n balls each of mass m impinge elastically each second on a surface with velocity u. The average force experienced by the surface will be

Options

A. mn u
B. 2 mn u
C. 4mn u
D. `mn u//2`

Explanation

To find the average force experienced by the surface when n balls, each of mass m, impinge elastically each second with a velocity u, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Momentum Change for One Ball:**
- When a ball of mass \( m \) strikes the surface with a velocity \( u \), it comes to a stop (assuming an elastic collision with the surface). The change in momentum (\( \Delta p \)) for one ball is given by:
\[ \Delta p = \text{Final Momentum} - \text{Initial Momentum} \]
Since the final momentum is 0 (the ball stops) and the initial momentum is \( mu \), we have:
\[ \Delta p = 0 - mu = -mu \]
- However, when the ball rebounds elastically, it will have the same speed \( u \) in the opposite direction. Thus, the total change in momentum for one ball is:
\[ \Delta p = mu - (-mu) = mu + mu = 2mu \]

2. **Calculate the Total Change in Momentum for n Balls:**
- If \( n \) balls strike the surface each second, the total change in momentum for \( n \) balls in one second is:
\[ \Delta p_{\text{total}} = n \times \Delta p = n \times 2mu = 2nmu \]

3. **Determine the Average Force:**
- The average force (\( F_{\text{average}} \)) experienced by the surface can be calculated using the formula:
\[ F_{\text{average}} = \frac{\Delta p_{\text{total}}}{\Delta t} \]
- Here, \( \Delta t = 1 \) second (since the balls are hitting the surface each second). Therefore:
\[ F_{\text{average}} = \frac{2nmu}{1} = 2nmu \]

4. **Final Answer:**
- The average force experienced by the surface is:
\[ F_{\text{average}} = 2nmu \]