A gun of mass M fires a bullet of mass m horizontally. The energy of firing is such that it is sufficient to project the bullet vertically to a height h. The velocity of recoil of the gun is

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$\sqrt{\frac{2 m g h}{(M + m)}}$

$\sqrt{\frac{2 m^{2} g h}{m (M + m)}}$

$\sqrt{\frac{2 m^{2} g h}{M (M + m)}}$

$\sqrt{\frac{2 M g h}{(M + m)}}$

answer is A.

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Detailed Solution

Let the recoil velocity of the gun be V

Applying conservation of momentum: 0 = mv − MV

mv = MV ⟹ v = $\frac{MV}{m}$ ............. (1)

Explosion energy must be equal to the total kinetic energy of the system after the explosion

∴ $E = \frac{1}{2} {mv}^{2} + \frac{1}{2} {MV}^{2}$

Using (1), $E = \frac{1}{2} m {(\frac{MV}{m})}^{2} + \frac{1}{2} {MV}^{2}$

⟹ $E = \frac{M (m + M)}{2 m} V^{2}$ ...........(2)

Also, the explosion energy is sufficient to project the shell to vertical height h

∴ E = mgh ..............(3)

From (2) and (3), we get mgh = $\frac{M (m + M)}{2 m} V^{2}$

⟹V= $\sqrt{\frac{2 m^{2} gh}{M (m + M)}}$

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