To solve this problem, we can use the concepts of momentum and force.
Step 1: Calculate the momentum of the bullet just before hitting the bag.
Momentum (\( p \)) is given by the formula: \[ p = m \cdot v \] where:
- \( m \) is the mass of the bullet,
- \( v \) is the velocity of the bullet.
Given:
- \( m = 20 \text{ g} = 0.020 \text{ kg} \) (since we convert grams to kilograms),
- \( v = 200 \text{ m/s} \).
Now, we calculate the momentum: \[ p = 0.020 \text{ kg} \cdot 200 \text{ m/s} = 4.0 \text{ kg m/s} \]
Step 2: Calculate the average force that stopped the bullet.
The average force (\( F \)) can be calculated using Newton's second law: \[ F = \frac{\Delta p}{\Delta t} \] where:
- \( \Delta p \) is the change in momentum,
- \( \Delta t \) is the time taken to stop.
The bullet comes to rest, so its final momentum is \( 0 \): \[ \Delta p = p_{\text{final}} - p_{\text{initial}} = 0 - 4.0 \text{ kg m/s} = -4.0 \text{ kg m/s} \]
The time taken to stop (\( \Delta t \)) is given as: \[ \Delta t = 0.011 \text{ s} \]
Now, let's calculate the average force: \[ F = \frac{-4.0 \text{ kg m/s}}{0.011 \text{ s}} \approx -363.64 \text{ N} \]
The negative sign indicates that the force is acting in the opposite direction to the bullet's motion, which is expected as it is stopping.
Summary of Results:
- The momentum of the bullet just before hitting the bag: \( 4.0 \text{ kg m/s} \).
- The average force that stopped the bullet: approximately \( 363.64 \text{ N} \).
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