Conservation of linear momentum :
$3\mathrm{u}=(3\mathrm{u}/2)+\mathrm{m}\mathrm{v} \Rightarrow \mathrm{m}\mathrm{v}=(3\mathrm{u}/2)\dots \left(1\right)$

Newton's experimental law :
$\mathrm{u}=–\mathrm{e}(\mathrm{u}/2–\mathrm{v})$

Putting $\mathrm{e}=1$ for elastic collision we obtain,
$\mathrm{v}=\frac{3\mathrm{u}}{2}\dots ..\left(2\right)$
Using (1) and (2), we obtain $\mathrm{m}=1 \mathrm{kg}$.