Question

A gun of mass M fires a bullet of mass m with a kinetic energy E. The velocity of recoil of the gun is
1. \[\dfrac{\sqrt{2ME}}{m}\]
2. \[\dfrac{\sqrt{2mE}}{M}\]
3. \[\dfrac{\sqrt{2mE}}{M+m}\]
4. \[\dfrac{\sqrt{2ME}}{\left( M+m \right)}\]

Answer 1

Hint: The recoil velocity is the result of conservation of linear momentum of the required system.
Initial momentum of the system including the gun and the bullet = 0
Final momentum of the system = momentum of the gun + momentum of the bullet
                                   \[=Mv'+mv\]
(Where v’ is the recoil velocity of the gun)

Complete step-by-step answer:
The velocity of recoil of the gun can be obtained as,
Applying the conservation of linear momentum,
\[\begin{align}
  & \text{Momentum of system initial = Momentum of system final} \\
 & 0=Mv'+mv \\
 & \Rightarrow v'=\dfrac{-mv}{M}\text{ }\left(i \right) \\
 & \therefore K.E=\dfrac{1}{2}m{{v}^{2}} \\
 & \Rightarrow v=\sqrt{\dfrac{2E}{m}}\text{ }\left(ii \right)\left[ E\text{ is the Kinetic energy} \right] \\
 & \text{From equation }\left( i \right)\text{ and }\left( ii \right),\text{ we get} \\
 & v'=\dfrac{-m\cdot \sqrt{\dfrac{2E}{m}}}{M}\text{ } \\
 & \text{v }\!\!'\!\!\text{ =}\dfrac{-\sqrt{2Em}}{M} \\
\end{align}\]
The negative sign indicates that the direction of velocity of the gun is opposite to the direction of velocity of the bullet.
Therefore, the velocity of recoil of the gun is \[\dfrac{\sqrt{2mE}}{M}\].

So, the correct answer is “Option 2”.

Note: Recoil velocity is the backward velocity experienced by the shooter when he fires or shoots a bullet. Due to recoil velocity the shooter experienced a backward jerk. The recoil velocity is the result of conservation of linear momentum of the required system.