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M=P^2/2KE: How is this derived?

  1. Jul 8, 2010 #1
    I am just curious as to how this fits in. If momentum P=mv, and kinetic energy KE=1/2mv^2, how would one combine, derive, switch and swap (whatever the process is called), these two equations to end up with the formula m=P^2/2KE. It seems like a no-brainer, but I can't seem to make sense of the algebra.

    Thanks,
    JHCreighton
     
  2. jcsd
  3. Jul 8, 2010 #2

    Pengwuino

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    Gold Member

    It's fairly simple, no trickery involved. What does [tex]P^2[/tex] equal? Now, that almost looks like something you have with your kinetic energy equation. Can you convince yourself that [tex]KE = \frac{m^2 v^2}{2m}[/tex] is the same as your original equation? If so, simply plug in [tex]P^2[/tex]. From there, simply solve for m.
     
  4. Jul 8, 2010 #3
    Hey, that's great! You're right, it is pretty simple. I almost feel foolish for not thinking to solve like that. Thanks for the speedy response.

    JHCreighton
     
  5. Jul 8, 2010 #4

    russ_watters

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    Staff: Mentor

    You can also easily derive it using f=ma, d=st and the definition of work, w=fd.
     
  6. Jul 8, 2010 #5
    By solving the system:

    [tex]
    p = m \, v
    [/tex]

    [tex]
    K = \frac{1}{2} \, m \, v^{2}
    [/tex]

    with respect to [itex]m[/itex] and eliminating [itex]v[/itex].
     
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