A gun of mass m₁, fires a bullet of mass m₂ with a horizontal speed vo. The gun is  fitted with a concave mirror of focal length f facing toward a receding bullet. Find the speed of separation of the bullet and the image just after the gun was fired

Let ${\mathrm{\nu }}_1$be the speed of gun (or mirror) just after the firing of bullet. From conservation of linear momentum,

${\mathrm{m}}_2{\mathrm{v}}_0={\mathrm{m}}_1{\mathrm{v}}_1$

${\mathrm{v}}_1=\frac{{\mathrm{m}}_2{\mathrm{v}}_0}{{\mathrm{m}}_1}$                     . . . .(i)

Now rate at which distance between mirror and bullet is increasing 

$\frac{\mathrm{du}}{\mathrm{dt}}={\mathrm{v}}_1+{\mathrm{v}}_0$               . . . . (ii)

Here, $\frac{\mathrm{du}}{\mathrm{dt}}={\mathrm{v}}_1+{\mathrm{v}}_0$

(as at the time of firing, the bullet is at pole).

$\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{du}}{\mathrm{dt}}={\mathrm{v}}_1+{\mathrm{v}}_0$           . . . . (iii)

Here, dv/dt is the rate at which distance between image (of bullet) and mirror is increasing. So, if ${\mathrm{\nu }}_2$ is the absolute velocity of image (towards right), then

$v_2-v_1=\frac{dv}{dt}=v_1+v_0$

or    ${\mathrm{v}}_2=2{\mathrm{v}}_1+{\mathrm{v}}_0$      . . .  (iv)

Therefore, speed of separation of the bullet and the image will be

$\begin{array}{l}{\mathrm{v}}_{\mathrm{r}}={\mathrm{v}}_2+{\mathrm{v}}_{\mathrm{o}}=2{\mathrm{v}}_1+{\mathrm{v}}_{\mathrm{o}}+{\mathrm{v}}_{\mathrm{o}}\\ {\mathrm{v}}_{\mathrm{r}}=2\left({\mathrm{v}}_1+{\mathrm{v}}_{\mathrm{o}}\right)\end{array}$

$\text{ Substituting value of }{v}_1\text{ from Eq. (i), we have }$

${\mathrm{v}}_{\mathrm{r}}=2\left(1+\frac{{\mathrm{m}}_2}{{\mathrm{m}}_1}\right){\mathrm{v}}_0$