How long does the contact between a free-falling rigid sphere and the ground last in a perfectly elastic collision? It seems to me that this time is finite, although it seems infinitely small, but if it is finite, is it also identical for any perfectly elastic collision?
What should I know about this time?
 A: In the case where both the ground and the sphere are rigid (which is impossible, but I assume it to be implied by the question.), the objects have no internal degrees of freedom to store the energy. The distance between the sphere and the ground is also constant during the collision, so no energy can be stored as gravitational potential energy either during the collision.
Therefore the Kinetic energy has to be constant during the collision, but since the ground and sphere are rigid their relative velocity during the collision must be 0. This means, that the kinetic energy is 0 (in the center of mass reference frame). Thus the ground and sphere can't be in contact for any amount of time. Which means the collision duration has to be 0 too.
A: If not interested in my comment on using material properties of the colliding body, skip straight to Nevertheless below
Answers that consider the material properties of rigid bodies are great to understand what happens in the real world. The fact that real world bodies aren't perfectly rigid and therefore deform upon hitting the ground, making contact for some finite time is a more correct way of modelling nature. By demanding bodies with small internal losses during deformation, we can impose approximate perfect elasticity of collision. In this way a freely falling damped spring is a better model for collisions than stiff rigid spheres.From the purely theoretic perspective too, rigid bodies ceased existing after GR.
This is not to say that using concepts of elasticity in analysis means only talking about experimental physics. Even in theory, by including and modelling elasticity, regardless of what happens in the real-world, is interesting and rewarding in and of itself.
Nevertheless,
I would like to offer a simple, theoretical argument that if a rigid body collides with a rigid ground, the collision must be instantaneous if it wishes to conserve energy. That is the premise of your question, and trying to answer it without using phenomenology, even in the slightest, though like asking to be blindfolded, is oddly satisfying.
Say a rigid body collides with a rigid ground at time $t=0$. At time $t>0$, the body can either:

*

*stay in contact with ground

*or rebound.

If the body stayed in contact, even for some arbitrarily small time post $0$, energy before and after can't be conserved. For at such time, the ball is at rest wrt. ground and so the kinetic energy from $t=0$ has vanished! Hence, the ball must rebound and hence rigid collisions that are perfectly elastic $must$ be instantaneous.
A: You can't deform a real material without losing some energy to heat (this is known as internal friction or mechanical hysteresis). In the ideal case of a perfectly elastic collision, if you're allowing this internal friction to exist, then zero deformation can occur, implying that the idealized materials are perfectly rigid—that is, that their elastic moduli are infinite. This in turn requires a contact time of zero, which is typical for introductory physics treatment of kinematics and collisions.
Alternatively, if you posit that the internal friction is zero, then you can have a nonzero contact time in which the materials squish together, storing strain energy, and then rebound. In fact, for very compliant materials, the contact time could be quite long. This problem is treated in the field of impact mechanics. Note, however, that compliant does not mean soft; no permanent deformation can occur (this is the soft–hard dichotomy), only recoverable deformation (this is the compliant–stiff dichotomy). Nonrecoverable deformation would preclude an elastic collision. Does this make sense?
As examples, rubber and steel balls both bounce quite well off a relatively stiff surface, which could be surprising because their stiffnesses differ by about six orders of magnitude. Elastomers such as rubber are compliant (with Young's modulus under 1 MPa, for example) but not soft (from a strain point of view; they are soft from an applied-force point of view). Thus, they provide a mostly-elastic collision (with a relatively long contact time) because they don't permanently deform much (compare with Silly Putty, for example) and don't waste a lot of deformation through friction. In contrast, the bounciness of steel balls arises from their relatively high stiffness and strength—which preclude lattice flexing and dislocation movement that would lead to hysteretic and plastic deformation losses—and the corresponding contact time is relatively short.
A: IMHO the right answer is: it depends ...
As noted before in order to have a perfectly elastic collision at least one of the colliding bodies must be able to store all the kinetic energy of the colliding bodies as potential energy and then via the elasticity of the body convert this energy again to kinetic.
In this case, the contact time depends on the rigidity of the ground, but it is not 0, I would guess, best case, a minimum value of one unit Planck time (Heisenberg's uncertainty theorem)  ;)
Assuming both bodies are non deformable -perfectly rigid- the energy can not be stored and it must be emitted as heat (principle of energy conservation), this means there is no "rebound" or elastic collision and both perfectly rigid colliding bodies remain in position.
In this case, there is nothing like an elastic collision and the bodies remain in contact "until an external force acts on the system"
A: It depends on the stiffness of the earth and the mass and velocity of the sphere. If the stiffness of the earth goes to infinity, the contact time goes to zero.
The formula for the contact time is relatively easy to derive by assuming the earth is a perfectly elastic spring, see this site:
https://isaacphysics.org/questions/drop_ball_spring
A: It lasts just long enough to reverse the velocity of the sphere - assuming elastic deformation.
The deformation has to provide an acceleration that stops the sphere, then accelerates it back the way it came. As the stiffness of the sphere and/or ground increases, the deformation becomes less, the accelerating force increases, and the contact time decreases.
As the material stiffness approaches infinity, the contact time approaches zero. In the mathematical limit, it becomes zero. In the practical case of real materials, the strength is eventually exceeded and plastic deformation occurs.
