Let

$\begin{array}{l}\mathrm{m}=50\mathrm{g}=50\times {10}^{-3}\mathrm{kg}\\ \text{ Side }=\mathrm{L}=1\mathrm{cm}=0.01\mathrm{m}\\ \text{ Speed }=\mathrm{v}=10\mathrm{cm}/\mathrm{s}=0.1\mathrm{m}/\mathrm{s}\\ \text{ Young's modulus }=\mathrm{Y}=2\times {10}^{11}\mathrm{N}/{\mathrm{m}}^2\\ \text{ Maximum compression }\mathrm{\Delta L}=?\end{array}$

In this case, all KE will be converted to PE
By Hooke's law,    
where A is the surface area and L is the length of the side of the cube If k is spring or compression constant. then

$\mathrm{Initial} \mathrm{KE}=2\times \frac{1}{2}{\mathrm{mv}}^2=5\times {10}^{-4}\mathrm{J}$
$\mathbf{Step}\mathbf{ }\mathbf{2}\mathbf{:} \mathrm{Find} \mathrm{the} \mathrm{expression} \mathrm{of} \mathrm{elastic} \mathrm{energy} \mathrm{stored} \mathrm{in} \mathrm{the} \mathrm{system} \mathrm{of} \mathrm{two} \mathrm{cube}.$
$\mathrm{By} \mathrm{Hook}\text{'}\mathrm{s} \mathrm{law}$
$\frac{\mathrm{F}}{\mathrm{A}}=\mathrm{Y}\frac{\mathrm{\Delta L}}{\mathrm{L}}$
$\mathrm{where} \mathrm{A} \mathrm{is} \mathrm{the} \mathrm{surface} \mathrm{area} \mathrm{and} \mathrm{L} \mathrm{is} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{cube}$
$\mathrm{If} \mathrm{k} \mathrm{is} \mathrm{spring} \mathrm{constant}, \mathrm{then}$
$\text{ force }\mathrm{F}=\mathrm{k\Delta L}$
$\therefore \mathrm{k}=\mathrm{Y}\frac{\mathrm{A}}{\mathrm{L}}=\mathrm{YL}$
$\mathrm{Final} \mathrm{PE}=2\times \frac{1}{2}\mathrm{k}(\mathrm{\Delta L}{)}^2$
$\mathbf{Step}\mathbf{ }\mathbf{3}\mathbf{:} \mathrm{Find} \mathrm{maximum} \mathrm{compression} \mathrm{of} \mathrm{each} \mathrm{of} \mathrm{the} \mathrm{cubes} \mathrm{by} \mathrm{equating} \mathrm{total} \mathrm{kinetic} \mathrm{energy} \mathrm{with} \mathrm{total} \mathrm{PE}.$
$\therefore \mathrm{\Delta L}=\sqrt{\frac{\mathrm{KE}}{\mathrm{k}}}=\sqrt{\frac{\mathrm{KE}}{\mathrm{\gamma L}}}=\sqrt{\frac{5\times {10}^{-4}}{2\times {10}^{11}\times 0.1}}=1.58\times {10}^{-7}\mathrm{m} [\because \mathrm{PE}=\mathrm{KE}]$