Calculate the recoil velocity of a gun having mass equal to 5 kg, if a bullet of 25 g acquires the velocity of $500m{s}^{-1}$ after firing from the gun.

The correct option is A $2.5m{s}^{-1}$

Given, mass of bullet ($m_1$) = 25 g = 0.025 kg
velocity of bullet before firing ($u_1$) = 0
velocity of bullet after firing $\left(v_1\right)=500m{s}^{-1}$
mass of gun ($m_2$) = 5 kg
velocity of gun before firing ($u_2$) = 0

Let velocity of gun after firing be $v_2$
From law of conservation of momentum, $m_1$$u_1$+$m_2$$u_2$ = $m_1$$v_1$+$m_2$$v_2$
$\Rightarrow$ $(0.025\times 0)+(5\times 0)=(0.025\times 500)+(5\times v_2)$
$\Rightarrow$$0=12.5+(5\times v_2)$
$\Rightarrow$$5\times v_2=-12.5$
$\Rightarrow$ $v_2=\frac{-12.5}{5}=-2.5m{s}^{-1}$

Thus, recoil velocity of gun is equal to $2.5m{s}^{-1}$. Here, the negative(- ve) sign shows that gun moves in the opposite direction of bullet.