Inelastic bouncing-ball impacts can accumulate in finite time

Start immediately after a bounce with upward velocity $v_0>0$. Under $q^{\mathrm{\prime }}=v$, $v^{\mathrm{\prime }}=-g$, the next impact occurs after $2v_0\mathrm{/}g$, and the incoming impact velocity is $-v_0$. Its guard derivative is $q^{\mathrm{\prime }}=v^{-}=-v_0<0$, so it has the declared downward orientation; the reset produces upward velocity $e{v}_0$ and does not immediately re-trigger that oriented guard.

Repeating the calculation, the time between the $k$th and $(k+1)$st impacts is $2e^k{v}_0\mathrm{/}g$ for $k=0,1,2,\dots$. Every such interval is positive, so the impacts are distinct; at each one the incoming guard derivative is $q^{\mathrm{\prime }}=v^{-}=-e^k{v}_0\mathrm{\ne }0$, so each crossing is transverse and has the declared orientation.

The impact times are ${\tau }_n={\sum }_{k=0}^{n-1}2e^k{v}_0\mathrm{/}g$ for $n\ge 1$, and converge to $T_Z=2v_0\mathrm{/}(g(1-e))<\mathrm{\infty }$. Concatenating the exact mode solutions and resets defines a hybrid execution on the bounded half-open interval $[0,T_Z)$: every $t<T_Z$ lies before only finitely many impacts, while the event times ${\tau }_n$ accumulate at the excluded endpoint. It therefore has infinitely many transverse impacts in a bounded interval contained in its time domain, refuting the statement. The event separations tend to zero, so the finite separated-event theorem does not apply.