A block of mass $1 kg$ is pushed towards another block of mass $2 kg$ from $6 m$ distance as shown in figure. Just after collision velocity of $2 kg$ block becomes$4 m/s.$

From conservation of linear momentum we can see that velocity of$1 kg$ block just after collision is $2 m/s$ leftwards. 

Now relative velocity of approach $= 6 m/s$ and relative velocity of separation$= 6 m/s$

$$\begin{gathered} e=\frac{relative velocity of separation}{relative velocity of approach}=1 \\ \mathrm{Initially} v_{cm}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}=\frac{1\times 6+2\times 0}{1+2}=2m/s \end{gathered}$$

During collision $v_{cm}$ will not change