<p>there seems to be a monster curve (at least, there was in '98, but things can’t have changed that much)…which makes me feel marginally better.</p>
<p>Scoring the AP Physics C Exam:
<a href=“http://faculty.trinityvalleyschool.org/hoseltom/handouts/Scoring%20the%20AP%20Physics%20C%20Exam.pdf[/url]”>http://faculty.trinityvalleyschool.org/hoseltom/handouts/Scoring%20the%20AP%20Physics%20C%20Exam.pdf</a></p>
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</p>
<p>I’m assuming no one has answered/explained this yet…</p>
<p>Part 1
Find center of mass. Set up coordinate system where moon is at x=0.
x<em>com = [ m(0) + 49m(r</em>a) ] / 50 m = 49<em>m</em>r<em>a / 50m = (49/50)r</em>a</p>
<p>Part 2
Conservation of angular momentum.
L<em>a = L</em>b
m<em>v_o</em>r<em>a = m5v</em>or<em>b
r</em>b = (1/5) r_a</p>
<p>ty ChemE!! so simple once I see the work…</p>
<p>Do you guys think its worth taking practice tests before 1998?! Or have the tests changed too much?</p>
<p>imo, they’ve changed a lot</p>
<p>No problem. As an additional note, always remember three things about planetary motion:</p>
<p>1) Angular momentum is conserved</p>
<p>2) Kepler’s second law: Equal Area is swept out in equal time intervals (never know… could be a multiple choice)</p>
<p>3) Kepler’s third law: (T<em>a)^2 / (r</em>a)^3 = (T<em>b)^2 / (r</em>b)^3, where A and B are two objects orbiting the same larger object (like Moon A and B orbiting Planet Z)</p>
<p>Also, if it’s a circular orbit, a = v^2/r, but how do we figure out acceleration if the orbit is not circular?</p>
<p>x (m) | F (N)
0 | 0
1 | 1
2 | 8
3 | 27
4 | 64</p>
<p>A specially designed spring is stretched from equilibrium to the distances x given in the table above, and the restoring force F is measured and recorded in each case. What is the potential energy of the spring when it is stretched 3 m from equilibrium?</p>
<p>A) 9/2 J
B) 9 J
C) 81/4 J
D) 27 J
E) 81/2 J</p>
<p>It’s pretty clear that the restoring force is F = -x^3, so integrate and multiply by -1 to get U(x) = (1/4)x^4, so at x = 3, U = 3^4 / 4 = 81/4 J</p>
<p>Wow… common sense fail D:
Thanks.</p>
<p>Chem, you’re a beast. Want to communicate telepathically tomorrow during the test? :)</p>
<p>I’ll try… not sure if it will work though. Also, I don’t know this stuff inside and out - I’ve got my own questions too (see above)! And now I’ve got to go back to studying (yes, my break from working physics problems is coming on here and working physics problems… ah, I can’t wait until I’m done with AP tests).</p>
<p>
</p>
<p>I’ve never seen an AP question like that so I’m hoping nothing like that will come up.</p>
<p>What are your [thread posters] thoughts on the FRQ’s tomorrow?</p>
<p>I think it’ll be:</p>
<p>M1: Potential/Kinetic Energy
M2: Spring System
M3: Rolling cylinder/T=Ia question</p>
<p>EM1: Gaussian surface
EM2: circuit
EM3: Electric/magnetic field</p>
<p>@ 134</p>
<p>a=g-bv
dv/dt=g-bv
∫dv/(g-bv)=∫dt
-1/b∫-bdv/(g-bv)=∫dt
-1/bln((g-bv)/(g-bvo)=t
ln((g-bv)/(g-bvo)=-bt
(g-bv)/(g-bvo)=e^(-bt)
g-bv=(g-bvo)(e^(-bt))
v=(g-(g-bvo)(e^(-bt)))/b</p>
<p>vo=0 so plug in 0 for vo and then factor out a “g”</p>
<p>v=g(1-e^(-bt))/b, which is answer “a”</p>
<p>For the released Physics C Mech test, number 29, anyone get it? Can you tell me where I went wrong?</p>
<p>E=K+U
K=U-E=.5kx^2-.5kA^2=.5k(x^2-A^2)
=.5k(A^2cos^2(2pift)-A^2)
=.5kA^2(cos^2(2pift)-1)</p>
<p>Also, if you could explain 32 and 35… heh</p>
<p>^ Which exam?!</p>
<p>Anyways, I just took the 1998 exam, and failed yet again (although it went better than before). Could someone please explain #'s 20, 27, 33, 34, 35? Thank you!!</p>
<p>EDIT nvm I got 35…but I could really use some help on the others!</p>
<p>The released practice one.</p>
<p>Someone earlier in the thread put up an awesome review for Mechanics, anyone have anything similar(like conceptual review stuff) for E & M?</p>
<p>I need MAJOR help on equipotential and electric field lines. I thought electric field lines pointed from high to low potential. But on the MCQ 2004 test for E&M, #59 says I’m wrong. It’s basically a picture of a bunch of equipotential lines and it asks for the electric field vector at a specific point. So I choose the one that would have it going from hight o low potential, but the answer is the complete opposite.</p>
<p>But then on FRQ 2005 #1 a-ii (<a href=“College Board - SAT, AP, College Search and Admission Tools”>College Board - SAT, AP, College Search and Admission Tools) it affirms that electric field lines go from highest to lowest potential…</p>
<p>if we replace a with g for acceelration, will we be doced any points for that??</p>
<p>
</p>
<p>A is consistent with the explanation. You’re at 5V and you point to 0V. </p>
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</p>
<p>Depends on the scenario.</p>