<p>I figured I would start this thread. I did out most of the questions and these are the answers I got. THEY ARE BY NO MEANS CORRECT, so let me know where I went wrong. I also didn't bother much with Mech 1 as (a) Left c blank and (b) there's probably a lot of leeway with the plotting. Also, I just didnt know how to do some parts so I'll leave that to you guys</p>
<p>Mech 2.
(a) Friction at bottom, upslope
Weight straight down
Normal perpendicular to surface
(b) Ff=8.41 N
(c) 5.29 m/s
(d) 2.29 m/s</p>
<p>E&M 1.
(a) Vb>Va=Vc
B is closer to the charge distribution (NOTE: I did this with an integral AND I did this the short way ie kQ/R and both answers matched up for me)
(b) V = kQ/R
(c) v=sqrt(2kqQ/mR)
(d) to the right (along the x-axis)
(e) E = rad(2)<em>Q / (2pi^2</em>R^2*e0)</p>
<p>E&M 2.
(a) I=0A
(b) Q=300uC
(c) 2.25 x 10^-3 J
(d) I = 2.25
(e) LOL (for giggles, I put 0 C)
(f) 600 J </p>
<p>E&M 3.
(a) Counterclockwise
By the RHR, B through the loop is out of the page but decreasing because I through the wire is decreasing. To compensate, the current will be ccw to oppose this changing flux by generating a B field that is out of the page</p>
<p>(b) Remains the same. The induced emf in the loop is the rate at which flux is changing, which is constant. Therefore, the induced current is constant?</p>
<p>Also, I’ll write out general stuff for Mech 1
(a) Cv^2 = mg at terminal speed
vT=sqrt(mg/C)</p>
<p>(b) i. graph vT^2 vs. m
ii. C = g/slope because slope = vT^2 / m</p>
<p>(c) I had something that was concave down and asymptotic to a terminal velocity. I also had absolutely no Idea what was going on with Y and T so I said roughly where the graph starts evening out is where t=T was -_-</p>
<p>Anyone wanna do a quick point distribution run-down?</p>
<p>Also, how many points would I lose in Mech2a if my normal vector wasn’t origination at the point of contact b/w the ball and surface, but all my other vectors are right?</p>
<p>I just did sum of forces was mgsin30 - f = ma, where f is frictional force and f and a are unknown, The second equation is sum of torques: fR = I(a/R). You just needed to solve for f, not the coefficient first, although I think if you used f=u*N you would’ve gotten the same answer anyway</p>
<p>For E&M 2d, I imagined the two charged capacitors as breaks in the circuit. This is equivalent to a circuit with the 30 V battery and 20 ohm and 40 ohm resistor in series. Therefore, by Ohm’s Law, the current is 0.5 A. </p>
<p>For E&M 2e, using Ohm’s Law again, the voltage across the 20 ohm resistor is 10 V since the current is 0.5 A. By using loop rule, the voltage across the 5 uF capacitor must be 20 V. The charge on a capacitor is capacitance*voltage, or 0.0001 C.</p>
<p>I remeber using the loop rule for two parts. When I had to find the charge, I just used the loop rule for the bottom cap, ressitor, and the battery.
V - IR - Q/C = 0, and solved for Q. omg</p>
<p>I honestly don’t think my answers for E&M 2 are correct, so i wouldn’t go by them. </p>
<p>I think for 2d I said that one loop was through the 40 ohm resistor, down through the switch, and through the 20 ohm resistor. So now that I do it, I get 0.5 A although I have absolutely NO idea how I got 2.25 on the test because I had the same concept. I might’ve done parallels by accident.</p>
<p>But then for some reason I switched to series for 2f because I did E=I^2<em>R</em>t = (.5)^2<em>40</em>60 = 600 -_-</p>
<p>Also, can someone offer insight on E&M1+3? Are my answers right?</p>
<p>I agree with all your answers for E&M 1 and 3! </p>
<p>The only tiny thing that caught my eye is the negative sign in your expression for voltage in your power expression. Applying chain rule when differentiating flux with respect with time gives you a -K. But then emf is -dflux/dt, so it becomes positive K. It won’t matter once you square the voltage though. Again, it’s a tiny thing that no one should worry about. :P</p>
<p>Haha, Arabidopsis, looks like you got all the E&M frqs right :o Do you think they’d actually take off for me leaving the negative in there? I thought the negative was for direction only so I just did the magnitude of the derivative, not realizing that the two negatives woul’ve just canceled out</p>
<p>Also, for 1b on E&M, my work was kind of stupid. I did</p>
<p>V = k∫dq/r
V = k∫(0–>Q) dq/r
V = kQ/R</p>
<p>It really didn’t make sense…would I lose a point for incorrect work? </p>
<p>Any idea if I could get a 5 on E&M with 7 omitted on MC, 28 answered, with maybe 6-8 or more wrong, but near full credit on the FRQs? I think my frq points will be something like 13 + 10 + 14. Do you think 2d and 2e were worth alot? I think that’s the only part I screwed up badly on, hoping it’s only like 5 points haha</p>
<p>Mech. 1
a-b) Same
c) i. I drew a concave-up function that was always decreasing and evened out at t=T
ii. I said just to take the area under the curve from t=0 to t=T to find Y.
