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

The correct option is 2 mnu.

To determine the force experienced by the surface, we can apply Newton's second law of motion, which states that the force is equal to the rate of change of momentum. Let us break down the derivation step-by-step:

Step 1: Calculate the change in momentum of a single ball.
Let the initial velocity of a ball approaching the surface be
$u$
along a chosen positive direction.
The initial momentum of one ball (
$p_{\text{initial}}$
) is given by:
$p_{\text{initial}}=mu$
Since the collision of the balls with the surface is elastic, each ball rebounds with the same speed but in the opposite direction. Therefore, the final velocity of the ball after collision is
$-u$
.
The final momentum of one ball (
$p_{\text{final}}$
) is:
$p_{\text{final}}=-mu$
The change in momentum (
$\Delta p$
) of a single ball is:
$\Delta p=p_{\text{final}}-p_{\text{initial}}$
$\Delta p=-mu-mu=-2mu$
Taking the magnitude of this change, the momentum transferred to the surface by a single ball is:
$\left|\Delta p\right|=2mu$

Step 2: Calculate the total force experienced by the surface.
According to Newton's second law of motion, the force (
$F$
) is the rate of change of total momentum:
$F=\frac{\Delta p_{\text{total}}}{\Delta t}$
Given that
$n$
balls strike the surface every second, the number of collisions per unit time is
$n$
(where
$\Delta t=1\text{second}$
).
Thus, the total force experienced by the surface is:
$F=n\times \left|\Delta p\right|$
$F=n\times 2mu=2mnu$
Therefore, the force experienced by the surface is 2 mnu.