In the case of maximum elongation, whole system will be moving with a common acceleration a, so for the system


$F=(m+m) a \Rightarrow a=\frac{F}{2 m}$
The centre of mass of the system will also move with the same acceleration $a$.

Let us suppose that both the masses elongate the spring by a distance $x_1$ and $x_2$, respectively. So, on taking the frame of centre of mass there will act a pseudo force equal to ma on the masses.

On applying the work-energy theorem, taking the centre of mass as a reference point.
$\begin{gathered}
W_{\text {all forces }}=\Delta K \\
(F-m a) x_1+(m a) x_2-\frac{1}{2} K\left(x_1+x_2\right)^2=0 \\
{[\because \text { w.r.t. centre of mass } \Delta K=0]} \\
\Rightarrow\left(F-m \times \frac{F}{2 m}\right) x_1+\left(m \times \frac{F}{2 m}\right) x_2 \\
\Rightarrow\left(F-\frac{F}{2}\right) x_1+\left(\frac{F}{2}\right) x_2-\frac{1}{2} K\left(x_1+x_2\right)^2=0 \\
\left.\Rightarrow \frac{F}{2} \times x_1+\frac{F}{2} \times x_2\right)^2=0 \\
\Rightarrow \frac{F}{2}\left(x_1+x_2\right)=\frac{1}{2} K\left(x_1+x_2\right)^2=0 \\
\Rightarrow \quad x_1+x_2=\frac{F}{K} \Rightarrow x_{\text {total }}=\frac{F}{K}
\end{gathered}$
Here, $F=10 \mathrm{~N}, K=2500 \mathrm{~N} / \mathrm{m}^2$
So, $\quad x_{\text {total }}=\frac{10}{2500} \mathrm{~m}=\frac{1}{250} \mathrm{~m}=\frac{100}{250} \mathrm{~cm}=0.4 \mathrm{~cm}$
So, the maximum distance between the blocks
$\begin{aligned}
\text { will be }= & \text { natural length }+x_{\text {total }} \\
= & 10 \mathrm{~cm}+0.4 \mathrm{~cm}=10.4 \mathrm{~cm} \\
& {[\because \text { Natural length }=10 \mathrm{~cm}(\text { given })] }
\end{aligned}$