A $10\text{kg}$ gun fires a bullet of $100\text{g}$ with a speed of $200\text{m/s}$ relative to the barrel of the gun, which is inclined at an angle ${45}^{\circ }$ with the horizontal. The gun is placed over a smooth horizontal surface. Find the recoil speed of the gun (in $\text{m/s}$). (Answer upto two decimal places)

Let the recoil velocity of the gun in negative direction = $-v\text{m/s}$


Horizontal velocity of bullet $V_1=200\cos {45}^{\circ }=\frac{200}{\sqrt{2}}=100\sqrt{2}\text{m/s}$

Given, mass of gun, $M=10\text{kg}$
and mass of bullet, $m=0.1\text{kg}$

Applying conservation of linear momentum in horizontal direction :-
$m(100\sqrt{2}-v)+M(-v)=0$
$0.1(100\sqrt{2}-v)=10\times v$
$v=\frac{100\sqrt{2}\times 0.1}{10.1}=1.4\text{m/s}$
$\therefore v=1.4\text{m/s}$ is the recoil velocity of the gun.