Question

A 20g bullet moving with a velocity of 200m/s hits a bag of SAN and comes to rest in 0.011sec. What is the momentum of the bullet just before hitting the bag? Find the average force that stopped the bullet

Answer 1

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 \( 363.64 \text{ N} \).