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Let the majority decide!

  1. Option 1 (look at the original post)

    0 vote(s)
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  2. Option2 (look at the original post)

    0 vote(s)
    0.0%
  1. Sep 17, 2004 #1
    This is the funniest decision you'll ever make.
    I say impulse is function of velocity alone i.e. [tex]p=f(V)[/tex]
    I use Tayler's differential order to develope that function near some V_0. I assume f(V) is n times differentiable and here is how the developemnet looks:
    [tex]p=p_0+\frac{dp}{dV}(V-V_0)+\frac{d^2p}{dV^2}\frac{(V-V_0)^2}{2!}+...+\frac{d^np}{dV^n}\frac{(V-V_0)^n}{n!}[/tex].
    Now I assume n=2,
    [tex]V_0=0[/tex],
    m such that [tex]m=\frac{dp}{dV}[/tex] is mass in kg,
    while g such that [tex]g=\frac{d^2p}{dV^2}[/tex] is I don't know what in kg^2.
    This is the final developement:
    [tex]p=p_0+m_0V+0,5gV^2[/tex]

    The decision you are about to make is what is going to be?

    Option 1: If for impulse we have
    [tex]p=mV[/tex]
    then for position we will have
    [tex]x=Vt[/tex]

    or

    Option 2: If for position we have
    [tex]x=x_0+V_0t+0,5at^2[/tex]
    then for impulse we will have
    [tex]p=p_0+m_0V+0,5gV^2[/tex]

    I don't think my account should be disabled because of this funny calculus.
    Thanks for voting!!
     
  2. jcsd
  3. Sep 17, 2004 #2
    if [tex]m=\frac{dp}{dV}[/tex] is a constant then [tex]g=\frac{d^2p}{dV^2}=0[/tex] and both answers are correct !
     
  4. Sep 17, 2004 #3
    There is no way I can vote ! You short-cuted my brain :surprised :bugeye:


    :biggrin:
     
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