Question

A machine gun fires $20$ bullets per second into a target. Each bullet weighs $150\mathrm{gms}$ and has a speed of $800\mathrm{m}/\mathrm{s}$. Calculate the force required to hold the gun in position?

Answer 1

Step 1: Given that:

Number of bullets fired per second by the gun= $20$

Mass($\mathrm{m}$) of each bullet = $150\mathrm{gms}$ = $150\times {10}^{-3}\mathrm{kg}$

Speed($\mathrm{v}$) of each bullet = $800\mathrm{m}/\mathrm{s}$

Step 2: calculation of force required to hold the gun:

The force applied to a body= rate of change in momentum

The force($\mathrm{F}$) applied to a body= $\frac{\mathrm{Change}\mathrm{in}\mathrm{momentum}(∆\mathrm{p})}{\mathrm{time}(∆\mathrm{t})}$

For $1\sec$ time,

$F=∆\mathrm{p}$

Thus,

Force on each bullet= Change in the momentum of each bullet

As the initial velocity of the bullet when it was in the gun is equal to zero, therefore change in momentum of the bullet will be calculated by the product of the mass of each bullet and its velocity of coming out from the gun.,

F_b= $\mathrm{Mass}\left(\mathrm{m}\right)\mathrm{of}\mathrm{each}\mathrm{bullet}\times \mathrm{velocity}\left(\mathrm{v}\right)\mathrm{of}\mathrm{bullet}$

F_b= $150\times {10}^{-3}\mathrm{kg}\times 800\mathrm{m}/\mathrm{s}$

F_b= $15\times 8$

F_b= $120\mathrm{N}$

Now total force on 20 bullets will be;

F= $20\times {\mathrm{F}}_{\mathrm{b}}$

F= $20\times 120\mathrm{N}$

F= $2400\mathrm{N}$

According to the third law of motion, action= -reaction

Therefore,

In order to hold the gun in its position the force required will be $\mathbf{2400}\mathbf{N}$.