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B Need help with algebra in proving kinetic energy is not conserved

  1. Mar 23, 2016 #1
    I'm trying to prove that kinetic energy is not conserved in inelastic collisions using the conservation of momentum. This is the set-up. An object A of momentum ##{m_1}{v_1}## collides inelastically with object B of momentum ##{m_2}{v_2}##
    using momentum conservation ##P_i = P_f##
    [tex] {m_1}{v_1} + {m_2}{v_2} = ({m_1}+{m_2}){v_f} [/tex] solving for ##v_f## we obtain
    [tex] v_f = \frac{({m_1}{v_1}+{m_2}{v_2})}{({m_1}+{m_2})} [/tex] now the fun part
    prove that [tex] KE_i \ne KE_f [/tex]
    [tex] \frac{{m_1}{{v_1}^2}}{2} + \frac{{m_2}{{v_2}^2}}{2} \ne \frac{1}{2}({m_1} + {m_2})(\frac{{m_1}{v_1} + {m_2}{v_2}}{{m_1}+{m_2}})^2 [/tex]

    I realize that the two quantities are no longer equal, especially if you try plugging in numbers. I was just wondering if this can be simplified to be seen more clearly.
    I multiplied out the numerator and denominator and got some pretty nasty algebra.
    [tex] \frac{{m_1}{{v_1}^2}}{2} + \frac{{m_2}{{v_2}^2}}{2} \ne \frac{1}{2}({m_1}+{m_2})\frac{({m_1}{v_1})^2 + 2{m_1}{v_1}{m_2}{v_2} + ({m_2}{v_2})^2}{{m_1}^2 + 2{m_1}{m_2} + {m_2}^2} [/tex]
    I'll continue working on it later but if anyone has any shortcuts to simplifying this I would appreciate it.
     
  2. jcsd
  3. Mar 23, 2016 #2

    Samy_A

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    You could look at the expression $$X=\frac{{m_1}{{v_1}^2}}{2} + \frac{{m_2}{{v_2}^2}}{2} - \frac{1}{2}({m_1} + {m_2})(\frac{{m_1}{v_1} + {m_2}{v_2}}{{m_1}+{m_2}})^2$$
    You want to see under what condition ##X \neq 0##.

    Simplify the expression by multiplying with ##2(m_1+m_2)##:
    You get $$2(m_1+m_2)X=({m_1}{{v_1}^2}+ {m_2}{{v_2}^2})(m_1+m_2) - (m_1v_1+m_2v_2)^2$$

    Now work out the RHS.
     
  4. Mar 24, 2016 #3
    Okay I continued and got this
    [tex] 2({m_1}+{m_2})X = ({m_1}{v_1}^2+{m_2}{v_2}^2)({m_1}+{m_2}) - ({m_1}{v_1} + {m_2}{v_2})^2 [/tex]
    [tex] 2({m_1}+{m_2})X = ({m_1}{v_1})^2 + {m_1}{m_2}{v_2}^2 + {m_1}{m_2}{v_1}^2 + ({m_2}{v_2})^2 - (({m_1}{v_1})^2 + 2{m_1}{v_1}{m_2}{v_2} + ({m_2}{v_2})^2) [/tex]
    Which simplifies to [tex]2({m_1}+{m_2})X = {m_1}{m_2}{v_2}^2 + {m_1}{m_2}{v_1} - 2{m_1}{v_1}{m_2}{v_2}[/tex]
    and if you take out an ##({m_1}{m_2})##
    [tex] 2({m_1} + {m_2})X = ({m_1}{m_2})({v_2}^2 + {v_1} + 2{v_1}{v_2}) [/tex] and we can conclude now that ##X \ne 0## when neither ##{m_2} \ne 0 ## nor ##{m_1} \ne 0## or if ##{v_2}^2 + {v_1} + 2{v_1}{v_2} \ne 0##

    Is this correct?
     
  5. Mar 24, 2016 #4

    Samy_A

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    No.
    The result has to be symmetric in ##m_1, m_2##, which it is.
    But also in ##v_1, v_2##, which your result is not.
    Check the term ##{m_1}{m_2}{v_1}## ...
    There also is a sign error in your last expression.

    I think you can take as given that the masses ##m_1, m_2## are strictly positive numbers.
     
  6. Mar 24, 2016 #5
    Yes I forgot a power of 2 [tex]2({m_1}+{m_2})X = {m_1}{m_2}{v_2}^2 + {m_1}{m_2}{v_1} - 2{m_1}{v_1}{m_2}{v_2}[/tex] here

    [tex]2({m_1}+{m_2})X = {m_1}{m_2}{v_2}^2 + {m_1}{m_2}{v_1}^2 - 2{m_1}{v_1}{m_2}{v_2}[/tex]

    and then [tex] 2({m_1} + {m_2})X = ({m_1}{m_2})({v_2}^2 + {v_1}^2 - 2{v_1}{v_2}) [/tex]

    If this is correct, ##X \ne 0## when either ##{m_1} > 0## or ##{m_2} > 0## as well as if ##{v_2}^2 + {v_1}^2 - 2{v_1}{v_2} \ne 0##
     
  7. Mar 24, 2016 #6

    Samy_A

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    "either ##{m_1} > 0## or ##{m_2} > 0##" is not correct, as the masses are both positive numbers.

    For ##X \ne 0##, the condition is ##{v_2}^2 + {v_1}^2 - 2{v_1}{v_2} \ne 0##.
    This last expression can be simplified, using the well known identity ##(a-b)²=a²-2ab+b²##.
    You then will get a nice condition for ##X \ne 0##, which will make sense if we remember that ##X## is the difference in kinetic energy (before and after the collision).
     
    Last edited: Mar 24, 2016
  8. Mar 24, 2016 #7
    so it can be written as [tex] 2({m_1} + {m_2})X = ({m_1}{m_2})({v_1}^2 - {v_2}^2) [/tex] and expressed such that ##X \ne 0## when (##{m_1} > 0## and ##{m_2} > 0##) and ##{v_2} \ne {v_1}##? is this correct or should I go further?
     
  9. Mar 25, 2016 #8

    Samy_A

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    No, this is not totally correct.
    From ##2({m_1} + {m_2})X = ({m_1}{m_2})({v_2}^2 + {v_1}^2 - 2{v_1}{v_2})##, you get ##2({m_1} + {m_2})X = {m_1}{m_2}(v_2-v_1)²##,
    ##(v_2-v_1)²## is not necessarily equal to ##{v_1}^2 - {v_2}^2##.

    The conclusion is correct. The kinetic energies will be different when ##(v_2-v_1)² \ne 0##, that is when ##v_1 \ne v_2##.
     
  10. Mar 25, 2016 #9
    I was very tired and wrote the tex incorrectly. I meant what you wrote and that's what I had on my scrap paper. Thank you for your guidance :)
     
  11. Mar 25, 2016 #10

    Samy_A

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    You are welcome.
     
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