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Conservation of Linear momentum and Kinetic energy

  1. May 14, 2013 #1
    Suppose that we are working with a horizontal string with two masses attached. There is no friction on the floor and we move those two attached masses away from the equilibrium position by an undefined distance.
    Once the string with those two attached masses passes through the equilibrium possition, we remove one of the attached masses.
    As there are no forces present in that moment (as we are in the equilibrium position), we should apply the conservation of the linear momentum such that:
    (m1 + m2 ) * v_slow = m1 *v_fast (as we remove the m2 with no speed)
    So, we move from an initial Kinetic energy = ((m1+m2)*v_slow^2)/2
    To a final Kinetic energy = (m1*v_fast^2)/2 from which we can derive using the conservation of the linear momentum --> ((m1+m2)^2 * v_slow^2)/(2*m1) and as ((m1+m2)^2)/m1 > (m1+m2), we have essentialy given energy to the system.
    Can someone explain me where does this "extra" energy come from?
    Thanks
     
  2. jcsd
  3. May 14, 2013 #2

    Doc Al

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    What do you mean by "equilibrium position"? Please define the scenario you have in mind more completely.
     
  4. May 14, 2013 #3
    https://ecourses.ou.edu/ebook/dynamics/ch10/sec101/media/d0121.gif

    We start from an original position and move the masses away from the origin to the maximum amplitude of the Simple Harmonic Motion. Once we get to that position, we let it contract by the spring force.
    Sorry for not including a simple diagram with the doubt.
     
  5. May 14, 2013 #4

    Bandersnatch

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    If I understand you correctly, you want to remove part of the moving mass from the system at some point?
    If so, then you can't use conservation laws - these work only for close systems.

    In other words, if you magically remove one of the masses, the other one will not change velocity.
     
  6. May 14, 2013 #5

    Doc Al

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    I see one mass attached to a wall by a spring. Your first post mentions two masses attached by a string. Are you changing the scenario?

    You pull the mass from the equilibrium position and let it go. So what? (I don't see what this has to do with conservation of momentum.)

    Please restate your question with reference to the diagram.
     
  7. May 14, 2013 #6
    Thanks, completely forgot that when talking about closed systems no matter can be exchanged.
    Sorry for the stupid question.
     
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