A gun fires a shell and recoils horizontally. If the shell travels along the barrel with speed $v$, the ratio of speed with which the gun recoils if
i) the barrel is horizontal
ii) inclined at an angle of ${30}^{\circ }$ with horizontal is

The correct option is D $\frac{2}{\sqrt{3}}$
Step 1:
Draw a labelled diagram.

Step 2:
Find the speed of barrel in horizontal direction.
Formula used:
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
Let mass of gun is 𝑀 and that of shell is $m$. The two cases are in this question.
Here $v_1$ and $v_2$ are the recoil speeds of the gun in two cases.
Using conservation of linear momentum in horizontal direction in first cases:
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
$m(v-v_1)=M{v}_1$
$v_1=\frac{mv}{M+m}$ .....(i)

Step 3:
Find the speed of barrel in inclined at an angle of ${30}^{\circ }$ with horizontal.
Using conservation of linear momentum in horizontal direction in second cases:
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
$M{v}_2=m(v\cos {30}^{\circ }-v_2)$
$v_2=\frac{\sqrt{3}mv}{2(M+m)}$ ....(ii)

From equations (i) and (ii)
$\frac{v_1}{v_2}=\frac{2}{\sqrt{3}}$
Final Answer: (b)