Show that the kinetic energy

\[
K . E=\frac{1}{2} m_{1} \dot{\mathbf{r}}_{1}^{2}+\frac{1}{2} m_{2} \dot{\mathbf{r}}_{2}^{2}
\]
of a system of two particles can be written in terms of their center-of-mass velocity \(\dot{\mathbf{R}}_{\mathrm{cm}}\) and relative velocity \(\dot{\mathbf{r}}\) as \[
K . E .=\frac{1}{2} M \dot{\mathbf{R}}_{\mathrm{cm}}^{2}+\frac{1}{2} \mu \dot{\mathbf{r}}^{2}
\]
where \(M=m_{1}+m_{2}\) is the total mass and \(\mu=m_{1} m_{2} / M\) is the reduced mass of the system.