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Calculating movement

  1. Nov 29, 2006 #1
    This problem (it's not homework, im looking for a concept) has stumped me.
    If a pool ball hits a a stationary pool ball how could I go about calculating how far the pool ball that was hit will travel until it stops? I have the masses and velocitys of both in motion but I can't seem to get a straight way to figure out its time to reach rest.
  2. jcsd
  3. Nov 29, 2006 #2
    The energy will be transferred to the new pool ball almost completely, so if you solve the time for rest, for the first pool ball, it will be close to the same for the second one (if struck directly).

    If not struck directly you have to use vectors to come up with answers.

    Note I say "almost completely" in the sense that in a virtual world, a direct pool ball - to pool ball hit will be 100% momentum transfer, but in the real world there will be some friction involved between the balls.

    Once again however even in a virtual world, the energy transfer to the second ball depends on the angle it is hit (have to take vectors into account).

  4. Nov 29, 2006 #3
    But, How would I calculate how far it will travel after it was hit?
  5. Nov 29, 2006 #4
    You're talking about the force of friction, which is normally dealt with by approximations. If you're concerned with accuracy you'll probably need to do some experiments.
  6. Nov 29, 2006 #5
    Even If I ignore friction, So hypothetically, is there a way to figure out how far the ball will roll after it has been hit?
  7. Nov 29, 2006 #6


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    There probably is, but it would be so complicated that it probably isn't worth the trouble. To start with, you can't ignore friction. That's what makes the balls behave as they do. For one thing, the object ball always travels in the same direction as the cue ball for a brief distance before cutting off at whatever angle. Secondly the amount of spin on the cue ball, and thus conferred oppositely to the object ball, determines how much 'legs' it has more than the power of the shot. Then there's the rolling resistance of the table itself.
    As a 'for instance', I once potted a shot in a corner pocket with a ton of backspin on the cue to suck it into position for the next shot. As I was lining up for that next shot, the ball that I had just sunk jumped back onto the table. It had picked up so much topspin from the cue ball that it climbed up the trough and out of the pocket. And that is not the only time that I've seen it happen.
  8. Nov 29, 2006 #7
    Hypothetically ignoring friction: easy, it will continue rolling in a straight line forever.
  9. Nov 29, 2006 #8
    So this this problem isn't possible under the contiditions given, right? Or how could I solve this? Any formulas known?
  10. Nov 29, 2006 #9
    Excellent question. That is a very good Dynamics problem. It requires a good sense of the details involved. Simplifying assumptions should be made if the error introduced is negligible.
    Clearly, Friction cannot be neglected - the ball will stop moving relatively quickly when moving atop water. Therefore, Conservation Of Energy does not apply.
    The Work-Energy principle is very useful here: Ti = Tf + U, where Ti, Tf = initial and final kinetic energy; U = Work exerted on the system (the ball).

    Kinematics of the ball itself can be challenging to deal with. We have angular velocity/acceleration as the ball rolls on the ground about the instantaneous point of contact on the floor; we also have translation of the ball, as it moves forward in a linear fashion.

    Notice that as ball 2 moves you have two possibilities:
    1. The ball rotates due to friction. No slipping occurs, and static friction is present
    2. The ball slips on the water, and kinetic friction is present. No rolling occurs while the ball is slipping.

    Additionally, because water is very deformable - and even the small deformations present during skillful ball-throwing may have a significant effect on the friction force developed. It's easy to understand Friction has a strong impact on the duration of movement, and thus on the total distance covered, since both are intimately related. However, for the accuracy you need, it seems this is one assumption you can ignore. It's wasteful to spend unnecessary time and effort to produce an unneeded accuracy level.

    All said, it is not as hard as you might be led to believe. The Work-Energy and the Impulse-Momentum principles are very useful in analyzing complex vectorial problems in a relatively simple (but not simplistic) way. Through them you can determine the direction and magnitude of the velocities of both balls after impact (and ball 1 does not necessarily immediately stop). Realize, though, that Dynamics is one of the most challenging Applied-Mechanics courses (THE most challenging in many college students' opinions), and it takes one full semester of continuous work to build the conceptual skills more complex problems require. That is to say that it may be tougher for somebody else to visualize the problem.

    Good question. Dynamics does not end with the kinematics equations we all study in a typical Physics class. Forces - such as Friction - must be considered. Conservation of energy does not hold, concepts such as the Work-Energy thm and the Impulse-Momentum thm become very useful.
    Last edited: Nov 29, 2006
  11. Nov 30, 2006 #10
    If you just need simple approximations, you can go for basic acceleration equations. You can assign the balls a "damping factor" which is basically friction.

    If you google "2D game physics" or such you will find a wealth of information on how to calculate physics which also deal with vectors (in this case you only probably need to worry about 2D vectors).

    If you are wanting a more 100% realistic solution (remember, an approximation can still look realistic, hence game usage) - anyway, for a more real world accurate solution, then yes you have to take into account all the factors which others have brought up.

  12. Nov 30, 2006 #11
    You can calculate the velocity from [itex]{m_1}{v_1}={m_1}\bar{v_1} + {m_2}\bar{v_2}[/itex]. Where [itex]m_1[/itex] and [itex]m_2[/itex] are the masses of the pool balls and the bars over velocities indicate that it is the "new value" of velocity after the collision.

    The 'easiest' solution to friction, in terms of a computer simluation, is to make the velocities decay exponentially. e.g., for a time interval [itex]\Delta t[/itex] we have that [itex]v_{t+\Delta t} = v_t k^{\Delta t}[/itex]. [itex]k[/itex] is some constant that will need tweaking depending on the surface that needs modelling.

    If we use this model it is quite easy to work out roughly how far the ball travels after a collison. We just need the velocity of the first ball during the collision, the masses of both the balls and an experimentally determined value of k (or one made up to look good).

    If you know calculus then use it to get an exact result, otherwise you will have to resort to using Euler steps or some such, e.g. work out [itex]v_{t+\Delta t}[/itex] for some constant [itex]\Delta t[/itex] and repeat [itex]n[/itex] times until the [itex]v_{t+n\Delta t}[/itex] becomes sufficiently small.
    Last edited: Nov 30, 2006
  13. Nov 30, 2006 #12


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    The answer is exactly what was already said: ignoring friction, the ball will roll forever due to Newton's first law.

    If you don't want to ignore friction, then you need to know what the coefficient of friction is and that can only be accurately determined experimentally.

    edit: and even then, since the surface of the table is not completely rigid, the coefficient of rolling friction will likely depend on the speed of the ball.
  14. Nov 30, 2006 #13


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    Not only that; in a real-world situation, it also varies from place to place on the same table. It's far more pronounced on something like a barroom unit that's had drinks spilled on it and such-like, but even a well-kept one will have irregularities due to normal usage. Just flipping a coin for break on the table (DON'T DO THAT!!!), for instance, crushes the fibres where it lands. Same for tapping a ball in the rack to keep it from moving, or doing jump-shots or heavy masses, or...
    Even simply breaking from the same spot repeatedly will wear a 'path' along the felt. Most of this stuff is on a microscopic level, but it's more than enough to influence the action the balls.
  15. Nov 30, 2006 #14


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  16. Nov 30, 2006 #15


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    I don't have time to read it all now, but that certainly looks like a great site. I've added it to my favourites for future reference. :cool:
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