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An inelastic collision changes total kinetic energy. In the dissipative case it decreases, transferring energy to internal energy, including deformation and microscopic motion. Isolated colliding bodies still obey momentum conservation; conservation of energy requires accounting for the transferred energy. Bodies need not stick in every inelastic collision. The decrease in translational kinetic energy drives dissipative spreading of a planetary ring.

For the incoming masses and velocities in the elastic collision example, stipulate that the bodies stick. This perfectly inelastic collision has a shared outgoing velocityThe outgoing momenta are $4$ and $2\mathrm{kg}m/s$, totaling $6\mathrm{kg}m/s$. The outgoing kinetic energies are $4$ and $2\mathrm{J}$, totaling $6\mathrm{J}$. Thus $9-6=3\mathrm{J}$ of kinetic energy becomes internal energy within the isolated pair. Nothing in momentum conservation requires this $3\mathrm{J}$ to remain as kinetic energy. ^[ref]

Initially every particle has $A=B=0$, so its Jacobi energy in a shearing sheet is $-3{\Omega }_0^2{x}_0^2/8$. During an inelastic collision, positions are fixed at the instant of impact, while momentum conservation preserves the sum of the tangential velocities. Hence the sum of $p_y=\dot{y}+2{\Omega }_{0}x$, and therefore the sum of the epicyclic guiding center positions $x_0=2p_y/{\Omega }_0$, is unchanged. Between collisions, these quantities are individually conserved.

The collision dissipates kinetic energy without changing the instantaneous tidal potential. Consequently the total Jacobi energy in a shearing sheet decreases. At any later time, writing $x_{0j}$ for the particles' current epicyclic guiding center positions givesIf $D\ge 0$ is the accumulated energy dissipated in the inelastic collisions, comparison with the initial circular orbits givesThus the mean guiding-center position stays fixed, while its variance grows. The initial ensemble has no preference for positive or negative $x$, and the local equations and collision law preserve the symmetry $(x,y,z)\mapsto (-x,-y,z)$. The spreading is therefore symmetric in the ensemble average: angular momentum transport moves some particles inward and others outward. A particular finite random realization need not be exactly symmetric. This is the microscopic energy argument for dissipative spreading of a planetary ring.