Question

A person is holding a rifle (mass of person and rifle together is $100kg$) stands on a smooth surface and fires $10$shots horizontally in $5s$ Each bullet has a mass of $10g$ with a muzzle velocity of $800m{s}^{-1}$. The final velocity acquired by the person and the average force exerted on the person are

Answer 1

Given, $Mass,M=100kg,m=10g,t=5s$

By the law of momentum conservation:-

$=>P\left(iniitial\right)=P\left(final\right)=>m{u}_1+M{u}_2=m{v}_1+M{v}_2=>0+0=nmu+(M-nm)v$

$=>10\times 10\times {10}^{-3}\times 800+(100-10\times 10\times {10}^{-3})\times v=>v=-0.80m/s$

{$-ve$ just indicating the direction is opposite to the direction of bullets}

Now, By $F=\frac{\Delta P}{\Delta t}=>F=\frac{10\times 10\times {10}^{-3}\times 800}{5}=16N$