Archimedes
Sep17-04, 10:22 AM
This is the funniest decision you'll ever make.
I say impulse is function of velocity alone i.e. p=f(V)
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:
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!}.
Now I assume n=2,
V_0=0,
m such that m=\frac{dp}{dV} is mass in kg,
while g such that g=\frac{d^2p}{dV^2} is I don't know what in kg^2.
This is the final developement:
p=p_0+m_0V+0,5gV^2
The decision you are about to make is what is going to be?
Option 1: If for impulse we have
p=mV
then for position we will have
x=Vt
or
Option 2: If for position we have
x=x_0+V_0t+0,5at^2
then for impulse we will have
p=p_0+m_0V+0,5gV^2
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. p=f(V)
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:
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!}.
Now I assume n=2,
V_0=0,
m such that m=\frac{dp}{dV} is mass in kg,
while g such that g=\frac{d^2p}{dV^2} is I don't know what in kg^2.
This is the final developement:
p=p_0+m_0V+0,5gV^2
The decision you are about to make is what is going to be?
Option 1: If for impulse we have
p=mV
then for position we will have
x=Vt
or
Option 2: If for position we have
x=x_0+V_0t+0,5at^2
then for impulse we will have
p=p_0+m_0V+0,5gV^2
I don't think my account should be disabled because of this funny calculus.
Thanks for voting!!