A careless driver driving at 25 m/s hit a tree while fumbling with his phone. Assume the driver was in contact with the air bag for 0.22 seconds, and his upper body has 37 kg in mass, then how much average force (absolute value) does the air bag exert on the driver's upper body? Ignore gravity since the force was horizontal. Hint: use impulse and the change of momentum.

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### Problem Statement

A careless driver driving at 25 m/s hit a tree while fumbling with his phone. Assume the driver was in contact with the airbag for 0.22 seconds, and his upper body has 37 kg in mass. How much average force (absolute value) does the airbag exert on the driver's upper body?

**Ignore gravity since the force was horizontal.**

**Hint:** Use impulse and the change of momentum.

### Solution Approach

To solve the problem, we need to follow these steps:

1. **Determine the initial and final velocities:**
   - Initial velocity (\(v_i\)) = 25 m/s
   - Final velocity (\(v_f\)) = 0 m/s (since the driver comes to a stop)

2. **Calculate the change in momentum (\( \Delta p \)):**
   - The momentum (\( p \)) of an object is given by \( p = m \times v \).
   - Change in momentum (\( \Delta p \)) = \( m \times (v_f - v_i) \).

3. **Use the impulse-momentum theorem:**
   - Impulse ( \( J \) ) = Change in momentum (\( \Delta p \)).
   - Impulse is also given by \( J = F \times t \), where \( F \) is the force and \( t \) is the time interval.

4. **Solve for the average force (\( F \)):**
   - Rearrange the impulse formula to solve for \( F \): \( F \) = \( \frac{\Delta p}{t} \).

Given:
- Initial velocity, \( v_i \) = 25 m/s
- Final velocity, \( v_f \) = 0 m/s
- Mass, \( m \) = 37 kg
- Time, \( t \) = 0.22 seconds

### Detailed Calculation

1. **Change in Momentum**
   \[
   \Delta p = m \times (v_f - v_i) 
   \]
   \[
   \Delta p = 37 \, \text{kg} \times (0 - 25 \, \text{m/s}) 
   \]
   \[
   \Delta p = 37 \, \text{kg} \times -25 \, \text{m/s} 
   \]
   \[
   \
Transcribed Image Text:### Problem Statement A careless driver driving at 25 m/s hit a tree while fumbling with his phone. Assume the driver was in contact with the airbag for 0.22 seconds, and his upper body has 37 kg in mass. How much average force (absolute value) does the airbag exert on the driver's upper body? **Ignore gravity since the force was horizontal.** **Hint:** Use impulse and the change of momentum. ### Solution Approach To solve the problem, we need to follow these steps: 1. **Determine the initial and final velocities:** - Initial velocity (\(v_i\)) = 25 m/s - Final velocity (\(v_f\)) = 0 m/s (since the driver comes to a stop) 2. **Calculate the change in momentum (\( \Delta p \)):** - The momentum (\( p \)) of an object is given by \( p = m \times v \). - Change in momentum (\( \Delta p \)) = \( m \times (v_f - v_i) \). 3. **Use the impulse-momentum theorem:** - Impulse ( \( J \) ) = Change in momentum (\( \Delta p \)). - Impulse is also given by \( J = F \times t \), where \( F \) is the force and \( t \) is the time interval. 4. **Solve for the average force (\( F \)):** - Rearrange the impulse formula to solve for \( F \): \( F \) = \( \frac{\Delta p}{t} \). Given: - Initial velocity, \( v_i \) = 25 m/s - Final velocity, \( v_f \) = 0 m/s - Mass, \( m \) = 37 kg - Time, \( t \) = 0.22 seconds ### Detailed Calculation 1. **Change in Momentum** \[ \Delta p = m \times (v_f - v_i) \] \[ \Delta p = 37 \, \text{kg} \times (0 - 25 \, \text{m/s}) \] \[ \Delta p = 37 \, \text{kg} \times -25 \, \text{m/s} \] \[ \
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