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Problem Involving Elastic Collision .

  1. Oct 30, 2006 #1
    Problem Involving Elastic Collision.....

    There are two ball that are both 10kg. The velocity of ball 1 is 10 m/s and the velocity of ball 2 is 40 m/s in the opposite direction of ball 1. They collide to form an elastic collision. Determine speed of both balls after collision???

    Am I starting off on the right foot?

    1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v3^2 + 1/2m2v4^2

    =1/2(10kg)(10m/s)^2 + !/2 (10kg)(40m/s)^2 = 1/2(10)V3^2 + 1/2(10)V4^2

    Solve for V3 and V4 and that should be my answers correct?
     
  2. jcsd
  3. Oct 30, 2006 #2

    radou

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    You can't solve for v3 and v4 without using the fact that momentum is conserved. Then you have two equations with two unknowns, which is solvable.
     
  4. Oct 30, 2006 #3

    Doc Al

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    So far, so good. But how can you solve for two variables with only one equation? What else is conserved that will give you a second equation?
     
  5. Oct 30, 2006 #4
    Ok.....Kinetic Energy is conserved and so is momentum......
     
  6. Oct 30, 2006 #5
    So..Conservation of Momentum yields me:

    m1v1 + m2v2 = m2vx + m2vy

    Where vx and vy are my unknowns
     
  7. Oct 30, 2006 #6
    Do I solve for vx in the momentum equation and then plug that anwer in for my V3 on my original or something like that?
     
  8. Oct 30, 2006 #7

    Doc Al

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    Why not use the same symbols that you used in your conservation of energy equation?
     
  9. Oct 30, 2006 #8
    yes... Than would be easier. I just named them different because they were different equations. But, I am going on the right track now?
     
  10. Oct 30, 2006 #9

    Doc Al

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    By using conservation of energy and conservation of momentum, you are on the right track. (Your notation of Vx and Vy bothers me a bit--I hope you are not thinking in terms of x and y components.)
     
  11. Oct 30, 2006 #10

    No. I am not thinking in terms of x and y components. I just simply changed the notations because I was looking at two different formulas.
     
  12. Oct 30, 2006 #11

    Doc Al

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    Cool. But the same quantities should have the same notation, even if they appear in different equations. That's how you'll end up with two equations and two unknowns--not four unknowns. :wink:
     
  13. Oct 30, 2006 #12

    radou

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    No problemo. But it's a bit more practical to name the velocities after the collision as v1' and v2'.
     
  14. Oct 30, 2006 #13
    So, is my answer correct. After doing calculations, I determined that ball 1 transfered it velocity to ball 2 and vise versa. Ball 2 is now at 10 m/s and ball 1 is now at 40 m/s.
     
  15. Oct 30, 2006 #14

    Doc Al

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    That's right--they swap speeds.
     
  16. Oct 30, 2006 #15
    Cool...Thanks for all your help everyone
     
  17. Oct 30, 2006 #16
    Just for general knowledge. If the collison were inelastic and the formula would be:
    Vfinal = (m1v1 +m2v2)/m1 + m2

    Would I need to set v1 or v2 to negative since they are going in the opposite direction?? When I set v2 to negative, the final V= -15m/s.
    When I leave it positive, Vfinal = 25 m/s
     
  18. Oct 30, 2006 #17

    radou

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  19. Oct 30, 2006 #18
    The link did not work...
     
  20. Oct 30, 2006 #19
    It should work for you.
     
  21. Oct 30, 2006 #20

    Doc Al

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    Yes. (See my added brackets.)

    Absolutely! If v1 is positive, v2 must be negative. Momentum is a vector--direction matters.
     
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