Question

A gun of mass $M$ fires a bullet of mass $m$, with a kinetic energy $E$. The total kinetic energy in the firing process is :

• A. $\left[\frac{m+M}{M}\right]E$
• B. $\frac{[m+M]E}{m}$
• C. $\left(\frac{M}{m+M}\right)E$
• D. $\left(\frac{m}{m+M}\right)E$

Answer 1

The correct option is A $\left[\frac{m+M}{M}\right]E$

Given,

Velocity of bullet is $V_B$

Velocity of gun is $V_g$

Kinetic Energy of bullet, $=\frac{1}{2}m{V}_B^2=E$

From conservation of momentum

Final momentum = initial momentum

$M{V}_g+m{V}_B=0$

$V_g=\frac{-m{V}_B}{M}$

Total Kinetic energy, $E_{total}=\frac{1}{2}M\times {\left(\frac{-m{V}_B}{M}\right)}^2+\frac{1}{2}m{V}_B^2$

$\Rightarrow E_{total}=\frac{m}{M}\frac{1}{2}m{V}_B^2+\frac{1}{2}m{V}_B^2$

$\Rightarrow E_{total}=\frac{m}{M}\times E+E=E\left(\frac{m+M}{M}\right)$

Hence, Total Energy is $E\left(\frac{m+M}{M}\right)$