Recoiling Of Gun MCQ - Practice Questions with Answers

For bullet and gun system $F_{\text {ext }=0 \text { ( as Force exerted by the trigger will be internal) }}$
So the Momentum of the system will be constant.
Initial momentum=0
Final momentum $=m_b V_b+m_g V_g$
So from momentum conservation, we get

$$
\overrightarrow{V_G}=-\frac{m_B}{m_G} \times \overrightarrow{V_B}
$$

Where  

$\overrightarrow{V_G} \rightarrow$ Recoil velocity
$m_B \rightarrow$ mass of bullet
$m_G \rightarrow$ mass of Gun
$\overrightarrow{V_B} \rightarrow$ velocity of Bullet
-ve sign indicates that $\overrightarrow{V_G}$ is opposite to of the velocity of the bullet
The higher the mass of the gun lesser 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

$$
\overrightarrow{V_G} \propto \frac{1}{m_G+m_{\operatorname{man}}}
$$

where, $m_{\operatorname{man}} \rightarrow$ mass of person holding gun

• If n bullet each of mass m is fired per unit time from a gun

Then 

$\begin{aligned} & F=V\left(\frac{d m}{d t}\right)=V(m n) \\ & F=m n v \\ & F=\text { force required to hold the gun } \\ & n=\text { no.of bullets }\end{aligned}$