Proof of König's theorem

The proof of König’s theorem is short and elegant, and shows where the vanishing of the cross term comes from.

Let’s write the velocity of each body as the sum of the CM velocity plus the relative one: $\vec{v}_i = \vec{v}_\text{CM} + \vec{v}_i'$. The total kinetic energy becomes

$E_\text{cin,tot} = \sum_i \tfrac{1}{2}m_i|\vec{v}_\text{CM}+\vec{v}_i'|^2$

Expanding the squared modulus, $|\vec{v}_\text{CM}+\vec{v}_i'|^2 = v_\text{CM}^2 + 2\,\vec{v}_\text{CM}\cdot\vec{v}_i' + v_i'^2$, we get three terms:

$E_\text{cin,tot} = \underbrace{\tfrac{1}{2}M_\text{tot}v_\text{CM}^2}_{E_\text{cin,CM}} + \underbrace{\vec{v}_\text{CM}\cdot\sum_i m_i\vec{v}_i'}_{\text{termine misto}} + \underbrace{\sum_i \tfrac{1}{2}m_i v_i'^2}_{E_\text{cin,int}}$

The cross term is zero, because $\sum_i m_i\vec{v}_i' = \vec{0}$: the momentum in the CMRF is always zero. What remains is just the CM term and the internal one, which is exactly the theorem’s claim.

Consequence in collisions

This decomposition is fundamental in collisions: $E_{\text{cin,CM}}$ never changes, because $\vec{v}_{\text{CM}}$ is constant (no net external force). In a perfectly inelastic collision, on the other hand, $E_{\text{cin,int}}$ drops entirely to zero, turning into thermal energy. This is why in an inelastic collision “kinetic energy is lost” but momentum — and hence the CM’s motion — stays intact: only the internal part disappears.

Links

Topics: Centre of mass Concepts: Kinetic energy · Inelastic collision Skills: Symbolic set-up

Related exercises: Perfectly inelastic collision · Collision between sledges · Ballistic pendulum (height)