What determines the bounce time of an elastic ball? Consider an elastic ball is bounced off a hard flat surface.
I would like to reconcile two different answers to the question "how does the contact time between the ball and surface depend on the speed of the ball?"
The first solution, which was the accepted solution here, is that the contact time is independent of the speed. It depends on the ball's diameter and the elasticity and density of its material. The signal for the ball to turn around is a compression wave that travels through the ball. The medium is non-dispersive.
The second solution, which seems to be more correct, is based on Hertzian contact mechanics. It claims that the bounce time is inversely proportional to the fifth root of the collision speed.
But what is the flaw in the first analysis? Is the speed of a compression wave through  a sphere homogeneous material dispersive? Is the relevant distance that the wave travels equal to the amplitude of the indentation, not the diameter of the ball? This seems to be what the authors of the second solution are asserting, but I don't see how this is justified.
 A: I believe the source of discrepancy in the second analysis comes in the line where they claim the time of impact is given by
$$\tau_{imp} \text{~}\frac{h_{max}}{v} $$
since that ignores the argument that the impact time must require the shock wave to travel to the other end of the ball and back again. When $h<<d$, then for values $v < \frac{h}{d}c$ where $c$ is the speed of sound in the ball, this equation is valid. But once it is greater, the time is too short for the entire ball to "know" that it is bouncing. The authors' argument becomes invalid at this point.
I think there are three different regimes:
First - treat the system as a mass + linear spring. Regardless of impact velocity, impact time will be the same. This works for very slow impacts, where deformation is very small but the natural frequency $\omega=\sqrt{\frac{k}{m}}$ is such that the shock wave of the impact would bounce many times - a quasi-static situation
Second - add the nonlinear behavior that results from the deformation of the ball: as the ball deforms more, it will have a higher spring constant. This affects the time - it will get shorter. This is shown by the $\tau\text{~}v^{-\frac15}$ relationship.
Third - the situation where the stiffness of the ball is such that the ball remains in contact while the pressure wave travels to the other end of the ball and back again. Once again, you have no dependence on velocity. You can think of this as a situation where the mass of the ball is not known at the time of impact, because parts of the ball have not yet been informed (by the pressure wave) that the ball is bouncing and so they can't yet contribute to the dynamics.
Which of these is the best regime depends on the actual properties of the ball: size, density, modulus - and to some extent, velocity.
A: I think you need to look at this problem using the concept of "Coefficient of Restitution". You can probably derive the equation using that concept.
For the period of deformation, the equation would be:
$$
M*V = -\int_0^T Pdt\,
$$
The restitution equation would be very similar. It's simply the impulse-momentum concept.
The compression of a ball as it strikes a wall is similar to a spring being compressed. The above equation clearly shows that the time of compression is dependent upon the mass and velocity of the ball as well as some average force P exerted by the wall. But I think P could also be expressed as P(t) or P(x).
The only way that time would not be dependent upon velocity is if P would always be equal to V. They would cancel. It doesn't make any logical sense that time would always equal mass. Time must be dependent upon velocity.
