How long does the contact between a free-falling rigid sphere and the ground last in a perfectly elastic collision?

Question

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?

Perfectly rigid spheres do not exist, nor do perfectly elastic collisions. Mathematically, the contact time in a collision between rigid bodies is zero in Newtonian mechanics, but that clearly contradicts special relativity, since the "information" about the collision at one point propagates through the whole finite-sized body at infinite speed. — alephzero, Jun 14 '21 at 11:25
If you put a rigid sphere on a glass table, then the pressure becomes infinite. The glass will break. — Display Name, Jun 14 '21 at 13:34
@TheShortestMustacheTheorem, the pressure only would be infinite if the table top also was perfectly rigid. Q: Do perfectly rigid bodies yield to infinite pressure? Unfortunately, nobody in my neck of the woods can tell me where I can obtain the necessary equipment and materials to try it out for myself. — Solomon Slow, Jun 14 '21 at 16:53
Impulse: $F\Delta t = \Delta mv$ In a given elastic collision, there is a fixed $\Delta mv$ (change in momentum). The shorter the contact time $\Delta t$, the higher the force $F$. The product is called impulse. — Kaz, Jun 15 '21 at 06:31
It would last for about as long as a perfect point is wide. No defineable magnitude whatsoever. — Stian, Jun 15 '21 at 13:23
The moment you inserted the word "perfectly", you threw out such considerations. To achieve perfectly elastic collision of a equally perfectly rigid sphere would require the instantaneous application of truly infinite impulse. The universe does not like "perfect" any more than it likes "infinite" forces. — PcMan, Jun 15 '21 at 20:08

Chemomechanics · Accepted Answer · 2021-06-15T18:11:43.350

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.

Is there an upper limit to the nonzero contact time in the frictionless example? Or could an infinitely squishy material spend an infinite amount of time on becoming an infinitesimally thin pancake? — Fax, Jun 14 '21 at 16:44
I don't know of an upper limit arising from materials science considerations; to my knowledge, structures can be made arbitrarily compliant and arbitrarily elastic (e.g., by decreasing the temperature to suppress creep). Often the astronomers and relativity physicists come in with interesting perspectives involving gravity, information density, the fabric of space, etc. — Chemomechanics, Jun 14 '21 at 16:55
the bounciness of steel balls arises from their relative rigidity Not really - "rigidity" (more correctly 'stiffness') has nothing to do with it. As you just said, a rubber ball is highly elastic but has very low stiffness. Wood can be quite stiff, but is not elastic at all. — J..., Jun 14 '21 at 18:01
@Chemomechanics No, stiffness varies wildly, so things have varying degrees of it (unlike pregnancy, which is binary - either you are or you are not). Steel is stiffer than wood which is stiffer than plastic which is stiffer than cheese. That's totally valid. Elasticity is not correlated to stiffness at all, however. As you correctly note later in the sentence the lack of allowable plastic deformation modes in the material microstructure is really what matters. — J..., Jun 14 '21 at 18:21
@Fax Your "material" could consist of a spring that stores the kinetic energy, secures itself with a latch right when the movement stops and then starts a 30 year timer to then release the latch again. But if you disallow such trickery and restrict yourself to simpler elastic materials, then you can derive a minimal time (for fixed velocity). Think about any point in the material. After contact, there is only a finite amount of distance for it to stop, along which the deceleration must increase, as the material gets more compressed. From this you get an upper bound on the time it has to stop. — mlk, Jun 15 '21 at 06:28

score 8 · Answer 2 · answered Jun 14 '21 at 01:57

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.

score 6 · Answer 3 · answered Jun 14 '21 at 20:24

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.

score 1 · Answer 4 · answered Jun 15 '21 at 15:39

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.

score 0 · Answer 5 · answered Jun 15 '21 at 10:43

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"

score 0 · Answer 6 · answered Jun 15 '21 at 13:19

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

How long does the contact between a free-falling rigid sphere and the ground last in a perfectly elastic collision?

6 Answers6