AP Physics Exam

Colleges and Universities A-Z Massachusetts Institute of Technology
21

<p>Synopsis:</p>

<p>I thought E&M MC was fairly good. For FR, I think I did ok: I aced the last one on Faraday's/Biot-Savart, but I couldn't for the life of me remember the derivation for charge in capacitor given emf and resistance (I don't think we went over it, though).</p>

<p>As for mech, FR was OK, I hardly remember what the questions were, but I skipped 7 on the MC: Mech took a lot more calculations, and I just skipped over the ones I couldn't get right away. After I went through, time was already up.</p>

<p>Anyway, just glad it's over. Regardless if I manage 5s, I don't want to skip any physics at MIT. I would've done a whole helluva lot better if they let us take the two exams on seperate days, or even seperate years (my school doesn't do that).</p>

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<p>I thought the Mechanics MC was much harder than any practice i've seen. It took forever...time was up before you knew it.</p>

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<p>I only did the mech. The MC was alright, but that 2 minute mini-section was just brutal. I mean, so many questions, I was rushing all the way through. Not short, straightforward questions either, long prompts and long choices. Gosh.</p>

<p>FRQ was better than I thought it was going to be, so hey. With any luck.</p>

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<p>So... Physics C killed me (like, destroyed me) but I took it as an independent study so I guess I forgive myself. I figured I'd do ok since I got a 5 on the B but... it's much harder. C'est la vie, and yay for 8.01 in the fall! </p>

<p>Chemistry, on the other hand, seemed far too easy. It was like, simplified versions of our normal chem tests. Which is really, really weird, at least for my normal experience with APs.</p>

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<p>How many did people people leave blank on the Physics C Mech MC? I felt really rushed the whole time and had to leave two blank - and lots of the ones I did answer I felt pretty iffy on. I especially disliked the purely conceptial questions in the I, II, III format (I think I got them all wrong...)</p>

<p>The FRQ weren't too bad - the second one was really confusing at first but it was actually fairly simple. The only part I'm not really confident on is the first part of the third question - acceleration of the center of mass of the hoop. Was it seriously just a = g sin theta? Or did I need to involve rotation some how?</p>

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<p>I'm taking the Physics C em and mech late. which gives me one week left to study! anyhow, have any of you used the Princeton Review AP Physics book? Do you think the problems in the sample test are harder than the actual AP?</p>

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<p>My prep book observations - barron's m.c. were a hell of a lot harder than the actual thing, but princeton review was great in that they grouped a lot of practice problems by topic/chapter in addition to the practice tests at the back. I just flat out didn't practice any free response ones (I knew it was hopeless) so I can't help you on that.</p>

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<p>odetojoy:</p>

<p>Hmm, I think that's exactly what I put. I was really suspicious also, and actually solved for Vcm at the bottom first (which was part B, which was the part that involved conservation of energy).</p>

<p>So, I might be wrong, but I figured: Fg on CM produces no torque, and the problem doesn't give you a mu-static of the ramp surface to calculate torque responsible for rotation. So force on CM is Mgsin(theta), F=ma, a=gsin(theta).</p>

<p>I'm sorry, I'm just glad more people got the same thing.</p>

<p>carmel:</p>

<p>I've noticed Barron's is always like that for anything.</p>

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<p>Yeah, I was pretty worried that they gave us all of that space for what amounted to just dividing m out of both sides of the equation!</p>

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<p>I believe there was more to it than that. If the hoop rolled down, then part of its energy was rotational, not translational. In addition, there had to be a frictional force which was providing a torque. You could use this fact, along with the rotational inertia, to solve for frictional force in terms of rotational acceleration, which is equal to a/r. Finding that allowed you to solve for the net force on the hoop (mgsin(theta) - frictional force), which you could then use to find the acceleration of the center of mass.</p>

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<p>lolllz</p>

<p>.</p>

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<p>The frictional force wasn't slowing down the hoop though, it was just causing it to rotate instead of slide. I don't see how friction would affect the acceleration of the center of mass.</p>

