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Relative velocity and conservation of momentum

  1. Dec 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Two particles of masses m and 3m are moving toward each other along the x-axis
    with the same speed v. They undergo a head-on elastic collision and rebound along the x-
    axis.

    I don't seem to be getting the right answer...I would really appreciate it if someone could help me figure out what I did wrong.

    I put the smaller mass as A and the bigger mass as B.
    MaVia + MbVib= MaVfa + MbVfb
    mv - 3mv= mVfa + 3MVfb
    -2mv= mVfa + 3MVfb

    then i used relative velocity formula
    Vaf-Vbf= -(Vai-Vbi)
    Vaf - Vbf= 2v
    Vaf= 2v+Vbf

    Then I substituted into the equation
    -2mv= m(2v+Vbf) + 3MVfb
    -2mv= 2mv + 2mVbf + 3MVbf
    -4mv= 5mvbf
    -4/5v= vbf
     
  2. jcsd
  3. Dec 10, 2008 #2

    Dick

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    Relative velocity formula? What's that? You want to use conservation of momentum (you got that equation right). Now you want to use conservation of energy to get a second equation for Vfa and Vfb. That's what 'elastic' means.
     
  4. Dec 10, 2008 #3
    my teacher said we could use relative velocity for head on collisions..
     
  5. Dec 10, 2008 #4

    Dick

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    Perhaps you can, but that would be a version of conservation of momentum. You need another relation here. You really have to use conservation of kinetic energy.
     
  6. Dec 10, 2008 #5

    mukundpa

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    I think you mean for elastic collision

    velocity of approach = velocity of separation
     
  7. Dec 10, 2008 #6

    Dick

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    No. I mean what I said. That's not right.
     
  8. Dec 10, 2008 #7

    mukundpa

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    I think we can get this using energy conservation law. I think from here only we define coefficient of restitution.
     
  9. Dec 10, 2008 #8

    Dick

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    I don't know what that is. I think 'elastic'=conserve energy and momentum. Is that the 'relative velocity rule' the poster is talking about? If it is equivalent to energy conservation then it will work. But I just don't know this jargon.
     
  10. Dec 10, 2008 #9

    mukundpa

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    The conservation of linear momentum and energy both leads to a relation

    u1 - u2 = v2 - v1

    I think this is called relative velocity of approach = relative velocity of separation and can be used with momentum conservation to avoid square terms.

    Sorry, the first reply was for student not for you.

    Regards
     
    Last edited: Dec 10, 2008
  11. Dec 10, 2008 #10

    Dick

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    Right, and it does avoid square terms. I'm sorry, I've just never seen this before. Thanks for the enlightenment. And I've got a doctorate in physics. Shows you can always learn new things. Thanks!
     
  12. Dec 10, 2008 #11
    Looking at this problem kinda helped me study for my physics final tomorrow as it is on my exam. Anyways getting back on topic. I think that because it is an elastic collision, we can assume that both objects are traveling at the same velocity after the collision in the -x direction.

    If we assume that's the case, then I think it would turn out like this instead:

    mVi - 3mVi = -(m + 3m)Vf
    mVi - 3mVi = -4mVf
    -2mVi = -4mVf
    Vi = 2Vf or Vf = .5Vi

    I'm not sure if this is how it should be conducted, but it seems to work and make sense.
     
  13. Dec 10, 2008 #12

    mukundpa

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    I think it should be

    Vbf - Vaf = 2v
     
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