Eureka!
My final mistake is a little embarrassing at this point: I was so sure the other formula was the only thing wrong that I neglected to double check my UT equation. This entire time I kept forgetting to divide KQQ by r before plugging it in... :rolleyes:
I'm getting 1.77 cm/s and...
That's the correct answer for the heavier of the two beads (technically B). And you used the formulas I worked out here? I'll have to try it again, I'm obviously missing something.
I used (Kq1q2)/r for Uelec of two point charges.
Are you getting something other than 1.77cm/s and 1.06cm/s? I will laugh (and cry) if it turns out I screwed it all up from the outset!
= Va2 = Ui/(Ma/2 + Ma2/2Mb)
= Ui/((2MbMa + 2Ma2)/4Mb)
Va = √<4MbUi/(2MbMa +2Ma2)>
I hope that's more legible. I tried using latex but I couldn't figure out how it worked.
I moved back a few steps, and tried calculating with:
Va2 = Ui/(Ma/2 + Ma2/2Mb)
Edit edit: I am going to post how I calculated this in a minute.
But the answer is still way off. I'm going to keep checking my algebra.
Edit: Yeah, it still comes up with around 6cm/s, when the answers are 1.77...
Math and I don't get along. I had 90% on all of my theoretical stuff in calculus, but barely passed because I make so many arithmetic errors. It's not that I rush, I just can't see them. I did those calculations three times and it looked perfect to me.
Process:
Ka + Kb = Ui
Vb = (MaVa)/Mb...
I'm subbing that in, but I am still getting the wrong answer.
(1/2)*(MaVa2) + (1/2)*(MbVb2/Mb) = Ui
Va = root(2MbUi/Ma+Ma2)
Comes to ~50 cm/s when I need 1.77.
I actually tried subbing in momentum earlier and nothing would work out correctly, which is why I asked.
Edit: Correcting typos.
Conservation:
1/2mava2 + 1/2mbvb2 = Ui
Momentum:
MaVa = MbVb
Tautology:
Ek = p2/m = 1/2mv2
I'm sure the fact that I haven't taken physics in a decade is probably contributing to my problems...
Update:
I'm still working on it, but so far I'm just ending up with equations that simplify back to the equivalent of Va = Va.
PS: Is there a reason the forums are timing me out after ten minutes? I don't have time to compose a reply before it starts throwing errors.
I haven't had a chance to try out the new info just yet. The class average for last week's quiz* on charge + fields was quite low, so our professors opted to do some review of that material over the past couple days. Since they wouldn't be able to finish the material this week, we were granted...
OH! So by combine you mean rearrange the equations to be equivalent to V and thus equivalent to each other? I'm not sure how that let's me solve for v but I'll try it!
Sorry, I've been rearranging things but I just don't understand how those are supposed to fit together; I don't even know where or how I am supposed to combine these equations.
I don't even definitively know what my "energy equation" is.
Do you mean I should multiply the final velocity by the bead mass? I tried that and it did not work out.
I am not sure I understand how to combine the two equations. I am posting from my phone but I will test a few things out after class and see what I get.
Hi tim,
I'm afraid I still don't understand. I grasp that momentum is conserved, and that this means the different masses will move at different speeds under an identical force.
I thought momentum was already accounted for by the mass component of the velocity equation. Since that isn't...