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Momentum and conservation of energy

  1. Oct 21, 2005 #1
    I two object collide, and there is no loss of kinetic energy, the following relationship is true:

    [tex]{v'_1}{m'} + {v_1}{m}= {v'_2}{m'} + {v_2}{m} [/tex]

    But also by conservation of energy,


    [tex]{v'_1^{2}}{m'}/2 + {v_1^{2}}{m}/2= {v'_1^{2}}{m'}/2 + {v_1^{2}}{m}/2 [/tex]


    [tex]{v'_1^{2}}{m'} + {v_1^{2}}{m} = {v'_2^{2}}{m'} + {v_2^{2}}{m} [/tex]

    That means for any two mass, and velocities that share the relationship


    [tex]{v'_1}{m'} + {v_1}{m}= {v'_2}{m'} + {v_2}{m} [/tex]

    they also share this relationship

    [tex]{v'_1^{2}}{m'} + {v_1^{2}}{m} = {v'_2^{2}}{m'} + {v_2^{2}}{m} [/tex]

    which is obviously not true. Why is there a paradox?
     
    Last edited: Oct 21, 2005
  2. jcsd
  3. Oct 21, 2005 #2

    Tide

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    Why do you think it is not true?
     
  4. Oct 21, 2005 #3
    If we start with the following:

    [tex]{v'_1}{m'} + {v_1}{m}= {v'_2}{m'} + {v_2}{m} [/tex]

    [tex]{v'_1}{m'} - {v'_2}{m'} = {v_2}{m} - {v_1}{m}[/tex]


    [tex]m'(v'_1 - v'_2) = m(v_2 - v_1)[/tex]


    Now with this relationship


    [tex]{v'_1^{2}}{m'} + {v_1^{2}}{m} = {v'_2^{2}}{m'} + {v_2^{2}}{m} [/tex]


    [tex]m'(v'_1^{2} - v'_2^{2}) = m(v_2^{2} - v_1^{2}) [/tex]


    Let m' = 4, m = 4, v'_1 = 4, v_1 = -3


    [tex]4(4 - v'_2) = 4(v_2 + 3)[/tex]

    [tex]4 - v'_2 = v_2 + 4[/tex]

    Let v'_2 be -2, so that v_2 = 3

    [tex]m'(v'_1^{2} - v'_2^{2}) = m(v_2^{2} - v_1^{2}) [/tex]

    [tex]4(4^{2} - (-2)^{2}) = 4(3^{2} - ((-3)^{2}) [/tex]

    [tex]12 = 0 [/tex]

    So it is not true
     
    Last edited: Oct 21, 2005
  5. Oct 21, 2005 #4

    HallsofIvy

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    No, you can't do that. If you have an elastic collision between two bodies, you can't select the final speed of either. Given your initial information:
    both masses= 4, v'_1= 4, v_1= -3,
    Then:
    conservation of momentum: 4(4)+4(-3)= 4v'_2+ 4(v_2)
    conservation of kinetic energy: 4(16)+ 4(9)= 4(v'_2)^2+ 4(v_2)^2.
    Of course, you can just cancel all of the masses (4s):
    v'_2+ v_2= 1 and v'_2^2+ v_2^2= 25.

    Now, you have two equations in two unknowns.
     
  6. Oct 21, 2005 #5
    Humm yes... they are consequent one of another so this makes more sense. Thank you!
     
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