This is the funniest decision you'll ever make.(adsbygoogle = window.adsbygoogle || []).push({});

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!!

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# Let the majority decide!

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