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 solve the problem of finding the average force experienced by a surface when \( n \) balls, each of mass \( m \), impinge elastically each second with a velocity \( u \), we can follow these steps:

### Step 1: Understand the concept of elastic collision
In an elastic collision, both momentum and kinetic energy are conserved. When a ball collides elastically with the surface, it rebounds with the same speed but in the opposite direction.

### Step 2: Calculate the change in momentum for one ball
The momentum of one ball before the collision is given by:
\[ p_{\text{initial}} = mu \]
After the collision, the momentum of the ball is:
\[ p_{\text{final}} = -mu \]
The change in momentum (\( \Delta p \)) for one ball is:
\[ \Delta p = p_{\text{final}} - p_{\text{initial}} = -mu - mu = -2mu \]
Since we are interested in the magnitude of the change in momentum, we take:
\[ \Delta p = 2mu \]

### Step 3: Calculate the total change in momentum for \( n \) balls
If \( n \) balls are colliding with the surface each second, the total change in momentum for \( n \) balls is:
\[ \Delta p_{\text{total}} = n \times \Delta p = n \times 2mu = 2nmu \]

### Step 4: Calculate the average force experienced by the surface
The average force (\( F_{\text{average}} \)) experienced by the surface is given by the change in momentum per unit time. Since the balls are colliding each second, the time interval \( \Delta t \) is 1 second:
\[ F_{\text{average}} = \frac{\Delta p_{\text{total}}}{\Delta t} = \frac{2nmu}{1} = 2nmu \]

### Final Result
Thus, the average force experienced by the surface is:
\[ F_{\text{average}} = 2nmu \]