Let the total energy of a system of $N$ particles be measured in an arbitrary frame of reference, such that $K=\Sigma \frac{1}{2} m_{n} v_{n}^{2} .$ In the center-of-mass reference frame, the velocities are $v_{n}^{\prime}=v_{n}-v_{\mathrm{cm}}$, where $v_{\mathrm{cm}}$ is the velocity of the center of mass relative to the original frame of reference. Keeping in mind that $v_{n}^{2}=\overrightarrow{\mathbf{v}}_{n} \cdot \overrightarrow{\mathbf{v}}_{n}$, show that the kinetic energy can be written
$$K=K_{\mathrm{int}}+K_{\mathrm{cm}}$$
where $K_{\text {int }}=\Sigma \frac{1}{2} m_{n} v_{n}^{\prime 2}$ and $K_{\mathrm{cm}}=\frac{1}{2} M v_{\mathrm{cm}}^{2} .$ This demonstrates
that the kinetic energy of a system of particles can be divided into an internal term and a center-of-mass term. The internal kinetic energy is measured in a frame of reference in which the center of mass is at rest; for example, the random motions of the molecules of gas in a container at rest are responsible for its internal translational kinetic energy.