A gun of mass M fires a bullet of mass m with a horizontal speed V...

Let the velocity of the bullet with respect to the ground be $V$ in the forward direction.

By the principle of conservation of linear momentum, the recoil velocity of the gun (and the mirror) is:
$v_{g} = \dfrac{m}{M}V$ (in the backward direction)

The velocity of the bullet with respect to the mirror (gun) is:
v_{O/M} = V - (-v_{g}) = V + \dfrac{m}{M}V = V \left(1 + \dfrac{m}{M}\right)<br><br>For a spherical mirror, the velocity of the image with respect to the mirror is given by:<br>\vec{v}_{I/M} = -m_{t}^2 \vec{v}_{O/M}$<br>where$m_{t} = $\dfrac{f}{f - u} is the transverse magnification.<br><br>Immediately after the gun is fired, the bullet is at the pole of the mirror, so the object distance$u = 0$.<br>Thus, the magnification is:<br>$m_{t} = $\dfrac{f}{f - 0} =$1$<br><br>Substituting this into the velocity equation, we get:<br>$\vec{v}_{I/M} = -(1)^2 \vec{v}_{O/M} = -\vec{v}_{O/M}$<br><br>This means the image moves with the same speed as the bullet relative to the mirror, but in the exact opposite direction.<br><br>The velocity of the bullet with respect to the image is:<br>$\vec{v}_{B/I} = \vec{v}_{O/M} - \vec{v}_{I/M} = \vec{v}_{O/M} - (-\vec{v}_{O/M}) = 2\vec{v}_{O/M}$<br><br>Therefore, the speed of separation of the bullet and the image is:<br>$v_{sep} = 2v_{O/M} = 2V \left(1 + \dfrac{m}{M}\right)$<br><br>Answer:$2\left(1 + \dfrac{m}{M}\right)V