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Collision of 2 balls on frictionless surface

  1. Jan 29, 2015 #1
    1. The problem statement, all variables and given/known data
    I am not asking you to help me solve this, so i won't put in any usefull relevant equations or attempts at a solution. The nature of my problem is more of an understanding this collision - it is more of a thinking thing to process here.

    The problem goes like this.
    Ball is rolling with some initial velocity and it collidies with the same ball (standing still at the moment). The surface is frictionless but there is a friction beetween 2 balls. The translation part of energy is conversed. Surfaces of the balls are sliding against each other while collision.
    The question is - what are the translational and rotational velocities of the second ball after the collision.

    But it is not about the question. I have a trouble understanding what is happening during the collision. Many quetions came to my mind.

    • The surface is frictionless and but the surfaces of the balls are not. Then how the ball is even rolling? Or was it intentionally rotated with the speed ωr that was the same as translational velicity
    • If they mean that table has static friction but not the kinetic friction my mental image of the problem is completly different.
    • the first ball is rotating in the negative direction (clockwise) and it it tries to make the second ball rotating in the positive direction (counter clockwise). However if we assume that they mean the table has in fact static friction (if not how the ball is even rolling) then the kineticfriction would try to slow down the rotation of the second ball in negative direction (the translational energy fromthe first ball is distribuated beetween the rotational an kinetic energy of the second ball)
    • if the rotation in negative direction is then lower than it should be the ball then would not be rolling - but in fact slipping. So the question is - it must have lower translational velocity too. And the first ball must collide back and have some translational velocity too (which makes sense in a way)
    • but how can i know how much the ball is slowing down the rolling of the second ball, if i don t have a time of the collision and force of friction?

    I am sorry. I am completly lost and i know this thought process is probably wrong in so many ways and really hard to follow. How do you understand the collision? What would happen with the first and the second ball in your opinion.

    Thank you for any kind of information regarding this painfull problem.

    2. Relevant equations
    No equations needed - it is a matter of thought, please read the intro if you want to report not following the template...

    3. The attempt at a solution
    Thinking beyond imaginable. My head will explode.
     
  2. jcsd
  3. Jan 29, 2015 #2

    PeroK

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    My guess is that the first ball is spinning around a vertical axis. And, not rolling but sliding along the frictionless table.

    The collision will then result in some of the linear and angular momentum being transferred to the second ball.
     
  4. Jan 29, 2015 #3

    haruspex

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    I interpret it more the way jojotank suggests, that it is rotating on a horizontal axis exactly as if it were rolling. (One could imagine it has come from part of the table where there was friction.)
    However, that has an interesting corollary: the first ball must become airborne.
    jojtank, I would assume the table is completely frictionless.
     
  5. Jan 29, 2015 #4

    DEvens

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    Somebody involved in writing the question plays pool. This is how one thinks, "first pass," about the collision of pool balls. The friction of the balls with the table is such that, for a little while under some conditions, the balls can slide. Eventually they will get caught by the friction and adjust. But the collision starts out with the transfer of momentum and angular momentum very nearly such that you can ignore the friction with the table to start with.

    The ball that starts out rolling (call it the cue ball) is going to impart a "kick" to the ball that starts out at rest (call it the red ball). Part of that kick will come from the collision. And part of the kick will come from the rotation. So the red ball is going to get some rotation and some linear momentum imparted to it. And, interestingly, the rotation the red ball gets is not going to start out being what it wants to roll naturally. The simplest motion of the cue ball would be ordinary rolling. But a good snooker player can impart a variety of motions to he cue ball before it hits. And each of these will give different behavior to the red ball. The usual terms are side, back, and top.

    These collisions are nearly always easiest to understand in the center of momentum frame. There the balls will start with equal but opposite momentums. They will collide by hitting at a point. Then, because translation energy is conserved, they will finish with the same linear momentum but "reflected." So what happens to angular momentum? Is it an elastic collision for the angular momentum? Can it be?
     
  6. Jan 29, 2015 #5

    BvU

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    Just to get the ball rolling (as a suitable figure of speech):

    The table can be frictionless all over. The tip of the queue (bud?) is definitely not -- never mind. Player can hit (I don't know the english word, sorry) with a lot of degrees of freedom. Let's assume the queue is horizontal and central in the left-right sense. One degree of freedom left for the height (and one for the amount of impact -- never mind that one either).

    The height of the impact determines spin around horizontal axis ##\perp## the motion. Nice exercise to evaluate the no-slip height (I forgot the answer, it's 42 years ago).
    ----
    That deals with the first dot. Second dot already over and done: no friction. And no however in the third.
    ----
    Third dot: collision is central in the left-right sense, I hope ? Exercise doesn't say. It's definitely central in the up-down sense, and that's well below the no-slip impact height for ball 2. So it slides, certainly in the no-friction-between-balls case - when there is no passing of angular momentum. Slips even more when there is friction and yes, ball 1 loses spin. Don't know if it can go airborne if coming in with no slip (no topspin).

    And with friction (which there is, according to the exercise tekst) the only thing you have is momentum conservation. linear and angular. And most of the time it's all happening in a plane. But with lots of degrees of freedom left over, so there's little telling what happens unless more restrictions are provided.

    Billiards physics isn't trivial physics at all
     
  7. Jan 29, 2015 #6

    haruspex

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    Think of it purely in terms of impulse. In an infinitesimal time, there is a non-infinitesimal impulse from the collision. This has a normal component and a frictional component (just as with normal and frictional forces). The will also be a normal impulse from the table. Solve for momentum and angular momentum change just as you would for forces and torques causing acceleration.
    'cue'
    'strike'
    ... slides on the table, right?
     
  8. Jan 31, 2015 #7
    Thanks to everyone here for help. :)
    Maybe i should ask more about the situation the proffesor who came up with this problem. I think the problem should have a few more statements.
     
  9. Jan 31, 2015 #8

    BvU

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    Well, there is a tidbit of given that has been left unused so far: "translation part of kinetic energy is conserved".

    For a central collison that makes things simple (I think...).

    For a non-central collision the balls still have to go out back to back in the center-of-mass system. That can be used to show that (going back to the table coordnate system) outgoing directions have a specific relationship. Perhaps that's what the exercise is aiming for, but then I can't understand the staging of the angular velocity complication.
     
  10. Jan 31, 2015 #9

    haruspex

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    Not so simple. As I mentioned, it is inevitable that the incoming ball becomes airborne. If the KE of its vertical motion is not counted in the "conserved translational energy" then (if my working is correct) total KE will increase! So it's not like a 1 dimensional elastic collision.
     
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