 Solution

Interpret $m$ as the rest mass of each initial ball and assume that no energy or momentum escapes the collision. Let $\gamma =(1-v^2/c^2{)}^{-1/2}$ be its Lorentz factor. The total initial energy and momentum areTheir conservation determines the final four-momentum. For a single lump, $P={\gamma }^{\prime }{m}^{\prime }{v}^{\prime }$ and $E={\gamma }^{\prime }{m}^{\prime }{c}^2$, henceThe final invariant mass follows from $m^{\prime 2}{c}^4=E^2-c^2{P}^2$:This is the rest mass of a coalescing relativistic collision. For $v>0$, $m^{\prime }>2m$: some kinetic energy becomes internal energy in the final rest frame. Setting $m^{\prime }=2m$ would discard that energy and contradict conservation. At small speeds $v^{\prime }\sim v/2$, while $v=0$ gives $m^{\prime }=2m$.