= Normal-mode impulse projection
{title2=$E_j=(\mathbf u_j^T\mathbf J)^2/(2\mathbf u_j^TM\mathbf u_j)$}
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 $\mathbf u_j^T\mathbf J/(\mathbf u_j^TM\mathbf u_j)$. 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.
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