 Solution

Put $\beta =v/c$ and $\gamma =(1-{\beta }^2{)}^{-1/2}$, the common Lorentz factor. The initial total relativistic energy and momentum areAssume that coalescence emits no additional radiation. By four-momentum conservation, the final rest mass $m_f$ is the rest mass of a coalescing relativistic collision:The final velocity is $V=c^2{P}_x/E=v/3$, the centre-of-momentum frame velocity.

For $v\ll c$,The leading term is the classical sum of masses. The first correction is the internal energy retained after sticking, divided by $c^2$: the initial classical kinetic energy is $3m{v}^2/4$, while the common-motion kinetic energy is $\frac{1}{2}(3m/2)(v/3{)}^2=m{v}^2/12$. Their difference is $2m{v}^2/3$, consistent with mass-energy equivalence.