In case 1:
Conservation of linear momentum: ${\mathrm{mv}}_0=3\mathrm{mv}$
$\Rightarrow \mathrm{ }\hspace{0.33em}\mathrm{v}=\frac{{\mathrm{v}}_0}{3}$
Conservation of mechanical energy: $\frac{1}{2}{\mathrm{mv}}_0^2=\frac{1}{2}3\mathrm{m}{\left(\frac{{\mathrm{v}}_0}{3}\right)}^2+{\mathrm{U}}_0$

Here ${\mathrm{U}}_0$ is the minimum energy required for the reaction to happen

$\Rightarrow \hspace{0.33em}{\mathrm{U}}_0={\mathrm{K}}_0-\frac{{\mathrm{K}}_0}{3}=\frac{2{\mathrm{K}}_0}{3}$
In case 2:
Conservation of linear momentum: $2{\mathrm{mv}}_0^{\mathrm{\text{'}}}=3{\mathrm{mv}}^{\mathrm{\text{'}}} \Rightarrow \hspace{0.33em}\mathrm{v}\mathrm{\text{'}}=\frac{2}{3}{\mathrm{v}}_0^{\mathrm{\text{'}}}$
Conservation of mechanical energy: $\frac{1}{2}2\mathrm{m}{\left({\mathrm{v}}_0^{\mathrm{\text{'}}}\right)}^2=\frac{1}{2}3\mathrm{m}{\left(\frac{2}{3}{\mathrm{v}}_0^{\mathrm{\text{'}}}\right)}^2+\frac{2{\mathrm{K}}_0}{3}$
$\Rightarrow \mathrm{ }{\mathrm{mv}}_0^{\mathrm{\text{'}}2}-\frac{2}{3}{\mathrm{mv}}_0^{\mathrm{\text{'}}2}=\frac{2{\mathrm{K}}_0}{3}$
$\Rightarrow \hspace{0.33em}\frac{1}{3}{\mathrm{mv}}_0^{\mathrm{\text{'}}2}=\frac{2{\mathrm{K}}_0}{3}$
$\Rightarrow \hspace{0.33em}\frac{1}{2}2{\mathrm{mv}}_0^{\mathrm{\text{'}}2}=2{\mathrm{K}}_0$