 Normal-mode impulse projection

For an impulse vector $\mathbf{J}=M\Delta \dot{\mathbf{q}}$ delivered at an unstrained equilibrium, project the velocity jump onto mass-orthogonal normal modes. Multiplying $M\Delta \dot{\mathbf{q}}$ by ${\mathbf{u}}_j^T$ gives modal velocity coefficient ${\mathbf{u}}_j^T\mathbf{J}/({\mathbf{u}}_j^{T}M{\mathbf{u}}_j)$. Its kinetic energy is the displayed expression, including a translational zero mode if one is present. A single endpoint kick in an equal-mass free three-mass spring chain puts fractions $1/3,1/2,1/6$ into translation and the two internal modes.