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1D Elastic Collision - Velocities in the CM/ZM Frame

  1. Oct 21, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove that, for any 1D elastic collision between two particles: as viewed from the centre of mass (or zero momentum) frame of reference, the velocity of each particle after the collision has the same magnitude but opposite sign to its velocity before the collision.

    2. The attempt at a solution
    For i=1,2, let [itex]m_i, u_i, v_i[/itex] be the mass, lab velocity before collision and lab velocity after the collision respectively.

    The velocity of the centre of mass is:

    [itex]V_{CM}=\frac{m_1u_1+m_2u_2}{m_1+m_2}=\frac{m_1v_1+m_2v_2}{m_1+m_2}[/itex] (1)

    Since the total momentum in the CM frame is zero, we have

    [itex]m_1(u_1-V_{CM})+m_2(u_2-V_{CM})=0[/itex] (2)

    [itex]m_1(v_1-V_{CM})+m_2(v_2-V_{CM})=0[/itex] (3)

    Conservation of kinetic energy:

    [itex]m_1u_1^2+m_2u_2^2=m_1v_1^2+m_2v_2^2[/itex] (4)

    I need to show that

    (i) [itex](u_1-V_{CM})=-(v_1-V_{CM})[/itex]

    (ii) [itex](u_2-V_{CM})=-(v_2-V_{CM})[/itex]

    I've been playing around with equations (1)-(4) (don't see much point copying my scribbles up on here) but can't seem to come up with (i) or (ii). Can anyone point me in the right direction to solving these equations? I'd very much appreciate it :)
     
    Last edited: Oct 21, 2013
  2. jcsd
  3. Oct 21, 2013 #2

    tiny-tim

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    Hi Zatman! :smile:

    Why aren't you using the centre of mass frame? :confused:

    (ie VCM = 0)​
     
  4. Oct 21, 2013 #3
    Hmm. Not sure I follow. VCM is the velocity of the centre of mass in the lab frame (which clearly isn't zero?). Then to transform into the centre of mass frame

    [itex]v_{CM}=v_{LAB}-V_{CM}[/itex]

    Guess I'm missing something pretty fundamental here...
     
  5. Oct 21, 2013 #4

    tiny-tim

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    the question requires you to use the centre of mass frame :wink:
     
  6. Oct 21, 2013 #5
    I'm really confused now! How, exactly, is what I've set up not the centre of mass frame?

    I've taken the velocities in the lab frame for each particle, and subtracted the lab velocity of the CM to get their respective velocities relative to the CM frame?
     
  7. Oct 21, 2013 #6

    tiny-tim

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    in the centre of mass frame, m1u1 + m2u2 = 0 :wink:
     
  8. Oct 21, 2013 #7
    I agree... ugh, that certainly throws a spanner in to the works, since that would mean my velocity of the CM relative to the lab would also be zero. I thought one could find the velocity of the CM relative to the lab by differentiating

    [itex](m_1+m_2)X_{CM}=m_1x_1+m_2x_2[/itex]

    giving

    [itex]V_{CM}=\frac{m_1v_1+m_2v_2}{m_1+m_2}[/itex] ?

    But if this doesn't work, how do I even transform into the CM frame? :confused:
     
  9. Oct 21, 2013 #8
    Wait hold on, what?

    Surely, in the CM frame, it is

    [itex]m_1(u_1-V_{CM})+m_2(u_2-V_{CM})[/itex]

    that is zero? And likewise for the velocities after collision.
     
  10. Oct 21, 2013 #9

    tiny-tim

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    yes :smile:

    but what is VCM in th CM frame? :wink:
     
  11. Oct 21, 2013 #10
    In the CM frame, the CM is stationary. BUT the "VCM" in that equation is a constant equal to the velocity of the CM relative to the lab frame.

    By your logic, m1u1 + m2u2=0 means that there is zero momentum in the lab frame, which isn't (necessarily) true?
     
  12. Oct 21, 2013 #11

    tiny-tim

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    i'm not using the lab frame

    the question requires you to use the centre of mass frame …
     
  13. Oct 21, 2013 #12
    I really hope I'm not coming across as awkward, but this makes no sense to me. :(

    You're saying that in the CM frame m1u1+m2u2=0?

    How is that possible? u1 and u2 are velocities that I defined in the lab frame, hence that expression is the momentum in the lab frame -- which isn't zero. When you transform it to the CM frame (which is what I thought I'd done) you get the velocities in the CM frame - then the momentum expression is zero, but this expression isn't m1u1+m2u2, it's what I wrote above? I really can't see the flaw in this reasoning.
     
  14. Oct 21, 2013 #13

    tiny-tim

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    yes :smile:

    don't :smile:

    work in the centre of mass frame​
     
  15. Oct 21, 2013 #14
    But I need to define the velocities somewhere!

    Are you saying that I can just forget the lab frame altogether and consider solely the CM frame -- i.e. do no transformation?
     
  16. Oct 21, 2013 #15

    tiny-tim

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    yes!

    the question requires you to use the centre of mass frame :smile:
     
  17. Oct 21, 2013 #16
    Well, yes, I know I'm required to! It just seemed natural to set it up in a lab frame and then transform it.

    Let me see what I can come up with. (thanks, tiny-tim)
     
  18. Oct 21, 2013 #17
    This is what I've set up in the CM frame - initial velocities u1 and u2, and final velocities v1 and v2 (NOT the same as the quantities defined in the first post!).

    [itex]m_1u_1+m_2u_2=0[/itex]

    [itex]\Rightarrow u_1=-\frac{m_2u_2}{m_1}[/itex] (1)

    [itex]m_1v_1+m_2v_2=0[/itex]

    [itex]\Rightarrow v_1=-\frac{m_2v_2}{m_1}[/itex] (2)

    Kinetic energy considerations:

    [itex]m_1u_1^2+m_2u_2^2=m_1v_1^2+m_2v_2^2[/itex] (3)

    Now I substitute (1) and (2) into (3) to eliminate u1 and v1. Going through the algebra I end up with

    [itex]v_2^2=u_2^2[/itex]

    [itex]v_2=\pm u_2[/itex]

    ...and v2 cannot be positive u2 because otherwise there would be no collision?
     
  19. Oct 21, 2013 #18

    tiny-tim

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    that's right :smile:

    can you see now that it's much simpler working in the centre of mass frame? :wink:
     
  20. Oct 21, 2013 #19
    Yes... when you don't over-complicate things :rolleyes:

    Thank you for your help, tiny-tim. :)
     
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