Question

Something i noticed, but havent really googled about.

Is momentum the derivative of kinetic energy; or is it just a coincidence that:

Dv [ 1/2 mv^2] = mv ??

Answer 1

kinetic is a part of that.

Answer 2

After about an hour of looking for the best answer (googling), I realized that I had not found a crystal clear explanation for the problem at hand, not even close. But still, I want to share the only 2 pieces of info that remotely concentrated on the problem:

The first one, i think, sorta confirms that its no coincidence: (read the last three paragraphs) http://van.physics.illinois.edu/qa/listing.php?id=1533

The second one is from a forum where this person says "It's important to realize that momentum is the derivative of K.E. only with respect to velocity.
K.E = (1/2)mv^2
d(KE)/dv=mv

Most people aren't good at visualizing things in the velocity domain - we usually use time or position.

When you say, "the rate at which something is changing" most people interpret this as referring to time, which isn't the case here.

In order to take the time-derivative, you have to express K.E. in terms of time, using v=dx/dt:

KE = .5mv^2 = .5m(dx/dt)^2

then take the time derivative:

d(KE)/dt = .5m[d(dx/dt)^2/dt]
d(KE)/dt = .5m[2(dx/dt)(d^2x/dt^2)]

replace dx/dt with v and d^2x/dt^2 with a (acceleration)

d(KE)/dt = mva
d(KE)/dt = pa

The time-derivative of KE is momentum times acceleration.

d(KE)/dv = p

The velocity-derivative of KE is momentum.

Because most people aren't good at thinking about things in the velocity domain, it is best to think of KE and P as two separate, but related properties. Different problems will be easier to solve in one or the other, so a person should be able to convert from one to the other - but don't concentrate on the relation between the two."

Answer 3

source of the second one: http://forum.woodenboat.com/archive/index.php/t-30070.html

Answer 4

definantly interesting stuff to me :) thanx for the research