AP Physics C 2008 Test is NOW on CB! Discuss FRQ's HERE!!!

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<p>Hey guys,</p>

<p>I haven't actually taken the Phys C Mech (I'm practicing for the late testing version I'm going to take), but I'll jump on the bandwagon:</p>

<p>Mech1.
1. a) There are 3 forces: normal force n, gravitational force Mg, and drag force bv.
b) Mgsin(theta)-bv=M(dv/dt). v is the unknown.
c) At terminal velocity, acceleration is 0. Thus, v=Mgsin(theta)/b
d) This is a separable differential equation. Separate variables, do indefinite integration, use the initial velocity to solve for the constant of integration. The answer I got is v=(Mgsin(theta)/b)(1-exp(-bt/M))
e) Differentiate the above equation, and sketch the graph. I got a=gsin(theta)exp(-bt/M). Thus, at t=0, a=gsin(theta).</p>

<p>2.
a) There are 4 forces: tension T, mg for mass, mg for rod, hinge force FH.
b) The hinge force is entirely horizontal. Thus, Tsin30 balances out the two masses, and T= 49N.
c) Moment of inertia of system about hinge = """ of rod about hinge + """ of block about hinge
Use parallel axis theorem and the definition of moment of inertia, I got I(total)=((1/3)M+mass of block)L^2=0.42kgm^2
d) Torque = Ia a= Torque/I=(mass of rod<em>g</em>L/2+mass of block<em>g</em>L)/I=21rads/s^2</p>

<p>Mech 3.
a) Standard graphing.
b) slope=25N/m
c) Use conservation of energy. Initially, there is GPE. In the final state, there is only SPE. Thus, solving the equation for m, I got 0.69kg.
d) Think about it this way: the mass accelerates all the way down until the cord stretches sufficiently enough such that the weight of the mass is balanced completely by the cord spring force, i.e. acceleration is zero. This is when maximum speed is attained.
Let extension of cord at max speed be x. Then, kx=mg, and x=0.27m. Thus, it dropped 0.87m.
To solve for the max speed, use conservation of energy. Initially, there is GPE. In the final state, there is maximum KE and SPE. v=3.80m/s</p>

<p>Did I make any stupid mistakes? :P</p>

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<p>here are mine</p>

<p>FRQ #1</p>

<p>a) three forces
b)....haha....i kinda did half of it with mgsin(theta)=ma, and then on the side i wrote the right one with the -(bv)....so maybe i might get some credit
c) mgsin(theta)=bv, solve
d) hehe.....intregated the gsin(theta)=m (dv/dt).....i hope i get a bit of credit for this
e) for me....i put accleration = ((mgsin(theta)-bv))/a, and at some point since the net force is zero, the accleration is zero too...right?, graph slopes down like a 1/x or was it 1/x^2?......i dont care....</p>

<p>frq #2
a) i only drew two forces....
b) haha......blah...blah
c) i got I= (1/3)ML^2....did you have to plug in numbers?
d).....i used torque = I(alpha) and set torque = force x distance......yeah....i hope i got partial credit....</p>

<p>FRQ#3
a) graph
b) k= 25 n/m
c)...i got 1.875 kg....i think i screwed up...by setting mgh=(1/2)(K)(x^2)
d)i, well.....i had a random number for this....i dont remember
ii....i said that since this is the point that the object is at like free fall and there is no tension from the cord that is providing a upward force....it that right?
iii).....i got like 3.3ish m/s......i set Vf=Vi+at.....i dont remember how i did it.....</p>

<p>so here is what i think i got for each question</p>

<h1>1- 9/15</h1>

<h1>2- 6/15</h1>

<h1>3- 8/15</h1>

<p>is that about right?</p>

<p>CAN ANYONE TELL ME WHAT I PROBABLY GOT????</p>

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<p>and another note.....for E&M FRQ #1
isnt r<a the E field is KQa/ r^3 right?....NOT zero....</p>

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<p>That would be true if the sphere isn't conducting, i.e. if it isn't metal. But in the question the sphere is metal. In conductors, the charges are at the surface, so any spherical gaussian surface within the sphere will enclose 0 charge, hence E=0.</p>

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<p>i only filled in like half of each,
didnt even get to the mech #1 graphs,
know i got a lot of it wrong, like the 1.5m-0.6m instead of 0.6m</p>

<p>I think i did even worse on the e&m...</p>

<p>now im freaking out.
any words of comfort for me?</p>

<p><em>cross my fingers, hope for a pass-or two</em></p>

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<p>shoeboy, I'm pretty sure that for #2 a) there is a reaction force having two components, one that is horizontal and one that is up
and thus for part b) I got T=30 N, setting the hinge as a reference point</p>

<p>I myself missed c) but I was consistent with d) so I hope I get some credit there...</p>

<p>as far as 1 and 3...1 is good. and im not sure about 3</p>

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<p>Ah right. Totally forgot about taking the torque in 2b).</p>

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<p>Also, here are my stuff for E/M. Since I'm merely practicing for late testing E/M, any pointing out of errors is appreciated.</p>

<p>E&M1.
a) i) Inner surface is -Q, outer surface is +Q.
b) i) E=0 because in conductor charges are on the surface.
ii) use spherical gaussian surface to get E=kQ/r^2
iii) E=0 because in conductor charges are on the surface.
iv) E=kQ/r^2 because the outer shell has 0 net charge.
c) The graph is basically the graph of kQ/r^2 with two chunks erased.
d) First, find potential difference between 10r and infinity through definite integration. Then, dU=qdV=-dKE. Solve this equation for v to get -Qq/20m(pi)(e0)(r)</p>

