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B Work/Impulse = Velocity?

  1. Mar 23, 2016 #1
    Since Work = F*delta(d) and Impulse = F*delta(t) then dividing work over impulse, through simple cancellation of F we can say that it is now equal to delta d / delta t which is equal to velocity right? My question is, does this make sense "physics-wise" since I onlu arrived at my answer through cancellations?
     
  2. jcsd
  3. Mar 23, 2016 #2
    Hi lonely:

    Your equations should be:
    delta(Work) = F*delta(d)
    delta(Impulse) = F*delta(t)​

    delta(Work) is an infinitesimal of Work = an infinitesimal of energy = ΔE
    delta(Impulse) is an infinitesimal of Impulse = an infinitesimal of momentum = Δp​

    From
    In empty space, the photon moves at c (the speed of light) and its energy and momentum are related by E = pc, where p is the https://www.physicsforums.com/javascript:void(0) [Broken] of the momentum vector p.

    Consider a photon which has momentum p and energy pc.
    The result of dividing the photon's energy by its momentum gives its velocity c.

    Hope this helps.

    Regards,
    Buzz
     
    Last edited by a moderator: May 7, 2017
  4. Mar 24, 2016 #3
    E = pc is the same as E = (mc)c --> E = mc^2 right? I still haven't encountered relativity so I'm not quite familiar with that famous equation but I think through your answer, it made sense. Thanks.
     
  5. Mar 24, 2016 #4

    PeroK

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    If you have ##KE = \frac{1}{2}mv^2## and ##p = mv##, then

    ##\frac{KE}{p} = \frac{1}{2}v##

    And:

    ##KE = \frac{p^2}{2m}##

    You might like to think about why

    ##v \ne Work/Impulse##
     
  6. Mar 24, 2016 #5

    jtbell

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    No, because p ≠ mc.

    The general relativistic relationship between energy, momentum and (rest) mass is E2 = (pc)2 + (mc2)2. Set m = 0 and you get E = pc.
     
  7. Mar 24, 2016 #6

    PeroK

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    Given this was a post on classical mechanics in the Classical Physics section, I'm not sure how we ended up talking about relativity!
     
  8. Mar 24, 2016 #7

    Vanadium 50

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    Because this is PF. You can't write down an inclined plane problem without someone chiming in about GR.
     
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