Recent content by Arm

$$ (5E-3)(v_1)=(5E-3+2)(v_2) $$ $$ \frac{1}{2} (5E-3+2)({v_2}^2)=(5E-3+2)(10)(0.2)(2) $$ $$ v_2 = 2.83 $$ $$ (5E-3)(v_1)=(5E-3+2)(2.83) $$ $$ v_1 = 1134 $$ thanks

$$ \frac{5E-3*v^2}{2} = (2 + 5E-3)(10)(0.2)(2) $$ v = 56.64 I just don't get how this is the wrong answer....it's just simple conservation of energy, right?

I have no idea what that means because I'm taking the equilivent of general college physics 2 right now but I'll look into it; if you have a recommended resource I'll read/watch it.

I see where I made the mistake sign now Ignoring variable of integration and +C for the umpteenth time has gotten me the wrong answer yet again; when will I learn my lesson? Here's the actual correct answer $$v^2 = \frac{-2k q_1 q_2}{mr} + C$$ The particle is released from rest ##x## meters...

I think the sign is correct

$$ \int v {\:} dv = \int \frac{b}{r^2} v {\:} dt $$ $$ \int v {\:} dv = \int \frac{b}{r^2} \frac{dr}{dt} {\:} dt $$ $$ \int v {\:} dv = \int \frac{b}{r^2} dr$$ $$ \frac{v^2}{2} = \frac{-b}{r}$$ $$ b = \frac{-k q_1 q_2}{m} $$ $$ \frac{v^2}{2} = \frac{(-)(-)k q_1 q_2}{mr}$$ $$ v^2 = \frac{2k q_1...

$$v \frac{dv}{dt} = \frac b {r^2} \frac {dr}{dt}$$ $$v {\:} dv = \frac{b}{r^2} \frac{dr}{dt}dt$$ $$\int v {\:} dv = \int \frac{b}{r^2} v {\:} dt$$ Not sure how I would take an integral with position and with velocity in it

also fixed, thank you

fixed and fixed, thank you But I pulled the dr out of nowhere, I don't know if this is allowed

Here is my epic fail at trying to derive the equation using Lagrange (this was my first time trying to use lagrangian mechanics except for when I memorized the derivation for a pendulum) $$L = \frac{m \dot r^2}{2} - \frac{k q_1 q_2}{r}$$ $$\frac{\partial L}{\partial r} = \frac{k q_1 q_2}{r^2}$$...

## \frac{-ba^2T^5}{5} ##

Yeah that's what I meant, mistyped the reply

My calculus is very bad which is why I don't follow the steps where you change domains, but I ended up solving it by using ##v(x(t))## Thank you

So is ##v(x) = \dfrac{d(x(t))}{dt}##? I'm confused right now about how to represent v(x) as a derivative.

That would make it ##\int v(x)*v(t) dt##, I don't know what to do from there. (I've only done a handful of integrals before)