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    Why is the derivative of kinetic energy (0.5mv2) wrt velocity equal to momentum (mv)? Does this have any physical or mathematical significance?
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    (Original post by Mathsandphysftw)
    Why is the derivative of kinetic energy (0.5mv2) wrt velocity equal to momentum (mv)? Does this have any physical or mathematical significance?
    I have always wondered this also. I believe it is a consequence of the fact that  \frac{dp}{dt} ds = \frac{ds}{dt} dp = v dp = mv dv , i.e for an object of constant mass m accelerated from rest to a momentum of P (at which point it has a speed V) though a displacement x, the KE gained is:

     \displaystyle \int_0^x F ds = \displaystyle \int_0^x \frac{dp}{dt} ds  

= \displaystyle \int_0^P v dp = \displaystyle \int_0^V mv dv = \frac{1}{2}mV^2

    This is really just two changes of variable; from s to p and p to v.

    But this is entirely speculative so I welcome any corrections.
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    (Original post by Mathsandphysftw)
    Why is the derivative of kinetic energy (0.5mv2) wrt velocity equal to momentum (mv)? Does this have any physical or mathematical significance?
    Again, Ive always wondered the same thing!
    I finally got around to asking one of the physics teachers at my college and she basically said it's just a coincidence.
    Aside from what you'd get from the integration/differentiation process, I don't think there's any other intuitive understanding to be had, unfortunately
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    (Original post by Mathsandphysftw)
    Why is the derivative of kinetic energy (0.5mv2) wrt velocity equal to momentum (mv)? Does this have any physical or mathematical significance?
    Physically, this only holds in Newtonian dynamics, where Galilean Invariance is assumed - i.e all inertial frames share the same universal time and space is absolute.
    In actuality relativistic effects feature and you get a somewhat messier form for the kinetic energy.

    Mathematically, we can see this by considering the definition of work and force.
    In the case where we accept that p=mv and we compute in one dimension, with a the change in kinetic energy of a particle of fixed mass m driven by a force F, with initial velocity we see that:
    \Delta E_k=\displaystyle\int F \ dx = \displaystyle\int F \dfrac{dx}{dt} \ dt = \displaystyle\int \dfrac{dp}{dt} v \ dt

\Rightarrow \Delta E_k = \displaystyle\int v \ d(mv) =\displaystyle\int mv \ dv = \dfrac{1}{2}mv^2+\mathcal{C}
    (\mathcal{C}=0 using the fact that we started at rest.)
    Hence \dfrac{dE_k}{dv}=p by the Fundamental Theorem of Calculus.

    EDIT: Lol, this is what happens if you take a shower before typing up stuff - you end up spouting the same stuff as everyone else.
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    Thanks for the clear up guys, much appreciated
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    (Original post by Mathsandphysftw)
    Why is the derivative of kinetic energy (0.5mv2) wrt velocity equal to momentum (mv)? Does this have any physical or mathematical significance?
    You have had good responses but I thought I should just to give a non-rigorous proof of the kinetic energy formula in case you are not convinced:

    W=Fs=K.E
    Using a suvat equation not containing time, v2=u2+2as, and rearranging to s we get: s=(v2-u2)/2a
    And from newtons second law: F=ma

    Substituting for s and F we get:

    W=ma(v2-u2)/2a=m(v2-u2)/2
    Assuming the initial velocity was zero, we arrive at:
    W=mv2/2=K.E

    Edit: Mathsandphysftw: just realised that you are not to derive K.E formula, please ignore me
 
 
 
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