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A snooker ball

  1. Dec 29, 2004 #1
    In snooker, I've noticed many times that it you hit some ball in right place the white ball will stop at the point of contact but the ball that was hit starts moving. Is this true that the white ball tranfers all of its momentum to the other ball and since there is no more momentum left it stops?
     
  2. jcsd
  3. Dec 29, 2004 #2
    Well, I think it does. But the reason i asked this STUPID question is this:
    If the white ball stops then it means all of its kinetic energy has been converted into the kinetic energy of the other ball(assuming that there are no energy losses), right?
    (in the following example I've assumed that both the balls have different mass)
    Now let the mass of white ball be 2Kg and its velocity be 1m/s. It means its has a momentum of 2Kg m/s. Now according to the conservation of momentum the other ball must have a momentum equal to 2Kg m/s. Let the mass of this ball be 1kg. Therefore its resultant velocity should be 2m/s.
    But now we find this velocity by using the forula for kinetic energy i.e. 1/2mV^2.
    The mass of white ball is 2kg and has a contant velocity of 1m/s. Therefore its kinetic energy is (1/2)2*1^2= 1 J. Since it has tranferred all of its energy to the other ball the kinetic energy of the other wall should be equal to it, right?
    Since it has a mass of 1 Kg it must move with a velocity of sqrt. 2 to have a a kinetic energy of 1J.
    Now my question is why are these to resultant velocities different?
    I know it the most stupid question but I love physics and I want to know as much as i can...
    Thanks in advance for any help.
    Abdullah
     
    Last edited: Dec 29, 2004
  4. Dec 29, 2004 #3

    dextercioby

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    1.U missed two elements.Taking them into consideration will explain the erroneous conclusion you reached.
    2.The pool/snooker balls definitely roll on the table which has a surface with nonzero (and which cannot be neglected either) rolling friction coefficient.So rolling friction definitely plays a role.Have u seen pool/snooker tables with the surface covered with glass or ice??I didn't.
    3.The collisions are not ideal,no matter how hard those snooker/pool balls are.A part of KE is converted into heat.
    4.Have u noticed that the white ball is lighter than the red/colored balls???So your little computation doesn't model reality...

    Daniel.
     
  5. Dec 29, 2004 #4
    Right but what will happen when this happens in space, without any board, only two balls and if we do assume that the collision is ideal?
     
  6. Dec 29, 2004 #5

    dextercioby

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    Then none of the balls would stop,unless their masses were equal.Obviously in this case (no friction+ideal collision) both the laws of momentum and kinetic energy conservation would apply and for collinear momentum/velocity vectors:
    [tex] m_{1}v_{1}+m_{2} 0=m_{1}0 +m_{2} v'_{2} [/tex] (1)
    [tex] \frac{m_{1}v'_{1}^{2}}{2}+\frac{m_{2}0^{2}}{2}=\frac{m_{1} 0^{2}}{2}+\frac{m_{2}v'_{2}^{2}}{2} [/tex] (2)

    From (1),(2) u get by simple division:
    [tex] v_{1}=v'_{2} \Rightarrow m_{1}=m_{2} [/tex]

    So,for equal mass snooker/pool balls colliding ideally in the absence of friction the total momentum and kinetic energy transfer is possible.

    This ideal picture is helpful coz it provides the expalanation for the fact that the the white & colored ball must have nonequal mass.This characteristic would compensate friction effect +KE loss in collision and would allow for this situation (total momentum transfer) to occur on a snooker/pool table.

    Daniel.
     
  7. Dec 29, 2004 #6
    Thanks alot! :)
     
  8. Dec 29, 2004 #7

    rcgldr

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    For both pool and snooker the cue ball is the same weight and size as the other balls. The only exception I'm aware of is those bar room tables you put coins in to play. On these the cue ball is bigger, so that if put into a pocket, a track with a certain width opening will return the cue ball, but allow the other balls to drop through to a collection area.

