Question

$n$ bullets strike per second elastically on a wall and rebound. Find the force exerted on the wall by the bullets if the mass of each bullet is $m$ .
A) ${\text{mnv}}$
B) ${\text{4mnv}}$
C) ${\text{2mnv}}$
D) $\dfrac{{{\text{mnv}}}}{{\text{2}}}$

Answer 1

As all the bullets are mentioned to have the same mass, each bullet will strike the wall with the same velocity. The collision of each bullet with the wall is elastic in nature and so the linear momentum of the bullet will be conserved. To determine the force exerted by all the bullets, we make use of the definition of force where it is described as the rate of change in momentum with time.

Formulas used:
-The linear momentum of a body is given by, $p = mv$ where $m$ is the mass and $v$ is the velocity of the body.
-The force exerted by a body is given by, $F = \dfrac{{\Delta p}}{{\Delta t}}$ where $\Delta p$ is the change in momentum of the body and $\Delta t$ is the time interval.

Complete step by step solution:
Step 1: List the known parameters of the striking bullets in the problem.

The mass of each bullet is given to be $m$ .
The total number of bullets is given to be $n$ .
It is mentioned that the bullets strike per second, so the time interval will be $\Delta t = 1{\text{s}}$ .

Step 2: Apply the conservation of momentum to obtain the change in momentum as the bullet collides with the wall.
According to the conservation of linear momentum principle, the linear momentum of the system (bullet and wall) before collision will be equal to the linear momentum of the system after the collision.

i.e., ${p_{before}} = {p_{after}}$ --------- (1)

If $v$ is the velocity of the bullet as it moves towards the wall and $v$’ is the rebound velocity, then the momentum before and after the collision will be ${p_{before}} = mv$ and ${p_{after}} = mv'$ .

Then equation (1) becomes, $mv = mv' \Rightarrow v = v'$

So the striking velocity and rebound velocity are the same.

Then the change in momentum will be $\Delta p = mv - \left( { - mv} \right) = 2mv$ .

Step 3: Express the relation for the force exerted by one bullet as it strikes the wall.

The force exerted by one bullet on the wall can be expressed as $F = \dfrac{{\Delta p}}{{\Delta t}}$ -------- (2)

Substituting for $\Delta p = 2mv$ and $\Delta t = 1{\text{s}}$ in equation (2) we get, $F = \dfrac{{2mv}}{1} = 2mv$

Then the total force exerted by the $n$ bullets will be ${F_{total}} = n \times 2mv = 2nmv$ .

$\therefore$ the force exerted by the bullets is obtained to be ${F_{total}} = 2nmv$ .

Hence the correct option is C.

Note: When the bullet strikes the wall its velocity is taken to be positive since the bullet is moving in the positive x-direction. But once the bullet rebounds, the bullet moves in the negative x-direction and the velocity will be negative. So we use $\left( { - mv} \right)$ while obtaining the change in momentum of the bullet. Here the wall is considered to remain stationary throughout the collision.