Two blocks $\mathrm{A}$ and  $\mathrm{B}$ of the same mass are connected to a light spring and placed on a smooth horizontal surface. $\mathrm{B}$ is given velocity ${\mathrm{v}}_0$ (as shown in the figure) when the spring is in natural length. In the subsequent motion.

In $CM$ frame the blocks will perform $SHM$ as

$*+\frac{\mathrm{T}}{\to }{\mathrm{v}}_0/2 \left[{\mathrm{v}}_{\mathrm{CM}}={\mathrm{v}}_0/2\right] \because {\mathrm{V}}_{\mathrm{CM}}$ is constant hence both will can have maximum velocity ${\mathrm{v}}_0/2$ towards right in $\mathrm{CM}$ frame.

$\therefore {\left({\mathrm{v}}_{\mathrm{B}}\right)}_{\max }={\mathrm{v}}_0$ in ground frame

In $\mathrm{CM}$ frame ${\left({\mathrm{v}}_{\mathrm{A}}\right)}_{\min }=-{\mathrm{v}}_0/2$

$CM$${\mathrm{v}}_{\mathrm{B}}={\mathrm{v}}_0/2$${\mathrm{v}}_{\mathrm{B}}={\mathrm{v}}_0/2$$CM$$\mathrm{CM}$$\mathrm{CM}$$\therefore {\left({\mathrm{v}}_{\mathrm{A}}\right)}_{\min }\stackrel{\mathrm{A}}{=}0$ in ground frame

i.e. A move only in right direction

In $\mathrm{CM}$ frame ${\mathrm{v}}_{\mathrm{B}}={\mathrm{v}}_0/2$ left in the case when spring is at natural length

$\therefore v_B=0$ in ground frame at that instant. 

When the extension of the spring is maximum , both $A$ and $B$ will move with same velocity in the same direction as the $COM$.