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2D elastic collisions with equal masses

  1. Jul 25, 2007 #1
    Hi everyone, i was just wondering, when two objects of the same mass collide (in an elastic collision) the angles that they bounce off from should combine to 90 degrees. right? so i'm trying to prove this, but i'm a little stuck. i used vector component equations, but i arrived at a little problem. i was just wondering if anyone knew that if the velocities of two objects after the collision multiplied to equal zero...what does that mean?
     
  2. jcsd
  3. Jul 25, 2007 #2
    i couldnt understand. could you show a diagram or something


    this too
     
  4. Jul 25, 2007 #3
    i don't know how to draw a diagram on this thing.
     
  5. Jul 25, 2007 #4
    Really? Let's say two the two objects (of same mass) head for each other. After an elastic collision, they retrace their paths. How would you calculate the angles and what do they add up to in my example?


    Do you have a more general problem at hand? If so, post that question, we will try to help.
     
  6. Jul 25, 2007 #5
    it's a two dimensional collision so they deflect at an angle off of each other...and i don't have a general problem...i have to prove that they bounce off at a combined angle of 90 degrees
     
  7. Jul 25, 2007 #6
    i havent heard of that, it depends upon the path of approach.
    if it was so as you said, people wouldnt be playing a game called snooker. try watching a game and see if the combined angle is 90
     
  8. Jul 25, 2007 #7
    okay...here's the situation. object a is moving with an intial velocity of whatever...object b is at rest. they collide elastically and in 2 dimensions and bounce off of each other...the combined angles that they bounce off each other should be 90 degrees...maybe that'll help clear things up?
     
  9. Jul 25, 2007 #8
    duh.. why wud it be 90 degrees???
    assuming that the path of approach of object a passes through the centre of object b, object b will go in the direction of motion of object a and object a will retrace its path(provided the masses of the objects allow this condition), change in the direction of object a is 180 degrees and object b is... duh what was the initial direction of motion of b???
     
  10. Jul 25, 2007 #9
    okay...i get what you're saying...but they're not colliding in a straight line...cause otherwise what you're saying would make sense. object b is at rest initially (before it's hit)
     
  11. Jul 25, 2007 #10
    Okay, I didn't quite notice the 2D-part. Sorry.

    It's an elastic collision. What is its definition? With the given information can you arrive at a simple relation between initial and final velocities of the objects?

    Next, use the conservation of momentum and the above result to show what you have been asked for. (It's just algebra)
     
  12. Jul 25, 2007 #11
    i did all that already...and then i get the final velocity of A and B equal to zero
     
  13. Jul 25, 2007 #12
    Could you show your work? We could make corrections at places where you may have done something wrong.
     
  14. Jul 25, 2007 #13
    ugh...nevermind...it's really hard for me to put the equations up...but thank you so much for your time and effort anyways...i have to get going to my class right now...hopefully a classmate can help me figure it out!

    thanks so much again!!!! THANK YOU THANK YOU THANK YOU THANK YOU!
     
  15. Jul 25, 2007 #14

    Doc Al

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    I presume you mean that you get that the dot (scalar) product of the final velocity vectors equal zero: [itex]\vec{V}_a \cdot \vec{V}_b = 0[/itex]. If so, you're done since that implies that the angle between the vectors is 90 degrees.
     
  16. Jul 25, 2007 #15
    Are You Serious!!!!
     
  17. Jul 25, 2007 #16
    but how does that work?
     
  18. Jul 25, 2007 #17

    Doc Al

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    That's a fact, jack. :smile:
     
  19. Jul 25, 2007 #18
    this is my first time taking physics, can you explain why the angle between those two vectors is 90 degrees, with the fact that vector quantities multiply to be zero
     
  20. Jul 25, 2007 #19

    Doc Al

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    The scalar product of the vectors [itex]\vec{A} \cdot \vec{B} = AB\cos\theta[/itex]. That can only be zero if theta = 90 (or A or B equals zero, which is not the case here).
     
    Last edited: Jul 25, 2007
  21. Jul 25, 2007 #20
    thank you so much sooooooooooooooooooooooooooo much Doc Al...you just helped me sooo omuch....take care and once again...THANK YOU!!!!!
     
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