If I were to drop a bouncy ball onto a surface, each successive bounce will be lower in height as energy is dissipated. Eventually, however, the ball will cease to bounce and will remain in contact with the ground. What happens during that small moment of time that is the transition period?
Here's my line of thought so far:
Let's say, for example, that the ball could be modeled as an "ideal" bouncy ball. Each time it hits the floor, it rebounds with one-half of the impact velocity, the other half of which is absorbed and dissipated by various things. Will this ideal bouncy ball ever stop bouncing? The answer, as far as I can tell, is yes (well, maybe). Since each bounce has half of the initial velocity of the previous bounce, the time between each bounce also halves. The first bounce might take $1$ second, the second takes $0.5$ seconds, the third $0.25$ seconds, etc. This infinite series converges to a finite amount of time, but with an infinite number of bounces.
Since it's obviously not possible for an actual ball to bounce an infinite number of times in finite time, it is clear that the above model doesn't work. Eventually, I decided that the problem must be in the fact that the model doesn't take into account the fact that ball spends time compressing and decompressing while in contact with the floor.
Even once the ball touches the ground, the center of gravity continues to move downward as the ball compresses. Soon, the ball expands, and the center of gravity moves upward. Lets $h_1$ be the height of the center of gravity when the ball is uncompressed, and $h_2$ the height of the center of gravity of the ball when compressed, and $D = h_1 - h_2$. Eventually, the height of the bounce will be smaller than $D$, so that center of gravity simply vibrates between those two locations, and the ball never leaves the ground.