# Collision of rolling billiard balls

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1. Jan 8, 2016

### AlonsoMcLaren

1. The problem statement, all variables and given/known data

There are two problems:
(A) Consider two identical billiard balls (spheres), each of mass M and radius R. One is stationary (ball 2) and the other rolls on a horizontal surface without slipping, with a horizontal speed v (ball 1).
Assume that all the frictional forces are small enough so as to be negligible over the time of the collision, and that the collision is completely elastic.

What is the final velocity of each ball after the collision? i.e. when each ball is rolling without slipping again

A link to the problem: http://physics.columbia.edu/files/physics/content/Quals2010Sec1.pdf , Problem 5. The solution is on page 20~21

(B) Two identical billiard balls of radius R and mass M rolling with velocities ±v collide elastically head-on. Assume that after the collision they have both reversed motion and are still rolling.
(a) Find the impulse which the surface of the table must exert on each ball during its reversal of motion
(b) What impulse is exerted by one ball on the other?

A link to the problem: http://physics.columbia.edu/files/physics/content/Section 1 - Classical Mechanics with Solutions.pdf , Problem 1. The solution is on page 7
2. Relevant equations
v=ωR L=Iω
3. The attempt at a solution
I can mostly follow the solution to (A), except:
- Why the solution claims "Just after the collision, v1'=0, v2'=v" before saying anything about angular momentum after the collision? I know this result is obviously correct when they is no rolling. But now he have rolling and therefore there are rotational kinetic energy terms in the conservation of energy equation, and it is not immediately obvious we can still use the familiar non-rolling result. Indeed, ω1 and ω2 did not change after the collision because the impact force vector goes through centers of the spheres and therefore there is no net torque on either ball but I just feel that something is a bit illogical with the solution.....

For (B), I don't really understand what "impulse which the surface must exert on each ball during its reversal of motion" means. After the collision, the angular velocity of each ball is unchanged for the same reason in (A), however the velocities are switched. Therefore friction by surface of the table is necessary to get the balls rolling without slipping again. And is this where the "impulse" in part (a) of this problem is coming from? Another interpretation is that the surface gives the ball an impulse during the collision. But I assume the only force involved in the collision is the impact between the two balls and the ground should not give the ball an impulse. Why would the surface want to give an extra lift to the ball?

Anyway, let's do something with this problem. Take the left ball as example. Before
the collision, v1=+Rω, ω1=ω (into the page). Immediately after the collision, v1'=-Rω, ω1'=ω(into the page). The angular momentum of the ball relative to a fixed point on the ground immediately after the collision is therefore L=MRv1'+Iω1'=MωR^2(out)+(2/5) MωR^2 (in) = (3/5)MωR^2 (out). Relative to the fixed point on the ground, the angular momentum of the ball is conserved. So (3/5)MωR^2 (out)=MRv_final+Iω_final, and this gives v_final=-3Rω/7

And what to do next?

By the way, I can't understand even a single word in the solution to problem (B).....

Is there a source that details the general theory of elastic collisions of rolling objects?

2. Jan 8, 2016

### TSny

Give an argument for why there can be no change in total translational KE during the collision. Also, what can you say about the total linear momentum during the collision? Combine these two observations to deduce that v1' = 0.

In this problem I believe they want you to assume that the direction of rotation of each ball changes during the collision so that immediately after the collision both balls are rolling without slipping.

3. Jan 8, 2016

### AlonsoMcLaren

I have not heard of any "law of conservation of translational kinetic energy"...

So in Problem (B), the balls are rolling without slipping immediately after collision, not the case in Problem (A)? How do I even know about that?
Therefore knowing the velocities, masses and momenta of inertia of the two balls are not enough to specify a rolling body elastic collision problem, and there are "hidden variables" which determine whether the balls will slip or not immediately after the collision?

4. Jan 8, 2016

### TSny

That is not a general law, of course. However, in problem (A) what can you say about the total rotational KE of the two balls during the collision?
Yes, I think that's the intended interpretation of the two problems based on the wording of the problems.
It would have helped if the problems were more explicit in the wording.
I think there are factors such as the coefficients of friction between the two balls and the coefficients between each ball and the table that might be part of determining what will happen.

5. Jan 9, 2016

### haruspex

A key parameter is how long the collision takes. Billiard balls are very stiff, so the collision is very quick. This means the normal force between the balls gets very large. Even if the table surface were highly frictional (which would be unusual for a billiard table), it is rather unlikely that rolling contact would be achieved immediately.
However, it does seem clear that is what the question is telling you to assume.