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Expressing Newton's 2nd law in terms of momentum

  1. Mar 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Show that ##\Sigma \vec F = \frac {d \vec p}{dt} ##

    2. Relevant equations
    ##\Sigma \vec F = m \vec a ##
    ## \vec a = \frac {d \vec v}{dt} ##
    ## \vec p = m \vec v ##

    3. The attempt at a solution
    We need to prove that ## \frac {d \vec p}{dt} = m \vec a ##. When I physicists correctly take the derivative of ## \vec p ##, they get ## m \vec a ##. How come taking the derivative doesn't affect ##m## ? If ##m## is constant, shouldn't it go to zero?
    I know that I am wrong somewhere, but I want to fully understand how this formula was derived.
     
    Last edited: Mar 15, 2015
  2. jcsd
  3. Mar 15, 2015 #2
    The method to find derivative of such a function is different.

    When derivative of 'only' the constant is taken, it is zero.

    When a constant is multiplied with some function, the derivative of the resulting function is the constant multiplied with the derivative of the function.
    F=dp/dt
    F=d(mv)/dt
    F=m dv/dt
    F=ma.
     
    Last edited: Mar 15, 2015
  4. Mar 15, 2015 #3
    Edit: the V is not to be here.
     
    Last edited: Mar 15, 2015
  5. Mar 15, 2015 #4
     
  6. Mar 15, 2015 #5
    Hope it helped
     
  7. Mar 15, 2015 #6
    Yes, that makes sense.
    Normally, if ##\vec v## is a variable and ##m## is a constant, the derivative of ## m \vec v ## will become ##m##. Is ##\vec v## not a variable here?
     
  8. Mar 15, 2015 #7
    Second law of motion states that, "The rate of change of momentum is directly proportional to the force applied. "

    Now p = mv

    We take m as constant, because rate of change of mass in situations involving second law of motion is negligible (you can consider rocket propulsion as an example in which mass varies significantly with time).

    taking derivative both sides with respect to time (t),

    ΣF ∝ dp/dt
    ΣFd(mv)/dt
    ΣF m.dv/dt
    ΣF ∝ ma (a = acceleration = d(v)/dt )
    ΣF = kma (k = proportionality constant = 1, in SI units)
    ΣF = ma

    Since, you are taking derivative of velocity with respect to time equals to 1 instead of acceleration, that is why you are getting on that result.
     
  9. Mar 15, 2015 #8
    V has nothing to do here. kick it out.
     
  10. Mar 15, 2015 #9
    V is the variable.
    But derivative of mv isn't m. It is m multiplied by derivative of v with respect to time.
    Derivative of velocity with respect to time is nothing but acceleration
     
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