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Momentum and Kinetic Energy

by Pseudo Statistic
Tags: energy, kinetic, momentum
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Pseudo Statistic
#1
Mar25-05, 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.
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PBRMEASAP
#2
Mar25-05, 04:53 PM
P: 190
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.
Pseudo Statistic
#3
Mar26-05, 06:45 AM
P: 390
Quote Quote by PBRMEASAP
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.
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.

selfAdjoint
#4
Mar26-05, 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.
PBRMEASAP
#5
Mar26-05, 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
dextercioby
#6
Mar26-05, 03:54 PM
Sci Advisor
HW Helper
P: 11,948
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.
Pseudo Statistic
#7
Mar27-05, 03:44 AM
P: 390
OK, I get it.
Thanks.
great_sushi
#8
Oct26-11, 08:59 AM
P: 31
whats the third integral? rate of change of KE?
ZealScience
#9
Oct26-11, 09:34 AM
P: 351
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.
great_sushi
#10
Oct26-11, 09:44 AM
P: 31
dont know what that means
ZealScience
#11
Oct26-11, 09:48 AM
P: 351
This is Lagrangian mechanics.
great_sushi
#12
Oct26-11, 12:02 PM
P: 31
so there is no third integral? nothing with the formula 1/3mv^3?
Naty1
#13
Oct27-11, 10:59 AM
P: 5,632
How are Momentum and Kinetic Energy related?
Some practical insights here:

http://en.wikipedia.org/wiki/Conserv...inear_momentum

Momentum has the special property that, in a closed system, it is always conserved, even in collisions and separations caused by explosive forces. Kinetic energy, on the other hand, is not conserved in collisions if they are inelastic. Since momentum is conserved it can be used to calculate an unknown velocity following a collision or a separation if all the other masses and velocities are known.

A common problem in physics that requires the use of this fact is the collision of two particles. Since momentum is always conserved, the sum of the momenta before the collision must equal the sum of the momenta after the collision.

Determining the final velocities from the initial velocities (and vice versa) depend on the type of collision. There are two types of collisions that conserve momentum: elastic collisions, which also conserve kinetic energy, and inelastic collisions, which do not.
jtbell
#14
Oct27-11, 11:43 AM
Mentor
jtbell's Avatar
P: 11,864
Necropost alert.


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