College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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![### Momentum and Impulse
Momentum \((p)\) and impulse \((J)\) are important concepts in physics, especially in the context of collisions and interactions between objects.
#### Equations
1. **Momentum Equation:**
\[
p = mv
\]
- \(p\) = momentum
- \(m\) = mass
- \(v\) = velocity
2. **Impulse Equation:**
\[
J = Ft
\]
- \(J\) = impulse
- \(F\) = force
- \(t\) = time
3. **Alternate Form of Impulse Equation:**
\[
Ft = m \Delta v
\]
- \( \Delta v \) = change in velocity
#### Example Problem
Calculate the change in momentum of a 0.5 kg ball that strikes the floor at 15 m/s and bounces back up at 12 m/s.
### Solution
1. **Identify Initial and Final Velocities:**
- Initial velocity \( v_i = 15 \, \text{m/s} \) (downward, so it can be considered negative if upward is positive)
- Final velocity \( v_f = 12 \, \text{m/s} \) (upward, positive direction)
2. **Calculate the Change in Velocity:**
\[
\Delta v = v_f - v_i
\]
Since the ball changes direction, we should consider the direction as well:
\[
\Delta v = 12 \, \text{m/s} - (-15 \, \text{m/s}) = 12 \, \text{m/s} + 15 \, \text{m/s} = 27 \, \text{m/s}
\]
3. **Calculate the Change in Momentum:**
\[
\Delta p = m \Delta v
\]
Given \( m = 0.5 \, \text{kg} \):
\[
\Delta p = 0.5 \, \text{kg} \times 27 \, \text{m/s} = 13.5 \, \text{kg} \cdot \text{m/s}
\]
So, the change in momentum of the ball is \( 13](https://content.bartleby.com/qna-images/question/bb95dca4-f6c7-46fa-bbcb-5cdbe2ca0e09/4f2ec098-bffe-48fe-81ae-2403379d46d5/x4yfibe_processed.jpeg)
Transcribed Image Text:### Momentum and Impulse
Momentum \((p)\) and impulse \((J)\) are important concepts in physics, especially in the context of collisions and interactions between objects.
#### Equations
1. **Momentum Equation:**
\[
p = mv
\]
- \(p\) = momentum
- \(m\) = mass
- \(v\) = velocity
2. **Impulse Equation:**
\[
J = Ft
\]
- \(J\) = impulse
- \(F\) = force
- \(t\) = time
3. **Alternate Form of Impulse Equation:**
\[
Ft = m \Delta v
\]
- \( \Delta v \) = change in velocity
#### Example Problem
Calculate the change in momentum of a 0.5 kg ball that strikes the floor at 15 m/s and bounces back up at 12 m/s.
### Solution
1. **Identify Initial and Final Velocities:**
- Initial velocity \( v_i = 15 \, \text{m/s} \) (downward, so it can be considered negative if upward is positive)
- Final velocity \( v_f = 12 \, \text{m/s} \) (upward, positive direction)
2. **Calculate the Change in Velocity:**
\[
\Delta v = v_f - v_i
\]
Since the ball changes direction, we should consider the direction as well:
\[
\Delta v = 12 \, \text{m/s} - (-15 \, \text{m/s}) = 12 \, \text{m/s} + 15 \, \text{m/s} = 27 \, \text{m/s}
\]
3. **Calculate the Change in Momentum:**
\[
\Delta p = m \Delta v
\]
Given \( m = 0.5 \, \text{kg} \):
\[
\Delta p = 0.5 \, \text{kg} \times 27 \, \text{m/s} = 13.5 \, \text{kg} \cdot \text{m/s}
\]
So, the change in momentum of the ball is \( 13
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