A proton collides with a stationary deuteron to form a He 3...

In case 1:
Conservation of linear momentum: ${mv}_{0} = 3mv$
$⇒ v = \dfrac{v0}{3}$
Conservation of mechanical energy: $\dfrac{1}{2}{mv}_{0}^{2} = \dfrac{1}{2}3m\left(\dfrac{v0}{3}\right)2 + {U}_{0}$

Here ${U}_{0}$ is the minimum energy required for the reaction to happen

$⇒ {U}_{0} = {K}_{0} - \dfrac{K0}{3} = \dfrac{2{K}_{0}}{3}$
In case 2:
Conservation of linear momentum: $2{mv}_{0}^{'} = 3{mv}^{'} ⇒ v' = \dfrac{2}{3}{v}_{0}^{'}$
Conservation of mechanical energy: $\dfrac{1}{2}2m\left({v}_{0}^{'}\right)2 = \dfrac{1}{2}3m\left(\dfrac{2}{3}{v}_{0}^{'}\right)2 + \dfrac{2{K}_{0}}{3}$
$⇒ {mv}_{0}^{'2} - \dfrac{2}{3}{mv}_{0}^{'2} = \dfrac{2{K}_{0}}{3}$
$⇒ \dfrac{1}{3}{mv}_{0}^{'2} = \dfrac{2{K}_{0}}{3}$
$⇒ \dfrac{1}{2}2{mv}_{0}^{'2} = 2{K}_{0}$