Why Does Second Collision in Ballistic Pendulum Lead to Initial State?

[The Head]

I was thinking about ballistic pendulums and the symmetry they exhibit. In the simplest case, you have one ball that begins at a certain height and collides with another ball at rest. You can calculate via conservation of momentum and energy the new velocities and max vertical displacements. Then, if you let the system continue and the second collision occurs, the first ball will be propelled to its initial height. This makes sense in terms of symmetry.

However, when I mathematically solve it, I technically get two solutions for what happens after both collisions. When solving for the velocity of the second ball, it factors and I get a zero and a non-zero solution. For the first collision I know we choose the non-zero solution because it moves. For the second, we choose the zero solution because of symmetry. But is there a better reason for this?

If it's useful, in the simple case where the balls are equal masses, after substituting my momentum equation into the energy one, I worked out that v_2(v_2 - sqrt(2gh))=0. I'm not concerned about the actual values of the velocity, but it appears to say that either v_2 = 0 or sqrt(2gh), where 'h' is the initial height that ball one was raised to. The same equation occurs after the second collision. How do we know to choose v_2 = sqrt(2gh) for the first collision and v_2=0 for the second?

 

[kuruman]

You can show in general that when balls of equal mass collide elastically, they exchange velocities. That's the key to understanding this.

 

[Nugatory]

We reject the solutions that have one ball moving through the other.

It's worth noting that this approach works because the balls are not point particles but rather solid objects that cannot both occupy the same volume of space. If they were point particles they would occupy no volume; the ambiguity you've found is the result of idealizing them as point particles; and we eliminate it by remembering at the last moment that they aren't really point particles.

Even then the solution only works because the balls are constrained to move in only one direction. If you take the same approach to the elastic collision of two identical point masses free to move in two dimensiosn (idealized billiard balls on a billiards table) you will find an infinite number of solutions because the balls can rebound in any direction, and you will have no way of rejecting any solution in favor of some other. Instead, we have to take on the much hairier problem of colliding balls; the rebound angle depends in complicated ways on the exact point of contact.

 

[The Head]

You can show in general that when balls of equal mass collide elastically, they exchange velocities. That's the key to understanding this.

Thanks for your reply and that does make sense. Though if they aren't equal, I get a similar equation with two solutions (one equal to zero and one non-zero), but in that case the velocities wouldn't have a clean exchange.

The simplest version of non-equal masses would be if the first ball were lighter than the second ball, and after the initial collision the first ball would swing back to some degree and the second would swing forward to some degree. Then they'd collide again and it would force the first ball back to it's initial position and the second would come to rest. How does one know to choose the non-zero velocity for collision #1, but not #2?

 

[The Head]

We reject the solutions that have one ball moving through the other.

It's worth noting that this approach works because the balls are not point particles but rather solid objects that cannot both occupy the same volume of space. If they were point particles they would occupy no volume; the ambiguity you've found is the result of idealizing them as point particles; and we eliminate it by remembering at the last moment that they aren't really point particles.

Even then the solution only works because the balls are constrained to move in only one direction. If you take the same approach to the elastic collision of two identical point masses free to move in two dimensiosn (idealized billiard balls on a billiards table) you will find an infinite number of solutions because the balls can rebound in any direction, and you will have no way of rejecting any solution in favor of some other. Instead, we have to take on the much hairier problem of colliding balls; the rebound angle depends in complicated ways on the exact point of contact.

Ohhh! Yes, of course. If the second ball remained at rest and the first ball went forward, that would make no sense! And after the second collision the signs reverse for the non-zero velocity, so that would also be a contradiction.

 

[kuruman]

We reject the solutions that have one ball moving through the other.

Strictly speaking, this "other" solution is not incorrect or unphysical and should not be rejected. It is the solution that satisfies momentum and energy conservation for the two-ball system when the balls are moving in one dimension in parallel tracks and go past each other. It would be unphysical if momentum and energy were not conserved in this case.