 Solution

For a particle of rest mass $m_r$, $E=\gamma m_r{c}^2$ and $p=\gamma m_{r}v$, with $\gamma =(1-v^2/c^2{)}^{-1/2}$. The low-speed expansion is $E=m_r{c}^2+\frac{1}{2}{m}_r{v}^2+O(m_r{v}^4/c^2)$, separating invariant rest energy from the nonrelativistic kinetic energy term.

Conserve the total laboratory energy and momentum at coalescence: $E_{\mathrm{tot}}=4m{c}^2$ and $p_{\mathrm{tot}}=2.4mc$. Since $V=c^2{p}_{\mathrm{tot}}/E_{\mathrm{tot}}$, the combined particle moves at $0.6c$. Its rest mass is $M=\sqrt{4^2-2.4^2}m=3.2m$. The incident particle is not of mass $m$: its stated energy and momentum imply $m_{\mathrm{incident}}=\sqrt{3^2-2.4^2}m=1.8m$. ThusThe extra $0.4m{c}^2$ is internal rest energy retained by the composite after a perfectly inelastic collision; total energy and four-momentum are conserved, while the sum of individual rest masses need not be.

The composite's Lorentz factor is $1.25$. For a rest-frame lifetime $\tau$, the corresponding laboratory time is $1.25\tau$ and its travel distance isIf $\tau$ denotes a mean radioactive lifetime, this is the mean flight distance, not a fixed distance for every decay. The separate frame diagrams below use the common positive collision axis.