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Derivative of kinectic energy , dK/dv 
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#1
Apr2412, 07:18 PM

P: 10

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



#2
Apr2512, 04:23 AM

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P: 26,160

hi c77793!
the (Newtonian) laws of physics are the same in any inertial framesuppose you have bodies with ∑m_{i}u_{i}^{2} = ∑m_{i}v_{i}^{2} in one frame now choose another frame with relative velocity w in the kdirection … ∑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_{i}u_{i}.k = w∑m_{i}v_{i}.ksince 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) 


#3
Apr2512, 10:40 AM

P: 69

The complete derivation you can find in this thread http://www.physicsforums.com/showthread.php?t=68682 


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