<p>E&M2.
a) i) 500V
ii) at t=0, it is as if the inductor has R=infinity. Thus, 900V.
iii) At t=0, it is as if the capacitor has R=0. Thus, 4500/11V.
b) I got a graph with a straight line, an exponential decay graph for C with y-int above the straight line, and a log-like graph for L with asymptote greater than the straight line.</p>

<p>E&M3.
a) i) up using right hand rule
ii) use Biot-Savart and algebra/trig to get B=4(u0)(I)/(25R^2)
b) Bnet=2B because the two Bs have same magnitude and direction.
c) Bnet<em>s^2
d) the dot product expression has coswt, so E = Bnet</em>s^2<em>sinwt</em>w</p>

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<p>Here are my answers for Mechanics:</p>

<p>I'm going to skip 1 because it's relatively self-explanatory, except for part D which I completely left blank.</p>

<p>M2:
a) Fg from rod, Fg from block, tension pulling up left, and normal pulling up right. Normal HAS to have a y component because torques don't balance otherwise. Plus, hinges usually have both y and x components.
b) Balancing torques is the ONLY way to do this problem because you don't know what normal force is. I got T in the late 20s. Around 27-29? Can't remember exact number.
c) I first doubled the mass and the radius of the rod and used 1/12 ML^2 to find the I of double the rod. I then divided that by two. The mass is just MR^2, and I think I got 0.42 for the final answer.
d) Torque = I(alpha). I think I got 21 rad/s^2?</p>

<p>I'll go through problem 3 tomorrow. Maybe I'll redo them.</p>

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<p>crap, how big of a deal would it have been if one wrong cos(theta) instead of sin(theta) on the mechanics #1?</p>

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<p>I didn't use torques for part B in the Mechanics #2 (I used FnetY... don't ask me why, I did a lot of dumb things on that exam), but I still got 28.9. How many points do any of you think I can get for that?</p>

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<p>Bojangles, your answers all sound correct, cool. I'm sure you got in the high 20's and not 30 b/c you used 9.8 for gravity and I used 10.</p>

<p>icie, that shouldn't be too much of a deal hopefully the points they take off for that won't cumulate...I'm not sure though</p>

<p>medio, how can you use Fnety? if you're trying to find tension, you can't really do it because the y component of the reaction force is unknown as well</p>

<p>Once again, do you guys think if I got the wrong answer for 2c and used it correctly for 2d I would get full credit for 2d?</p>

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<p>Here is my solution to the differential equation, if any of you need to see it.</p>

<p>M(dv/dt) = Mg<em>sin(θ) - bv => dv/(Mg</em>sin(θ) - bv) = (1/M)dt</p>

<p>let u = Mg<em>sin(θ) - bv => du = -b</em>dv</p>

<p>Integrate to get</p>

<p>(-1/b)<em>ln(u) = t/m + C, which is equivalent to ln[(Mg</em>sin(θ) - bv)^(1/b)] = -t/m + C </p>

<p>I decided to move the negative sign to the left, to make computations easier. Then I used a well-known property of logs.</p>

<p>Exponentiate to get</p>

<p>(Mg*sin(θ) - bv)^(1/b) = Ke^(-t/m), where K is simply another constant due to e^C.</p>

<p>Apply the initial condition, v = 0 at t = 0 => K = [Mg*sin(θ)]^(1/b), which gives us</p>

<p>(Mg*sin(θ) - bv)^(1/b) =[(Mg*sin(θ))^(1/b)][e^(-t/m)]</p>

<p>Finally, raise both sides to the b-th power, to get</p>

<p>Mg<em>sin(θ) - bv = Mg</em>sin(θ)*e^(-bt/m). Rearranging gives</p>

<p>v = Mg*sin(θ)[1-e^(-bt/m)] / b</p>

<h2>Now check the limiting case. As t -> +inf, v -> Mg*sin(θ) / b. This agrees with the terminal velocity in part c, so we're done. </h2>

<p>When it asked you to find the graph for acceleration, I pretty much knew it was going to look like 1/x, so I just drew that curve. So will that be a point off for not actually finding the y-intercept?</p>

<p>Also, for 2c, I also ended up getting (1/3)ML^2 by integration. Dexter, I think even if you got the wrong answer to that (which I think I did as well), you will get full credit on 2d if you knew the equation but had the wrong I value.</p>

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<p>Nice.</p>

<p>So my estimates:</p>

<p>-probably raw score of 30~ on MC
-1. I prolly got ~12 points (do you think they'll take off a point if I put -bv on the free body, instead of just bv?)
-2. I'm guessing at least 9
-3. I think I got at least 6</p>

<p>So assuming I did as good as I think on MC, and I didn't make any stupid mistakes on FR, I might get a 5 still?</p>

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<p><a href="https://dl.getdropbox.com/u/6871/PhyC_M2008_FRQ.pdf%5B/url%5D"&gt;https://dl.getdropbox.com/u/6871/PhyC_M2008_FRQ.pdf&lt;/a&gt;&lt;/p>

<p>My two solutions to 1 and 2.
I think my hinge force is incorrect for number two. Can anyone tell me why it has to be completely horizontal?</p>

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<p>I forgot to add the moment of the block to 3D. Funny, that's one of the things I actually remembered to do on the actual test!</p>

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<p>For #3 part D (3D) on Mechanics, it's simple. Until it hits equilbrium, it has the behavior of a falling object with only gravity downwards.
Just use equations of motion.</p>

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<p>is the latter part of question 3 dealing with projectile motion...how it falls and behaves. Since the mass is adjacent, it will fall in a projectile arc...right?</p>

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<p>hi does anybody know how to do 1e, i completely blanked out on this one. Also, i know we are not allowed to discuss part , but does anyone remeber any hard ones on mechanics and e and m, i totally forgot.</p>

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<p>so is the answer to 3c) .24 or .42?</p>