Question

A machine gun fires $n$ bullets per second and the mass of each bullet is $m$. If the speed of the bullets is $v$, then the force exerted on the machine gun is:
A. $m \times n \times g$
B. $m \times n \times v$
C. $m \times n \times g \times v$
D. $\dfrac{{m \times n \times v}}{g}$

Answer 1

Use the formula for the momentum of the bullet and determine the initial and final momentum of the bullet to calculate the change in momentum of the bullet. Use the relation between the change in momentum, time interval and the force exerted to determine the force exerted on the machine gun.

Complete step by step solution:
The momentum $P$ of an object is given by
$P = mv$ …… (1)
Here, $m$ is the mass of the object and $v$ is the velocity of the object.
The relation between the force exerted $F$ on an object is given by
$F = \dfrac{{\Delta P}}{{\Delta t}}$ …… (2)
Here, $\Delta P$ is the change in momentum and $\Delta t$ is the time interval.
We can see that a machine gun fires $n$ bullets per second and the mass of each bullet is $m$. The speed of the bullets is $v$.
Initially, the bullet is at rest. Hence, the initial velocity of one bullet is zero. So, the initial momentum ${P_i}$ of the bullet is also zero.
${P_i} = 0\,{\text{kg}} \cdot {\text{m/s}}$
Let us determine the final momentum and change in momentum of one bullet.
The mass of the bullet is $m$ and the speed of the bullet is $v$.
Therefore, according to equation (1), the final momentum ${P_f}$ of one bullet is $mv$.
The change in the momentum $\Delta {P_1}$ of one bullet is the subtraction of the final momentum ${P_f}$ of the bullet and the initial momentum ${P_i}$ of the bullet.
$\Delta {P_1} = {P_f} - {P_i}$
Substitute $mv$ for ${P_f}$ and $0$ for ${P_i}$ in the above equation.
$\Delta {P_1} = mv - 0$
$\Rightarrow \Delta {P_1} = mv$
Hence, the change in momentum of one bullet is $mv$.
Determine the change in momentum $\Delta P$ for $n$ bullets.
$\Delta P = n\Delta {P_1}$
Substitute $mv$ for $\Delta {P_1}$ in the above equation.
$\Delta P = nmv$
Determine the force $F$ exerted by the bullets on the machine gun.
The machine gun fires $n$ bullets per second.
Substitute $nmv$ for $\Delta P$ and $1\sec \,\,$ for $\Delta t$ in equation (2).
$F = \dfrac{{nmv}}{{1\sec \,\,}}$
$\Rightarrow F = m \times n \times v$
Therefore, the force exerted on the machine gun is $m \times n \times v$.

Hence, the correct option is B.

Note: One may get confused about why the initial momentum of the bullet is zero. Initially the bullets when they are not fired from the machine gun are at rest. Hence, their initial velocity is zero and hence, the initial momentum of the gun is also zero.