
#1
Mar2505, 04:32 PM

P: 390

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
Mar2505, 04:53 PM

P: 190

Not dumb at all. The object that has kinetic energy got it from having work done on itthat 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. 



#3
Mar2605, 06:45 AM

P: 390

"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. 



#4
Mar2605, 10:12 AM

Emeritus
PF Gold
P: 8,147

Momentum and Kinetic Energy
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.




#5
Mar2605, 03:49 PM

P: 190

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 



#6
Mar2605, 03:54 PM

Sci Advisor
HW Helper
P: 11,866

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. 



#7
Mar2705, 03:44 AM

P: 390

OK, I get it.
Thanks. 



#8
Oct2611, 08:59 AM

P: 31

whats the third integral? rate of change of KE?




#9
Oct2611, 09:34 AM

P: 343

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.




#10
Oct2611, 09:44 AM

P: 31

dont know what that means




#11
Oct2611, 09:48 AM

P: 343

This is Lagrangian mechanics.




#12
Oct2611, 12:02 PM

P: 31

so there is no third integral? nothing with the formula 1/3mv^3?




#13
Oct2711, 10:59 AM

P: 5,634



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