Question

A gun fires a shell and recoils horizontally. If the shell travels with speed $\text{v}$ with respect to the barrel, 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

• A. $1$
• B. $\frac{2}{\sqrt{3}}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $\frac{2}{\sqrt{3}}$

Let mass of gun is $\text{M}$ and that of shell is $\text{m}$. The two cases are shown in the figure as below.

Here ${\text{v}}_1$ and ${\text{v}}_2$ are the recoil speeds of the gun in two cases. As there are no net external forces in horizontal direction, therefore using conservation of linear momentum in horizontal direction in both cases:
For case 1
$\text{m}(\text{v}-{\text{v}}_1)={\text{Mv}}_1$
$\therefore {\text{v}}_1=\frac{\text{mv}}{\text{M+m}}...\left(i\right)$
For case 2
From F.B.D, horizontal velocity of shell w.r.t gun is $\text{v}\cos {30}^{\circ }$
Applying PCLM,
$\text{m}(\text{v}\cos {30}^{\circ }-{\text{v}}_2)={\text{Mv}}_2$
$\therefore {\text{v}}_2=\frac{\sqrt{3}\text{mv}}{2\left(\text{M+m}\right)}...\left(ii\right)$
From equations $\left(i\right)$ and $\left(ii\right)$,
$\frac{{\text{v}}_1}{{\text{v}}_2}=\frac{2}{\sqrt{3}}$