- #1
Archimedes
- 2
- 0
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!
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!