d) Same</p>
<p>Mech. 2
a-d) Same</p>
<p>Mech. 3
a-d) Same
e) ma = T - mgsin(theta), so graph:
F<em>1 = ma</em>max<em>sin(pi</em>t/T) + mgsin(theta) <– 1/2 period of sine with endpoints at mgsin(theta) and a max at m<em>a<em>max + mgsin(theta)
F</em>2 = m</em>a<em>max<em>e^(-pi</em>t/2/T) + mgsin(theta) <– negative exponential function that starts at ma</em>max + mgsin(theta) and ends a little bit above mgsin(theta)</p>
<p>Haha, awesome that we got so many same answers. Do you think the graph questions will be worth that much? I mean in mech3, it just seems like a-d were so calculus/algebra “intensive” that those parts will be worth way more than the graph. </p>
<p>I think I should be safe for a 5 on mech but not sure. 6 omitted (ran out of time), maybe 4-6 wrong, but like 10+15+12 on the FRQs.</p>
<p>I know for 1998, 55/90 and 49/90 respectively were 5s but is that extreme is typical?</p>
<p>Wait, wow, I have no idea what I was thinking. My graph starts at a positive number - not zero - and it decreases as it approaches v_t. I can’t believe I did that. Can someone explain to me what I was thinking during that section??? Maybe I was thinking it was velocity-time graph (as opposed to speed-time), but I new I was graphing speed… wow. Oh well, hopefully it’s just one point.</p>
<p>EDIT: My guess is that one point for starting at zero/concave down, another for having an asymptote at v_t.</p>
<p>It’s alright, I just looked at c-d of #1 and was like “oh wow…hmm, ok draw the graph, then skip to 2.” Also, I think c-i will also have a point for labeling t=T where the distance fallen was Y. What did we have to do for that? </p>
<p>I prob lost points for c-ii considering I left it completely blank. I don’t know why but when I get nervous I have trouble reading coherently and I’ll just read like every other word, double back, do it again, then finally read it slowly. </p>
<p>Any ideas as to the curve? I’m hoping it won’t be something ridiculous like 60+/90 because I’m at around a 57ish/90 in the worst case right now</p>
<p>Yeah, the negative sign is only there to indicate direction in Faraday’s Law. I put the negative because Faraday’s Law on the test’s equation sheet had it. Once again, it’s insignificant and you will get full credit.</p>
<p>For Mech. 1ci, if you drew a decreasing concave-up function, did you have your asymptote for terminal velocity below the x-axis? v(t) with drag force usually resembles a k-e^-t curve. For example: <a href=“http://www.gcsescience.com/Velocity-Time-Graph-Rocket.gif[/url]”>http://www.gcsescience.com/Velocity-Time-Graph-Rocket.gif</a> I finished a bit early and tried to solve the differential equation to confirm my graph. I gave up because using partial fractions to integrate 1/(v^2 - mg/C) dv was too much work, but at least it told me that it was still an e curve.</p>
![http://www.gcsescience.com/Velocity-Time-Graph-Rocket.gif[/url]](http://www.gcsescience.com/Velocity-Time-Graph-Rocket.gif%5B/url%5D){kind=link}
