Two blocks of equal masses m are connected by a relaxed spring with a force constant k. The blocks rest on a smooth horizontal table. At t = 0, the block on the left is given a sharp impulse J towards the right, and the blocks begin to slide along the table. The velocity of the left block as a function of time is

When impulse acts on block A, initially velocity of block A and B are $u_A=J/m$ and $u_B=0$  respectively. Velocity of CM of system is J/2m.
 

$u_B=0,u_A=J/m,v_{CM}=J/\left(2m\right)$
We will solve this problem in CM frame because in CM frame motion is purely simple harmonic. Initial velocity of block in CM frame is given by
 $$\begin{gathered} u{\text{'}}_A=u{\text{'}}_A-v_{CM}=J/\left(2m\right) \\ u{\text{'}}_B=u{\text{'}}_B-v_{CM}=J/\left(2m\right) \end{gathered}$$
 
In CM frame blocks are doing SHM with angular frequency
 $\omega =\sqrt{\frac{k}{mm/(m+m)}}=\sqrt{\frac{2k}{m}}$
At initial instant spring is relaxed therefore
 $x_{A/CM}=A\sin \omega t\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}{x}_{B/CM}=A\sin (\omega t+\pi )$
Thus velocity of A in CM frame
 $v{\text{'}}_A=u{\text{'}}_A\cos \omega t=\frac{J}{2m}\cos \omega t$
And velocity of A in ground frame
 ${\overrightarrow{v}}_{A/g}{\overrightarrow{v}}_{A/CM}+{\overrightarrow{v}}_{CM}$
Thus  $v_A=v{\text{'}}_A+v_{CM}$
 $=\frac{J}{2m}(1+\cos \omega t)$
Thus,  ${\left(v_B\right)}_{\max }=v_0$ in ground frame
${\left(v_A\right)}_{\max }=0$  in ground frame when A has velocity  $v_0/2$ towards left in CM frame.
Also, in CM Frame $v_B=V_0/2$ left in the case when spring at natural length
Thus, $v_B=0$  in ground frame at that instant.