    When reverse spin is put on a cue ball when it is struck, the friction of the felt slows this reverse rotation as the cue ball moves forwards. If just the right amount of reverse spin is applied, the cue ball ends up with almost no rotation as it strikes a target ball, and the result appears to be a near elastic collision. Even when there is spin on the cue ball, the collision itselft has near elastic properties, but the direction of the cue ball will be affected by any rotation of the cue ball at the time of collision. If there's forwards rotation, the cue ball will continue forward, if there's reverse rotation, the cue ball will move backwards after the collision.

    Imparting side spin on a cue ball cause it to curve a slight amount, but the main reason for using side spin is to affect how the cue ball will behave when there's a second collision with one of the cushions on the sides of a pool table.

    There is little friction between the balls, but the impact / normal force is large enough that some of the spin from the cue ball will be imparted onto the target ball, which will end up some spin in the opposite direction. A good example of this is a straight shot with two target balls lined up near a corner. Put a good amount of reverse spin on the cue ball, and it collides with the first ball, imparting enough forward spin on that target ball that, so that after the target ball strikes the secondary ball, the target ball will not stop, but also continue forwards into the pocket.

    In case anyone is wondering, it's illegal to jump a cue ball by striking it below the center of mass with a cue. However it is legal to get the same affect by striking downwards on the cue ball from above, so that it flexes the surface of the pool table downwards a bit, which then returns upwards, imparting an upwards force on the cue ball. You're bascially causing the cue ball to bounce. If the airborne cue ball then "lands" on a target ball, the target ball will also go airborne (not much use for this secondary effect, except for trick shots).
     
  9. Dec 29, 2004 #8

    krab

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    Such questions are never stupid. I imagine Newton asked himself the exact same question. It is how physics progresses and also how a beginner learns physics.

    So your reasoning can be considered a proof that in an elastic collision, with one ball hitting another stationary ball, the first ball cannot come to a stop unless the masses of the 2 balls are equal. There are many other neat things related to the physics of collisions. A non-obvious example is that in the above scenario when the balls are equal in mass and neither ball stops, their trajectories are exactly 90 degrees apart.
     
  10. Dec 29, 2004 #9
    just wondering, would some of the energy be lost as sound as well as heat when the two balls hit? I'm a gr 11 physics student, and wanted to confirm this.
     
  11. Dec 29, 2004 #10

    dextercioby

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    It would.Actually i neglected the internal structure of the snooker/pool balls and assumed the friction with air was to be neglected.Definitely that interaction would imply loss of KE due to collisions between layers of molecules,but that's way too complicated to describe and analyze,so it's included among corrections.Actually any macroscopical process of collision would imply using the techniques of deformable solid mechanics (equations of linear elasticity and plasticity).But,again,for simplicity,we assume the collision to occur between rigid solids.

    Daniel.
     
  12. Dec 29, 2004 #11

    rcgldr

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    Only if the cue ball has no spin on it at the point of collision (just the right amount of reverse spin put on it when it's struck by a cue). If there is forward spin on the cue ball, the angle is less than 90 degrees, if there is backward spin on the cue ball, the angle is greater than 90 degrees.

    Pool players take advantage of the spin reaction to position the cue ball for suceeding shots.
     
  13. Dec 29, 2004 #12

    rcgldr

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    Another interesting feature of pool and snooker is the way cushions react. When a ball collides with a cushion, the direction it goes afterwards depends on how hard it hits the cushion. The harded the impact on a cushion the more perpendicular to the cushion the path of the ball will be after impact. If a ball impacts a cushion lightly at a 60 degree angle, it will bounce off with just a bit less than a 60 degree angle. If hits the cushion hard, then it will bounce off with much more than a 60 degree angle, maybe as much as an 80 degree angle.

    Side spin on the cue-ball greatly affects the way it bounces off a cushion as well.
     
  14. Dec 29, 2004 #13

    rcgldr

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