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3 Body Elastic Collision

  1. May 9, 2014 #1
    This may be intuitively obvious to you and I'm just missing it.

    Say you had 3 balls each with a mass of 1g. Each moving at velocity of 10m/s on a 2D Cartesian plane. Ball 1 & 2 are moving toward the origin from opposite sides, so they are approaching each other at 20m/s. Ball 3 moving from negative Y values to positive values (so Up) at 10m/s. The 3 balls collide simultaneously at the origin so that Balls 1 & 2 contact Ball 3 at a 90° angle. Balls 1 & 2 dont collide with each other but rather each side of Ball 3.

    Similar to hitting the cue ball in pool directly in the middle of 2 side by side balls, the cue ball would keep trajectory and the angle between the other 2 would be 90°. Only in reverse.

    What would be the speed of ball 3 after the collision? I dont know if it would remain the same or not. I tried combining balls 1 & 2 to use it in the 2D collision with 2 objects equations but I dont know if that works or if I did it right.

    My goal with this long winded explanation is to figure out if theoretically since the kinetic energy would be conserved in the system and the closing velocity between 1&2 is 20m/s could ball 3's velocity ever be above 10m/s.
     
  2. jcsd
  3. May 9, 2014 #2

    Nugatory

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    Staff: Mentor

    Remember that momentum also has to be conserved.
     
  4. May 10, 2014 #3
    Thats just a fn or kinetic energy and mass. So I'm still not sure. Is the answer obvious that it couldnt possibly increase the velocity of ball 3 or ?
     
  5. May 10, 2014 #4

    UltrafastPED

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    Science Advisor
    Gold Member

    The momentum of balls 1 & 2 is exchanged - they reflect from ball 3.

    The momentum of ball 3 is unchanged in your scenario: equal but opposite impulses occur simultaneously from balls 1 & 2. Thus ball 3 continues to roll along after the collision, as though nothing happened.

    Of course this is all wrong if the balls are spinning, etc.

    But you can reproduce this with pucks on an air table if your setup is just right.
     
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