Block A of mass 2kg moving with  10 m/s strikes a spring of constant  ${\text{π}}^2\text{\hspace{0.17em}N}/\text{m}$ attached to another 2kg block  B at rest kept on a smooth floor. The time for which rear moving block  A remain in contact with spring will be

According to the statement
$m_1=m_2=2\text{\hspace{0.17em}kg}$ 
We have a relation for the time period of two spring mass system
$t=\left(2\text{π}\right)\sqrt{\left(\frac{\mu }{k}\right)}\mathrm{.....}\left(1\right)$ 
Where  k=spring constant= ${\text{π}}^2\text{\hspace{0.17em}N}/\text{m}$
$\mu =$ reduced mass= $\frac{m_1{m}_2}{m_1+m_2}$
The time period of both oscillations are same
$\left(2\text{π}\right)\sqrt{\frac{m_1}{k_1}}=\text{π}\sqrt{\frac{m_1}{k_2}}\mathrm{....}\left(2\right)$ 
Now calculations for  $\mu =\frac{m_1{m}_2}{m_1+m_2}$
Substituting the values
$\mu =\frac{2\times 2}{2+2}=\frac{4}{4}=1$ 
For  $\mu =1$
Substituting the value of $\text{'}\mu \text{'}\&\text{'}k\text{'}$ in (1)
$\text{T}=\left(2\text{π}\right)\sqrt{\frac{1}{{\text{π}}^2}}$ 
 $\text{T}=2\text{\hspace{0.17em}s}$
But it takes  $\frac{\text{T}}{2}$ time for first block to detach, implies time $\frac{T}{2}=0.5\times 2=1\text{\hspace{0.17em}s}$
The time for which rear moving block remain in contact with spring will be 1 second.