Question

A bullet of mass $10\mathrm{g}$ is fired at a speed of $400{\mathrm{ms}}^{-1}$from the gun of mass $4\mathrm{kg}$. What is the recoil of the gun ?

Answer 1

Step 1: Given

Mass of bullet, $m=10g=0.01\mathrm{kg}$
Initial velocity of bullet$u=0$
Mass of gun $M=4{\mathrm{kg}}_{\text{,}}$
Initial velocity of gun $\mathrm{U}=\mathbf{0}$
Final velocity of bullet $\mathrm{v}=\mathbf{400}{\mathrm{ms}}^{-1}$

Step 2: Finding the recoiling speed
Let $V$ be the velocity of recoil of the gun
Then by the law of conservation of linear momentum, we have
$mu+MU=mv+MV$
we have
$0=0.01\times 400+4\times V$
$V=-4/4=-1\mathrm{m}{\mathrm{s}}^{-1}$

Therefore the recoiling velocity is $-1m{s}^{-1}$. the negative direction shows the direction of recoil.