Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Work Power Energy

Question:

A gun of mass M fires a bullet of mass m and the total energy released in the explosion so caused is found to be E. What is the kinetic energy of the bullet ?

Options:

$\frac{EM}{M+m}$

$\frac{2EM}{M+m}$

$\frac{Em}{2M+m}$

$\frac{EM}{M+2m}$

Correct Answer:

$\frac{EM}{M+m}$

Explanation:

By Conservation of Momentum : 

Initial momentum : $p_i = 0 \text{ As both the gun and the bullet are at rest}$

Final momentum : $p_f = -MV + mv \Rightarrow MV = mv$

$\Rightarrow V = \frac{mv}{M}$

Also, total energy : $E = \frac{1}{2}MV^2 + \frac{1}{2}mv^2$

$E = \frac{1}{2}M\frac{m^2}{M^2}v^2 + \frac{1}{2}mv^2$

$E = \frac{1}{2}mv^2 (\frac{m}{M}+1)$

$\frac{1}{2}mv^2 = \frac{EM}{M+m}$