Derivative of kinectic energy , dK/dv

c77793

i'm in my freshman year and i'm starting to learn Derivatives in Calculus, and I was wondering, once Ek (kinectic energy) = 1/2mv², then, the derivative of Ek in term of velocity would be mv, which is equal to the linear momentum..... I'm finding hard to understand the idea that the variation of Kinect over a variable velocity is the linear momentum...can someone explain me this in a didactic way? I was just playing around deriving physics formulas haha

tiny-tim

hi c77793!
it's because of Newtonian Relativity …

the (Newtonian) laws of physics are the same in any inertial frame

suppose you have bodies with ∑m_iu_i² = ∑m_iv_i² in one frame

now choose another frame with relative velocity w in the k-direction …

∑m_i(u_i - wk).(u_i - wk) = ∑m_i(v_i - wk).(v_i - wk)

cancelling (and dividing by 2) gives us …

w∑m_iu_i.k = w∑m_iv_i.k

since w is arbitrary, we can divide by w, and get conservation of momentum in in the k direction

however, instead of cancelling, we could have differentiated wrt w (in other words, exploiting the translational symmetry of Newtonian space), giving …

w∑m_i(u_i - wk).k = w∑m_i(v_i - wk).k

(and then cancelled, giving the same result)

(btw, this also works with Einsteinian energy and momentum, in Minkowski space)

yoelhalb

Well I was also wondering about that and I can't still find the physical relationship between the two, however both are related to force (since force is the derivative of momentum and work is (force)(distance)).
The complete derivation you can find in this thread http://www.physicsforums.com/showthread.php?t=68682