n bullets per sec each of mass m moving with velocity v strike normally on a wall. The collision is elastic then force on wall is-

Options

A. zero
B. mnv
C. 2 mnv
D. 4 mnv

Explanation

To solve the problem of finding the force exerted on a wall by n bullets per second, each of mass m and moving with velocity v, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Collision**:
- We have n bullets striking a wall per second, each with mass m and moving with velocity v. The collision is elastic, meaning that the bullets will bounce back with the same speed but in the opposite direction.

2. **Calculating Change in Momentum for One Bullet**:
- The initial momentum of one bullet before striking the wall is \( p_{initial} = mv \).
- After the elastic collision, the bullet bounces back with the same speed but in the opposite direction, so the final momentum is \( p_{final} = -mv \).
- The change in momentum (\( \Delta p \)) for one bullet is:
\[ \Delta p = p_{final} - p_{initial} = (-mv) - (mv) = -2mv \]

3. **Calculating Force Exerted by One Bullet**:
- The force exerted by one bullet on the wall can be calculated using the formula for force, which is the change in momentum per unit time. If we consider the time taken for the bullet to strike and bounce back as \( t \), then:
\[ F_{bullet} = \frac{\Delta p}{\Delta t} = \frac{-2mv}{t} \]
- Since we are interested in the magnitude of the force, we take the absolute value:
\[ F_{bullet} = \frac{2mv}{t} \]

4. **Calculating Total Force for n Bullets**:
- If n bullets strike the wall per second, the total force exerted on the wall by all bullets is:
\[ F_{total} = n \cdot F_{bullet} = n \cdot \frac{2mv}{t} \]
- Since \( t = 1 \) second (as we are considering n bullets per second), we can simplify this to:
\[ F_{total} = n \cdot 2mv \]

5. **Final Result**:
- Therefore, the total force exerted on the wall by n bullets per second is:
\[ F_{total} = 2nmv \]

### Conclusion:
The force on the wall is \( 2nmv \).