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 ? |
$\frac{EM}{M+m}$ $\frac{2EM}{M+m}$ $\frac{Em}{2M+m}$ $\frac{EM}{M+2m}$ |
$\frac{EM}{M+m}$ |
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 : $E = \frac{1}{2}MV^2 + \frac{1}{2}mv^2$ $E = \frac{1}{2}M\frac{m^2}{M^2}v^2 + \frac{1}{2}mv^2$ $E = \frac{1}{2}mv^2 (\frac{m}{M}+1)$ $\frac{1}{2}mv^2 = \frac{EM}{M+m}$ |