AP Physics C: Post-exam FRQ solutions (Mech and E&M)

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<p>When I was doing the free response, I didn’t even read most of the questions more than half-way through. I just saw a key word or two and solved for it. Thankfully, I didn’t miss anything important by doing that.</p>

<p>Your probably good for a 5 as far as the curve goes (assuming you didn’t fail the multiple choice part).</p>

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<p>No, I understand what I SHOULD HAVE done (and what I WOULD HAVE done if I were not completely stupid for a 5-minute time span). It’s hard to explain what happened, but I think what happened was in my head I was thinking velocity-time graph, and thought the x-axis was at the top of the graph for some reason. Just don’t ask. I don’t even know what I was thinking!</p>

<p>EDIT: Wait, I think I figured out what I was thinking. I set initial speed somewhere at the top for no reason at all and… wait, nevermind. I still don’t know what I was thinking. I think I’ll still get credit for c. ii. though.</p>

<p>EDIT2: My graph looked like an |acceleration|-time graph that approached a positive value. Maybe I was trying to combine the a vs t graph with the v vs t graph. Oh well, not as bad as my terrible mistake on the MC section.</p>

<p>For E/M 3, they should have really put the “brightness of the bulb” question directly after the “find the power dissipated by the bulb” question. Usually the FRQs follow in a logical order in which you use previous answers to find the next. </p>

<p>It’s not intuitive that the decreasing current in the wire creates a constant current in the loop. The current in the loop was constant only because the current in the wire was a function of t^1. If it were t^2 or t^3, then the bulb would be getting brighter as intuition suggests. You only knew the power (and therefore brightness) was constant if you completed part e, and not everyone got that far. Just my two cents about why the majority of people didn’t put that the bulb stays the same brightness. :P</p>

<p>Yeah, usually that’s the way they do it. Also, I just realized #3 from the 2004 exam was almost exactly the same as this one. No light bulb but still had to find the flux and power, and direction of induced current.</p>

<p>Any ideas what the curves will be like? I’m not familiar enough with the AP exams to guess</p>

<p>bump for tonight</p>

<p>Consolidated solutions:</p>

<p>Mech. 1
(a) vT=sqrt(mg/C)
(b) i. graph vT^2 vs. m
ii. C = g/slope
(c) concave down, starts at (0,0) and asymptotic to terminal velocity
(d) delta E = 1/2 mvT^2 - mgy</p>

<p>Mech. 2
(a) Friction at bottom, upslope
Weight straight down
Normal perpendicular to surface
(b) Ff=8.41 N
(c) v(cm)=5.29 m/s
(d) vf=2.29 m/s</p>

<p>Mech. 3
(a) v=a(max)T/pi * (1 - cos(pi<em>t/T))
(b) 2m(a(max)</em>T/pi)^2
(c) F=mgsin(theta)
(d) 2ma(max)T/pi
(e) F2 is a negative exponential function, F1 is a sine curve</p>

<p>E&M 1.
(a) Vb>Va=Vc
B is closer to the charge distribution
(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) U=2.25 x 10^-3 J
(d) I=0.5 A
(e) Q=0.0001 C
(f) E=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>(c) B=u0<em>I0 / (2pi</em>r)
(d) FluxB = u0<em>b</em>(I0 - Kt) / 2pi * ln((d+a)/d)
(e) P = (1/R)<em>(-u0</em>bK/2pi * ln ((d+a)/d))^2</p>

<p>Please pardon me for my stupidity but how did you calculate a numerical value for the force of friction when they didn’t give a number for the coefficient of friction?</p>

<p>for mech FR number 2.
i got friction is 11N haha.
i did what u guys said i just plugged in wrong i guess…</p>

<p>but for E&M number 1.
for finding electric field
i got</p>

<p>E=[2sqrt(2)kQ]/[pi(R^2)]</p>

<p>i thought i did it right?</p>

<p>Ah, nevermind I see above that someone else had my same question and I understand now. </p>

<p>Also I am not getting 2.29 for part d, I used the formula m1v1cos30 = (m1+m2)v, with that I got 1.53. I noticed that to get 2.29 you would just divide m1v1cos30 by only the mass of the wagon and box, but you should also add to that the mass of the ball. Am I wrong?</p>

<p>^^^ still talking about question 2 btw.</p>

<p>That formula should work. m1 is 6 kg and m1+m2 is 12kg</p>

<p>And now I think that part c of #2 is wrong as well, using conservation of energy:
U = K + Work done by friction. I get 5.29 m/s without considering the rotational energy (damn why couldn’t have I thought of this during test):, but considering the rotational energy I am getting v equals 4.48 m/s. Just to show my work:</p>

<p>mgh = .5mv^2 + .5(.4)mv^2 + 8.4*4 (work done by friction), when you do all those computations you get v = 4.48 m/s. I don’t know how you guys got your answers but were you taking into account the speed of rotation of the ball?</p>

<p>I took it to mean that m2 alone is 12 kg, it says the mass of the wagon and box total is 12, so (m1 + m2) =6 +12 = 18kg</p>

<p>For E/M 3a. Shouldn’t the answer be clockwise? d(phi)/dt is positive after you take the derivative of the flux. </p>

<p>From you guys’ answers, I think I got around 65%-70% on the E/M FRQ, what’s the curve for this? In 1998, it was 49/90, does that stay around there? If you have 2004 or 2009 released exams for E/M, it would be great if you can tell me the curve for that particular year. </p>

<p>P.S. I do not need Mech, I took that last year and got a 5.</p>

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<p>I don’t believe friction does any work. Since the ball is rolling without slipping, there is only static friction. (<------ right? wrong?)</p>

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<p>And thank you for making me realize I made a second horrible mistake on this test. I said in another thread somewhere that I wasn’t even reading the questions closely through the whole test and that it didn’t hurt me in the end. Wow, was I wrong - I thought it said “wagon and ball” - I guess all I saw was “wagon and b”. I didn’t even realize there was a box on this truck. That makes an error in the page 1 answers.</p>

<p>Oops, I made the same mistake. Didn’t even realize there was a box</p>

<p>And for E&M3a, it’s counterclockwise because the current int he wire produces a B field out of the page through the loop that is decreasing because I=I0 - Kt is a decreasing function. Thus, magnetic flux through the loop is decreasing. Lenz’ law tells us that current in the loop will oppose this change in flux - namely that a B field will be produced out of the page by the current in order to combat the decreasing flux.</p>

<p>And I’m almost positive no work is done friction. I’ve done the hoop/sphere/disk rolling down an incline without slipping so many times to remember that we never factored in work done by friction</p>

<p>nvm.
i got the same thing it’s just i used k instead of 1/[4pi(e0)]</p>

<p>Ah, yes Chem you are right static friction does no work, it is interesting that you get 5.29 m/s when you factor in what I did but forget the rotational energy.</p>

<p>well now I am just feeling horrible about the frq</p>

<p>I left 2 mc blank, but am expecting a pretty low score for the free response. Do you think the curve will be nicer b/c of the free response being so hard this time?</p>