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A system of particles

  1. Dec 15, 2004 #1
    I need some help on how to solve this question, I don't really know how to do this at all.

    A vessel at rest explodes, breaking into three pieces. Two pieces having equal mass, fly off perpendicular to one another with the same speed of 30 m/s. The third piece has three times the mass of each other piece. What are the magnitude and direction of it's velocity immediately after the explosion?

    I was told that the sum of the potential energy has to equal the final kinetic energy, but isn't there no potential energy at the beginning? :confused:
  2. jcsd
  3. Dec 15, 2004 #2


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    You don't need to know anything about the potential energy. Whatever chemical potential energy there was has been expended (the problem states immediately after the explosion) so all you have to concern yourself with is the initial kinetic energy and momentum of the fragments and, of course, both of those quantities will be conserved.
  4. Dec 16, 2004 #3
    So the sum of the energy and momentum has to equal zero? Since the initial kinetic energy was zero?
  5. Dec 16, 2004 #4


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    In the initial situation,the KE of the body was zero.The momentum of the body was zero.

    Hopefully u can project the momentum conservation law on the 2/3 axes of coordinates and with use of the KE conservation law,u can find your answers.


    EDIT:OOOOOOOOOOOOOOOOOPSSSSSSSSSSSS!!!!!!!!!!!!The KE cannot be zero in the initial case,since it would be zero at the end,too,therefore it would be no moving around...
    I'm an idiot!!!!!!
    Last edited: Dec 16, 2004
  6. Dec 16, 2004 #5


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    The initial kinetic energy is NOT zero! However, the momentum vector is.
  7. Dec 16, 2004 #6
    The Kinetic Energy will increase as you can see... the initial energy is 0. It probably has chemical potential energy but that's not relevant. MOMENTUM IS CONSERVED. Which means that:

    m_1v_1' + m_2v_2' + m_3v_3' = 0
  8. Dec 16, 2004 #7
    When I try to solve for V3 in that equation I keep getting the square root of a negative number...
  9. Dec 16, 2004 #8
    nevermind I figured it out
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