Momentum and Kinetic Energy

  1. Probably dumb question asked before...
    How are Momentum and Kinetic Energy related?
    I've noticed P = mv and KE = 0.5mv^2 indicating that KE is just taking the integral of momentum with respect to velocity, is that a coincidence or is there a reason for such a relation?
    Which discovery came about first?
    Thanks.
     
  2. jcsd
  3. Not dumb at all. The object that has kinetic energy got it from having work done on it--that is, a force exerted over a distance.

    KE = Work Done = [tex]\int F dx[/tex]

    The force is just the rate of change of momentum: F = ma = m dv/dt. Put this into the integral to get

    KE = [tex]\int m \ \frac{dv}{dt} \ dx[/tex]

    Now use the chain rule to write dv/dt = (dv/dx)(dx/dt) = v (dv/dx):

    KE = [tex]\int m v \ \frac{dv}{dx} \ dx = \frac{1}{2} m v^2[/tex]

    This is equivalent, like you pointed out, to just integrating p=mv with respect to v.

    edit: I'm not sure which formula came first. I would think the formulas originated about the same time, but I really don't know. The concepts were known before Newton, but I think he made them precise.
     
    Last edited: Mar 25, 2005
  4. What I don't understand is this bit:
    "Now use the chain rule to write dv/dt = (dv/dx)(dx/dt) = v (dv/dx):

    KE = [tex]\int m v \ \frac{dv}{dx} \ dx = \frac{1}{2} m v^2[/tex]"
    How'd you get 1/2mv^2 after integrating mv dv/dx with respect to x? If you did the integral of mv you get 1/2mv^2... but I'm kinda tripped up with the dv/dx.
    Care to explain?
    Thanks.
     
  5. selfAdjoint

    selfAdjoint 8,147
    Staff Emeritus
    Gold Member

    Because of the Fundamental Theorem of Calculus, that integration reverses differentiation and vice versa, plus the linearity of integration, you can pretend the dv/dx is really a quotient and cancel the dx's.
     
  6. Right. The easiest way to see it is by working backwards from the answer. For example, the chain rule for derivatives gives us

    [tex]\frac{d}{dx} (\frac{1}{2} v^2) = v \frac{dv}{dx}[/tex]

    and the fundamental theorem of calculus says

    [tex]\int \ \frac{d}{dx} (\frac{1}{2} v^2) \ dx = \frac{1}{2} v^2[/tex]

    We can use the first equation to see that

    [tex]\int \ v \frac{dv}{dx} \ dx = \frac{1}{2} v^2[/tex]


    edit: took out some stray parentheses
     
    Last edited: Mar 26, 2005
  7. dextercioby

    dextercioby 12,293
    Science Advisor
    Homework Helper

    As for the history of the concepts of KE & momentum,i think u'll find interesting discussions in bigraphies of G.W.Leibniz,R.Descartes & I.Newton.

    Daniel.
     
  8. OK, I get it.
    Thanks.
     
  9. whats the third integral? rate of change of KE?
     
  10. What about considering Lagrangian? For principle of least action (with Lagrangian function) to hold, partial derivative of it must be momentum, so that action can be minimized using Newton's Law.
     
  11. dont know what that means
     
  12. This is Lagrangian mechanics.
     
  13. so there is no third integral? nothing with the formula 1/3mv^3?
     
  14. Some practical insights here:

    http://en.wikipedia.org/wiki/Conservation_of_linear_momentum#Conservation_of_linear_momentum

     
  15. jtbell

    Staff: Mentor

    Necropost alert.
     
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