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Suppose we have two balls, $A$ and $B$, of radius $1$ with equal mass. Ball $B$ is initially at (center is at) two on the $x$-axis, i.e. $(2,0,0)$, and has velocity $0$. Ball $A$ is initially at (center is at) some point on the $x$-axis to the left of ball $B$ (not touching ofc) and is moving right with speed $s > 0$. Some time later the balls collide, and following the collision, ball $A$ is now stationary and ball $B$ is moving right with speed $s$.

If an impulse model is used then the collision between $A$ and $B$ is treated as occurring instantly and following the collision the center of ball $A$ is at the origin.

If the collision is modeled using forces, will the center of ball $A$ end up being exactly at the origin? Or might ball $A$ end up with center at $(\epsilon, 0, 0)$ for some small nonzero $\epsilon$, or even at some point $c$ not on the $x$-axis with $||c||$ small but not zero? If the center of ball $A$ does not end up exactly at the origin, what might be some of the major properties that determine where it does end up? What about if such a collision occurs in reality?

From my limited understanding, during the collision there is temporary deformation and spring-like forces at work. For some small proper interval of time ball $A$ (center of mass) decelerates while ball $B$ (center of mass) accelerates, but even if $A$ ends up stationary it's not clear to me that it's center will end up where it was when the collision began.

Any help or information is very much appreciated. Thank you for your time and have a great day.

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  • $\begingroup$ Nice question. "or even at some point c not on the x-axis" For ideal spherical balls, no motion is possible in other directions than perpendicular to the impact surfaces (think of an impact as a 1-dimensional incident). "What about if such a collision occurs in reality?" Then things are different and the slightest bump of unevenness can cause lots of unexpected results. $\endgroup$ – Steeven Jun 20 '18 at 12:04
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An issue that is glossed over in most introductory material—how you define a collision, and in particular when are the initial and final states—plays a roll here.

We could define a collision by considering the interaction(s) between the affected bodies. We look for a situation when the interactions (i.e. forces) go from negligible to non-negligible and back to negligible.

The collision is the entire time of non-negligible interactions.

And that tells us that the initial and final states are taken when the interaction is negligible.


Now in the case of colliding billiard balls this is easy: while they are in contact the interaction is non-negligible.

If we consider, instead, gravitational scattering the issue is hard. The force has an "infinite" range, so we have to define negligibility by reference to other forces acting on the bodies: when the force between them is less than the differences of external forces on the two bodies we can treat it as negligible. Yuck!1


So coming back to your question, the collision itself occupies a short, but non-zero period of time2 (while that deformation that you are worrying about takes place), and you would not expect to find the final condition applying at any point during the collision.

But the final state can be expected to apply immediately as the balls come out of contact. So, you would expect the initial position of the "target" ball and the final position of the "incident" ball to be slightly closer together than $2R$ ($R$ the radius of the balls), but you could rig that so that the incident ball still ends at the origin, or so that the target ball really starts at $2R$ or split the difference and have them both just slightly displaced.

In the case of real billiard balls the size of deformation is not larger than the grain of the felt, so it is probably safe to ignore it for real-world purposes.


1 Actually, in many cases this rather convoluted definition has very simple application, but the general case is nasty.

2 Here we are looking at the process in more detail than the idealization of "instant" interaction that you discuss in your question. Congratulations, you've moved on to a deeper level of understanding.

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  • $\begingroup$ Thank you! I thought that might be the case, but I wasn't really sure. Don't suppose you know how to estimate (ballpark) the final position of the incident ball or the "time length" of the collision based on say initial relative velocity and maybe a few "important" (to the question) material properties? e.g. for billiard balls am I looking at centiseconds, milliseconds, microseconds, etc..., and how does that compare to say a super (rubber) ball? Or should I make a new question? Thanks again! $\endgroup$ – DAS Jun 23 '18 at 22:37
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It is easier to visualize this from an inertial frame of reference that is moving to the right with half the initial speed of the left ball. From this frame of reference, both balls will initially be moving toward each other with equal speed.

Upon contact, each ball will experience a deformation, and the spatial extent of the deformation in each ball will not be limited to the region in close proximity to the contact plane. A compression zone will start to develop at the contact plane and will travel, and the compression wave will travel at the speed of sound in the metal across each ball until it arrives at the far end of each ball. At this point, both balls will have fully stopped, but will be under compression. Then a dilation wave will result from the reflection at the far end; the dilation wave will travel back at the speed of sound until it reaches the contact plane; at this point, the balls will separate.

So during the initial part of the collision, the back end of each ball remains traveling at the initial velocity while the from end of each ball will be under compression and will have stopped. Later, as the compression is released, the back end of each ball will be traveling in the opposite direction while the front end remains stationary. As the compression is released, more of each ball is moving in the opposite direction until the balls separate.

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