Recoiling Of Gun - Practice Questions & MCQ

Let us consider a case of a gun held at rest. Let the mass of gun be m_g, mass of bullet be m_b, velocity of bullet after firing the gun be V_b and, the recoil velocity of gun be V_g.

Assuming the net external force acting on the bullet and gun systembe zero.

$
\Rightarrow F_{\text {ext }}=0
$

Since net external force on the gun and bullet system is zero the total momentum of the system will be constant.
Assuming the gun and bullet to be at rest initially, Initial momentum=0
Final momentum $=m_b V_b+m_g V_g$
So from momentum conservation, we get-

$
\begin{aligned}
& 0=m_b \vec{V}_b+m_g \vec{V}_g \\
& \Rightarrow \vec{V}_G=-\frac{m_B}{m_G} \times \overrightarrow{V_B}
\end{aligned}
$

- -ve sign indicates that $\vec{V}_G$ is opposite to that of the velocity of the bullet
- Higher the mass of gun lesser will be recoil velocity i.e

$
* \overrightarrow{V_G} \propto \frac{1}{m_G}
$

- When the body of the shooter and the gun behave as one body/system

Then

$
\vec{V}_G \propto \frac{1}{m_G+m_{\operatorname{man}}}
$

where, $m_{\text {man }} \rightarrow$ mass of person holding gun
- If $n$ bullet each of mass $m$ is fired per unit time from a gun

Then

$
\begin{aligned}
& F=V_{\text {rel }}\left(\frac{d m}{d t}\right)=V(\mathrm{mn}) \\
& F=m n v \\
& F=\text { force required to hold the gun }
\end{aligned}
$
 

n = no. of bullets