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Kinetic energy, the integral of vdp

  1. Dec 2, 2004 #1
    I read that kinetic energy is:
    where v is velocity and p is momentum
    I'd like to see a calculation of this, I can't really get it right
    Is this right? If so then how do I continue?
  2. jcsd
  3. Dec 2, 2004 #2
    [tex]\int v dp = \frac{1}{m} \int mv dp = \frac{1}{m} \int p dp = \frac{p^2}{2m}[/tex]

    Which is one of the familiar forms of KE.

  4. Dec 2, 2004 #3
    bah that's a mean definition, is that 1/m understandable or is it just how it is
    I mean if you hadn't seen it before would you be able to solve it?

    u=p du=dp dv=dp v=p
    What am I doing wrong?
    how is [tex]\int(pdp)=\int(p)=\frac{p^2}{2}[/tex] ??
    Still can't solve it...
  5. Dec 2, 2004 #4


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    The idea is to get consistent variables. You either have to integrate:

    [tex]\int (p) dp[/tex] or [tex]\int (v) dv[/tex]

    Since p=mv, it's easy to convert v into p. Except you don't want to change the value of your equation, so you can only multiply by 1.

    [tex]\frac{m}{m}\int (v) dp = \frac{1}{m} \int (mv) dp[/tex]
  6. Dec 2, 2004 #5
    Oh my god, that's a good way to forget the definition of the basic integration theorem
    I thought of [tex]\int(p)dp[/tex] as [tex]\int(pdp)dp[/tex] totally forgot the meaning of dx in [tex]\int(f(x))dx[/tex] well at least I feel a bit better now and it is good that I got this sorted out before my cambridge interview, thank you!

    Could someone explain why I get
    Why the left side does not equal the right.
    I believe I have done a correct integration by parts...
    This integral is a hypothetical [tex]\int(pdp)dp[/tex]
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