A gun of mass M fires a bullet of mass m with a horizontal speed v. The gun is fitted with a concave mirror of focal length f facing towards the receding bullet. Then

If  $v^1$  is the recoil velocity of the gun, then by conservation of linear momentum,
$O=mv+M{v}^1$  
 $v^1=-\frac{mv}{M}$ .
   
The velocity of the bullet with respect to the mirror
 $\begin{array}{l}{\left[{\overrightarrow{v}}_{bullet}\right]}_{mirror}={\left[{\overrightarrow{v}}_{bullet}\right]}_g-{\left[{\overrightarrow{v}}_{mirror}\right]}_g\\ =v-v^1\end{array}$ 
  $\begin{array}{l}=v-(-\frac{mv}{M})\\ =(1+\frac{m}{M})v\end{array}$       
 Or  $\frac{du}{dt}=(1+\frac{m}{M})v.$
By the definition the image velocity,
   $\frac{dv}{dt}=-\frac{v^2}{u^2}\left(\frac{du}{dt}\right).$
At the instant of firing bullet and its image are at the pole of the mirror, so  . Thus
    .$\frac{dv}{dt}=-\left(\frac{du}{dt}\right)$
Or the velocity of image w.r.t mirror = velocity of bullet w.r.t mirror. The speed of  separation between bullet and its image
   
$\begin{array}{l}=2\left(\frac{du}{dt}\right)\\ =2(1+\frac{m}{M})v.\end{array}$