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 ?
Answer & explanation
Correct answer: option 1
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 : $E = \frac{1}{2}MV^2 + \frac{1}{2}mv^2$
$E = \frac{1}{2}M\frac{m^2}{M^2}v^2 + \frac{1}{2}mv^2$
$E = \frac{1}{2}mv^2 (\frac{m}{M}+1)$
$\frac{1}{2}mv^2 = \frac{EM}{M+m}$