# Let the majority decide!

## What is it going to be?

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

0 vote(s)
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1. Sep 17, 2004

### Archimedes

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!!

2. Sep 17, 2004

### humanino

if $$m=\frac{dp}{dV}$$ is a constant then $$g=\frac{d^2p}{dV^2}=0$$ and both answers are correct !

3. Sep 17, 2004

### humanino

There is no way I can vote ! You short-cuted my brain :surprised