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Reduced Mass

  1. Apr 29, 2009 #1
    1. The problem statement, all variables and given/known data
    2 bodies move towars each other M1 WITH Velocity V1 and M2 with velocity V2 ,on a frictionless surface with a spring between them(attached to M1),what will be the maximum contraction(I mean You know, shrink or W\E)?


    2. Relevant equations
    reduced mass =[tex]\bar{M}[/tex]=M1M2/(M1+M2)

    3. The attempt at a solution
    now the solution says that in Center of Mass frame , all the kinetic energy is transferred to potential energy:
    0.5[tex]\bar{M}[/tex]*V(1,2)=0.5K*A(max)^2

    now why is that? I can't see it! in C.O.M after the spring has shrunk A(max ) the bodies are @ rest in C.O.M frame? how is that?
    Thanks in advanced.
     
  2. jcsd
  3. Apr 29, 2009 #2

    berkeman

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    Staff: Mentor

    Not sure what you mean with the reduced mass thing. The masses have different velocities initially -- that doesn't seem to show up in your equation... what is V(1,2) ?
     
  4. Apr 29, 2009 #3
    V(1,2)=v1+v2 relative velocity .
    forgot to square it, should be
    0.5*[tex]\bar{M}[/tex]V(1,2)^2=0.5K*A(max)^2

    this is the solution from the book, what I don't understand is why in the C.O.M frame all Kinetic energy is converted when A is max.
     
  5. Apr 29, 2009 #4

    berkeman

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    Staff: Mentor

    When the bounce happens (relative and absolute velocities go to zero), that's the instant of the max compression on the spring -- at least that's the way I interpret the question.
     
  6. Apr 29, 2009 #5
    So how is that differ from a frame of reference that is @ rest?
    this question is to show how elegenat is the solution from the center of mass frame ,
    if when the spring shrinks to its maximum the absolute velocities go to zero than in frame @ rest we will get different answers.
    still confused :S
     
  7. Apr 29, 2009 #6

    berkeman

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    Staff: Mentor

    Maybe if the masses have a velocity component that is orthogonal to their approach component... Those velocities aren't seen in the COM frame? Dunno beyond that. Sorry.
     
  8. Apr 29, 2009 #7
    Yap this is a tough one,
    It doesn't make sense :S in c.O.m which moves with speed of (m1v1+m2v2)/(m1+m2)
    and lets say it's directed right and M1 is going left initially, after the spring was compressed it has to go in the right direction(in the C.O.M) so that some how relative velocity will be 0.
    arghh stupid question, thank You that You've tried to help!
    any one?
    thanks.
     
  9. Apr 29, 2009 #8
    Using the ground as a frame of reference is hardly rocket science:
    When the relative velocity is 0, the masses have the same velocity V
    (actually the velocity of the C of M).

    M1V1 - M2V2 = (M1+M2)V

    Final KE = 1/2(M1+M2)V^2

    You can check that the initial KE referred to the ground exceeds
    that in the C of M frame by precisely this amount.
     
  10. Apr 29, 2009 #9
    M1V1 - M2V2 = (M1+M2)V
    ^^^^^
    This rox!
    It actually means that V(final) =V of center of mass, thus In C.O.M both bodies are @ rest!!
    Awesome thanks!
     
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