Two identical block A and B each having mass  m, are connected with a spring of force constant  k. The floor is smooth and A is pushed so as to 
compress the spring by x₀. The system is released  from this position
If the maximum speed of the centre  of mass of the system during subsequent  motion is $V_m$ and the
 acceleration of the centre of mass  at the instant it acquires half its maximum  speed is $a_c$ , then
 

(a) The normal force of wall on B is the reason for acceleration of the COM. Just when B is about to leave the wall  (i.e. when the spring is relaxed) let the speed of A be v. 
$$\begin{gathered} \frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}{\mathrm{kx}}_0^2\Rightarrow \mathrm{v}=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}{\mathrm{x}}_0 \\ \text{ Speed of COM is }{\mathrm{v}}_0=\frac{\mathrm{v}}{2}=\frac{1}{2}\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}{\mathrm{x}}_0 \end{gathered}$$

This is the final maximum speed.
(b) Let compression in the spring be x when speed of A  is $\frac{\mathrm{v}}{2}$
$$\begin{gathered} \frac{1}{2}{\mathrm{kx}}^2+\frac{1}{2}\mathrm{m}{\left(\frac{\mathrm{v}}{2}\right)}^2=\frac{1}{2}{\mathrm{kx}}_0^2 \\ {\mathrm{kx}}^2+\frac{1}{4}{\mathrm{kx}}_0^2={\mathrm{kx}}_0^2 \\ {\mathrm{x}}^2=\frac{3{\mathrm{x}}_0^2}{4};\mathrm{x}=\frac{\sqrt{3}{\mathrm{x}}_0}{2} \end{gathered}$$
$\therefore \text{ Normal force by the wall }=\frac{\sqrt{3}{\mathrm{kx}}_0}{2}$
${\mathrm{a}}_{\mathrm{cm}}=\frac{\sqrt{3}{\mathrm{kx}}_0}{2\left(2\mathrm{m}\right)}=\frac{\sqrt{3}{\mathrm{kx}}_0}{4\mathrm{m}}$