Question

A machine gun fires a bullet of mass $40g$ with a velocity of $1200m/s$. The man holding it can exert a maximum force of $144N$ on the gun. How many bullets can be fired per second at the most?

Answer 1

Step 1: Given data

The mass of a bullet fired by the machine gun is given as: $m=40g=40\times {10}^{-3}kg$

Velocity is given as:$v=1200m/s$

The maximum force is given by :$F=144N$

Step 2: Formula applied

We have to determine the number of bullets that can be fired per second at the most.

From Newton’s second law of motion, we come to know that

$F=\frac{dp}{dt}$

Where,

$p=$ momentum, which can be given as; $p=mv\dots \left(1\right)$

$\mathrm{t}=$time in seconds

Let us assume the total number of bullets fired be $=n$

Hence, the above equation can be written as:

$F=n\times \frac{dp}{dt}\dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, the equation for force can be written as $F=n\times \frac{d\left(mv\right)}{dt}$. Since the initial velocity of the bullet and the time at which it was fired both are zero, the momentum will be $\frac{d\left(mv\right)}{dt}=\frac{mv-m\times 0}{t-0}=\frac{mv}{t}$. Thus, force will be $F=n\times \frac{mv}{t}$.

Step 3: Calculation of the number of bullets that can be fired

Substitute $144N$ for $F$, $40\times {10}^{-3}kg$ for $m$, $1200m{s}^{-1}$ for $v$ and $1s$ for $t$ in the in the equation $F=n\times \frac{mv}{t}$.

$$\begin{gathered} F=n\times \frac{mv}{t} \\ \Rightarrow 144=\frac{n\times (40\times {10}^{-3}\times 1200)}{1} \\ \Rightarrow n=\frac{144}{48000\times {10}^{-3}} \\ \Rightarrow n=3 \end{gathered}$$

Thus, the number of bullets that can be fired per second at the most $n=3$.