A gun weighing $4kg$ fires a bullet of $40g$ with a velocity of $500m{s}^{-1}$. The velocity with which gun recoils is

The correct option is A $5m{s}^{-1}$
Mass of gun, $m_1=4kg$.
Let recoil velocity of gun be $v_1$, initial velocity of gun be $u_1$, initial velocity of bullet be $u_2$
Mass of bullet, $m_2=40g=0.04kg$
Velocity of bullet after firing be $v_2=500m{s}^{-1}$.

Using law conservation of momentum :
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$ ...$\left(1\right)$
Total momentum of gun and bullet initially is zero, as the bullet is in rest before firing, i.e initial velocities of bullet and gun are $0m{s}^{-1}$
$\Rightarrow$$(m_1+m_2)\times 0$ = $m_1{v}_1+m_2{v}_2$
$v_1=-\frac{m_2{v}_2}{m_1}$
Here the negative sign suggests that this velocity is in the opposite direction to the motion of bullet.
$\therefore$Recoil velocity, $v_1=\frac{m_2{v}_2}{m_1}$$=\frac{0.04\times 500}{4}=5m{s}^{-1}$