A small block of mass m and a concave mirror of radius R fitted with a stand, lie on a smooth horizontal table with a separation d between them. The mirror together with its stand has a mass m. The block is pushed at t = 0 towards the the mirror so that it starts moving towards the mirror at a constant speed V and collides with it. The collision is perfectly elastic. Find the velocity of the image (a) at a time $\mathrm{t}<\mathrm{d}/\mathrm{V}\text{ (b) at a time }\mathrm{t}>\mathrm{d}/\mathrm{V}$ 

Part - I 
After time ‘t’, distance travelled by block = Vt
and for the concave mirror, f= –R/2, u = –(d-Vt) 
${\mathrm{V}}_{\mathrm{OM}}=\mathrm{V}$ (Velocity of object w.r.t mirror) 

Velocity of image w.r.t. mirror =V_IM
$$\begin{gathered} {\mathrm{V}}_{\mathrm{IM}}=-{\left(\frac{\mathrm{f}}{\mathrm{f}-\mathrm{u}}\right)}^2{\mathrm{V}}_{\mathrm{OM}} \\ {\mathrm{V}}_{\mathrm{I}}-{\mathrm{V}}_{\mathrm{M}}=-{\left[\frac{-\mathrm{R}/2}{-\frac{\mathrm{R}}{2}+(\mathrm{d}-\mathrm{vt})}\right]}^2\left({\mathrm{V}}_0-{\mathrm{V}}_{\mathrm{M}}\right) \\ {\mathrm{V}}_{\mathrm{I}}-0=-\mathrm{V}{\left[\frac{-\mathrm{R}}{-\mathrm{R}+2(\mathrm{d}-\mathrm{Vt})}\right]}^2 \\ \Rightarrow {\mathrm{V}}_{\mathrm{I}}=\frac{-{\mathrm{R}}^2\mathrm{V}}{\left[2\right(\mathrm{d}-\mathrm{Vt})-\mathrm{R}{]}^2} \end{gathered}$$

Part - II As the collision is elastic, the masses of block and system (mirror + stand) are same, the velocities are interchanged. After collision,
the block comes to rest, the mirror moves away with velocity V. Collision takes place after a time d/V from t = 0. Let us consider the
situation after a time interval t₀ after collision. Total time upto this instant from the beginning is $\mathrm{t}=\frac{\mathrm{d}}{\mathrm{V}}+{\mathrm{t}}_0\Rightarrow {\mathrm{t}}_0=\mathrm{t}-\frac{\mathrm{d}}{\mathrm{V}} \mathrm{Now}$
$\begin{array}{l}\mathrm{u}=-{\mathrm{Vt}}_0=-\mathrm{V}\left(\mathrm{t}-\frac{\mathrm{d}}{\mathrm{V}}\right)=-(\mathrm{Vt}-\mathrm{d})\\ \therefore {\mathrm{V}}_{\mathrm{I}}-{\mathrm{V}}_{\mathrm{M}}=-{\left[\frac{-\mathrm{R}/2}{-\frac{\mathrm{R}}{2}+(\mathrm{Vt}-\mathrm{d})}\right]}^2\left({\mathrm{V}}_0-{\mathrm{V}}_{\mathrm{M}}\right)\\ {\mathrm{V}}_{\mathrm{I}}-\mathrm{V}=-{\left[\frac{-\mathrm{R}}{-\mathrm{R}+2(\mathrm{Vt}-\mathrm{d})}\right]}^2(\mathrm{O}-\mathrm{V})\\ \Rightarrow {\mathrm{V}}_{\mathrm{I}}=\mathrm{V}\left[1+\frac{{\mathrm{R}}^2}{\left[2\right(\mathrm{Vt}-\mathrm{d})-\mathrm{R}{]}^2}\right]\end{array}$