<p>Obviously you have to take into account the rotational energy and the translational energy when finding the velocity (as in the second part of the problem.) So how could you find the acceleration without using conservation of energy / finding v first?</p>

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<p>Capacitors on pulleys.</p>

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<p>AHHHHHHH the pain! THE PAIN!!</p>

<p>why oh WHY?!</p>

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<p>The last question was hilarious.</p>

<p>My physics teacher always makes jokes about that.</p>

<p>He gives us a free response on every test we take. Often he'll unfold the board with the question written on it, and it'll have like 4 capacitors in a random arrangement with like springs, pulleys, ramps, induced currents. And he'll start describing the problem...everyone's like "CRAP"...</p>

<p>Then he's like "Just kidding..." and erases parts that are irrelevent.</p>

<p>But the AP was like "Wait...they're serious...springs and currents and magnetic fields?!?!?!"</p>

<p>Just glad there wasn't any motion or sins or anything like that.</p>

36

<p>odetojoy, dseisenberg:</p>

<p>dseisenberg might be right, then again I dont really know: that's why we consider mu-static as opposed to mu-kinetic in this circumstance. Can you give full work? </p>

<p>Torque=Ialpha
Ff(R)=(MR^2)(a/R)=(MRa),
Ff=Ma, a here is a(CM), Ff=mu-static(N)</p>

<p>Where do we go from there?</p>

<p>Spartan Pho3nix:</p>

<p>Ah, I know what you mean; I got the question back today, so here's what I put:</p>

<p>A.) Right hand rules, B into the page, fingers toward current, so two lengths experience equal, opposite force, bottom width experiences force downward, no force on top width, as it is not in the B-field.</p>

<p>B.) This means that the spring returns to equilibrium after some time (SHM), which means the force on the bottom width was opposing the restoring force of the spring when the circuit was complete (ie., current flowing). So you set the two equations equal:
Fs=kx (Hooke's Law)
Fb=ILB0
kx=ILB0
{B0=kx/IL}</p>

<p>C.) Since area of B-field enclosed by loop is decreasing, dflux/dt is decreasing into the page. As such, induced EMF will produce current I such that I generates flux into the page (and hence B into the page). By RHR, current is CW.</p>

<p>ii. Induced EMF = dflux/dt; B-Flux=BAcos(theta) [Theta=pi/2, dflux/dt=BdA/dt, A=hw, dA/dt=(dh/dt)w=v0w], dflux/dt=B0v0w=Induced EMF; EMF=IR, {I=B0v0w/R}</p>

<p>D.) P=IV; P=(B0v0w)^2/R</p>

<p>E.) F=ILB, if B magnetude increases, both I and B increases, thus F increases, and F opposed must increase to maintain v0.</p>

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<p>I solved for acceleration using conservation of energy and I got a = ½ g sin θ, not a = g sin θ. <em>sigh</em> so much for that five…</p>

<p>H = L sin θ</p>

<p>U = K(r) + K(t)</p>

<p>M g L sin θ = ½ I (v/R)^2 + ½ M v^2 = M v^2 </p>

<p>Half of the potential energy is going into rotational energy and half into translational.</p>

<p>v = √(g L sin θ )</p>

<p>To find acceleration of the center of mass:</p>

<p>v^2 = 2 a L</p>

<p>a = ½ g sin θ</p>

<p>There must be a way to do it with forces but I don't know.</p>

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<p>odetojoy:</p>

<p>That is most wonderful! I wonder why they ask us to find v,cm after a,cm.</p>

<p>Ah, five shmive, I don't want to skip any physics.</p>

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<p>
[quote]
One of my teachers whose going through law school told me that the whole "contractual agreement" where you're not allowed to talk about the m/c ever isn't binding because contracts aren't legally allowed to be set for an indefinite amount of time.

[/quote]
</p>

<p>I hope the teacher learns some more law before taking a bar exam.</p>