Question

n bullets per second each of mass $m$ moving with velocity $v$ strike normally on a wall. The collision is elastic then force on the wall is-
(A) $zero$
(B) $mnv$
(C) $2mnv$
(D) $4mnv$

Answer 1

We have to find the new velocity of each bullet after the collision. Since the collision is elastic, the velocity would be the opposite of the initial velocity as the wall is an immovable and unbreakable body. Next, find the change in momentum of a single bullet using this new velocity and multiply it with the number of bullets hitting the wall per unit time to get the force on the wall.
Formula used: $Momentum = mass \times velocity$

Complete step by step solution:
The number of bullets striking the wall per second is $n$.
The mass of each bullet striking the wall is $m$.
The velocity of each bullet of mass $m$ striking the wall is $v$.
The net impulse from the bullets on the wall is the product of net mass and velocity.
Thus the momentum of each bullet of mass $m$, and velocity $v$ is $mv$.
This is the impulse or momentum of a single bullet.
To find the impulse of all $n$ bullets on the wall, we multiply the momentum of a single bullet with the total number of bullets hitting the wall,
Thus the net momentum of the bullets before the collision is $mnv$.
It is given that the collisions are elastic collisions. In elastic collision, the kinetic energy of the system is constant. In this case, since the wall will have $0$ momentum before and after the collision, due to the fact that it is an immovable object, thus the velocity of the bullets will be reversed after they collide with the wall. In that way, the kinetic energy is conserved, and the momentum is reversed since the direction of the velocity vector is reversed.
Thus the new velocity after collision is $- v$, where $v$ was the initial velocity
Thus the momentum change of the bullet is,
$mv - mu = mv - ( - mv) = 2mv$
This momentum change multiplied with the number of bullets hitting the wall per unit time will give the force on the wall.
Thus, ${F_{wall}} = 2mv(n)$, where $n$ is the number of bullet hits on the wall per second.

Thus, the correct option is option (C)$2mnv$.

Note: The value of the velocity of the bullets after the collision can be found by conserving kinetic energy and momentum; i.e. if the new velocity of the bullets are $u$, then $\dfrac{1}{2}m{v^2} = \dfrac{1}{2}m{u^2}$, such that $u = \pm v$. We take the negative value since the positive value would mean it breaks the wall which is not the case.