 \, \text{kg} \cdot \text{m/s} \) - \( p_{\text{after}} = 0.625 \, \text{kg} \cdot 2.50 \, \text{m/s} = 1.5625 \, \text{kg} \cdot \text{m/s} \
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