<p>The frq's are now on collegeboard:
<a href="http://apcentral.collegeboard.com/apc/public/repository/ap08_physics_c_mech_frq.pdf%5B/url%5D">http://apcentral.collegeboard.com/apc/public/repository/ap08_physics_c_mech_frq.pdf</a></p>
<p>discuss answers, explanations here</p>
<p>:)</p>
<p>bump </p>
<p>This shouldn't be on page 2.
Anyone have anything?</p>
<p>I was completely dominated by this exam!
Here's what I put though for Mech 1</p>
<p>Mech. 1:</p>
<p>a) Fg downward, Fn Perpendicular to surface, Fd along surface.
b) ??? (All I know is that I'm going to have to change a to dv/dt)
c) Set Fgx equal to Fd.</p>
<p>F=ma = 0
F= Fgx - Fd
0 = mgcosθ - bv
bv = mgcosθ
v = (mgcosθ)/b</p>
<p>d) ??? (I couldn't write a diffeq, so therefore, no equation to solve)
e) Line at ao curving downward toward 0. (I missed this one!)</p>
<p>lemme see...</p>
<p>1.a) 3 forces: n, mg, and bv, the directions are pretty obvious
b)dv/dt = gsin(theta)-bv/m
c)v(t)=mgsin(theta)/b
d)really complicated I think I messed up some steps but in the end I'm sure it was...v=sin(theta)g(1-e^(-bt/m))...something like that
e)easy, just a curve that looks like y=1/x...cept the y intercept is gsin(theta)</p>
<p>2.a) 5 forces(including the components of R): Rx, Ry, 2g, .5g, and T...pretty easy I think the only thing to note is that Ry is upward
b)i got t=30N
c)I THINK I DID THIS WRONG...but I got .24....and are the units kg * m^2?
d)using .24 I got -37.5 rad/s^2</p>
<p>3.a)easy graph
b)k = 25 N/m
c) I THINK I DID THIS WRONG TOO...I got 3.75kg
d)DUNNO</p>
<p>So if I got 2b) wrong, do I get credit for plugging it in right to 2c)?</p>
<p>Hm, Im trying these now (Im in physics B and taking C in a week), and I dont think Ive learned how to solve differential equations like this one from BC calc...or maybe I just dont remember. I got what dexter got for 1b though. </p>
<p>Actually I think I just figured it out. You have to leave the equation as dv/dt = (mgsin(theta) - bv) / m, then you can manipulate it to dt = m dv / (mgsin(theta) - bv) and integrate that, right? O_O</p>
<p>correct.</p>
<p>tough, huh?</p>
<p>oh and I made a mistake on 1d) above. It's not sin(theta)g in front...it's msin(theta)g/b</p>
<p>The differential equation wasn't too bad; Electricity and Magnetism students shouldn't have had a problem with it because it was very similar to the equation for RL/RC-circuits.</p>
<p>bump</p>
<p>cmon ppl post your answers isn't there anyone who can confirm?
I think we got #1 down but what about 2 and 3?</p>
<p>I think I just figured out how to to 3D right now... </p>
<p>Hold on, lemme try this out</p>
<p>Oh, I think you're supposed to use the same kx=mg equation for 3C that you used before. Even if the mass starts at the top of the thing, it still won't change those force. However, keep in mind that when you measure x, it has to be 1.5-.6, not just 1.5, because .6 is equilibrium.</p>
<p>25(1.5-.6)=m(9.8)</p>
<p>m= 2.2-ish</p>
<p>3D is meant to be done as a simple harmonic equation. We already know that the amplitude is .9, because 1.5 is .9m from equilibrium, so the equation so far is Y=.9sin(wt) Now we have to find w.</p>
<p>w= 2pi/t (I can't find pi anywhere on my keyboard)</p>
<p>t=2pi (m/k)^(-1/2)= about 1.9</p>
<p>w=2pi/1.9= 3.3709
y= .9sin(3.3709t)
v= Awcos(wt)= .9(3.3709)cos(3.709t)
v=3.033899382cos(3.3709t)
v/3.033899382=cos(3.3709t)</p>
<p>V varies directly with cos, so it's at its max when cos is at its max, and cosmax=1. vmax/3.033899382=1 vmax=3.03 m/s</p>
<p>There's the first part to D.</p>
<p>Oh wait, that's D3. My bad.</p>
<p>For E&M these are my answers:
(a) (i.) -Q
(ii.) +Q
(b) (i.) E = 0
(ii.) E = kQ/r^2
(iii.) E = 0
(iv.) E = kQ / r^2
(c) first interval 0,
second interval 1/r^2,
third interval 0 ,
fourth interval 1/r^2
(d) don't remember, pretty sure I got it wrong.</p>
<ol>
<li>(a).(i) 500 V, I think. Can't remember for sure.
(ii.) don't remember, but I acted like the branch with R3 didn't exist, since the inductor resists current, right?
(iii.) don't remember the answer, but I acted like there was no gap between A and B, since the capacitor acts as a closed circuit, hopefully.</li>
</ol>
<p>(b.) guessed. I put straight line for (i.), sqrt(x)-looking graph for (ii.), and a 1/x looking graph for (iii.) probably wrong</p>
<ol>
<li><p>(a.) (i.) up
(ii.) Biot-Savart (I used ampere though, but I think you're supposed to use Biot.)
B = uI/R</p>
<p>(b.) don't remember</p>
<p>(c.) flux = BA = Bs^2
(d.) I got an answer but I'm pretty sure it's wrong cus it's not in terms of t...</p></li>
</ol>
<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>The phase constant wasn't a problem in solving D3, but I'll get D1 wrong without it. It's -3pi/2, because SHM usually starts with the object moving upwards toward the crest. </p>
<p>so... cos(wt-3pi/2)=v/Aw Now, we're trying to find the value of t at cosmax, because cosmax is when v is maximized and we're trying t at that moment of vmax, and cosmax is 1, so we need to find what value of t gives us a cos of 1.
cos(wt+pi/2)=1
wt-3pi/2= cos^-1(1)
wt-3pi/2= 0
3.3709t= 3pi/2
t=1.397...</p>
<p>Plugging that back into the original equation...</p>
<p>y= .9sin(3.3709t-3pi/2)
y= .9sin(3.3709(1.397)-3pi/2)
y= .9sin(0)
y= 0</p>
<p>So... that means that it would've reached vmax at equilibrium.</p>
<p>Wow. I just did a lot calculations for no reason. I could've solved this much more easily just by finding vmax to be at the equilibrium point for the spring, .6.</p>
<p>I just wasted a lot of calculation.</p>
<p>Well, it looks like I was on the right track all along with the conservation of energy thing when I was making my feeble attempts on that problem...</p>
<p>I hope that I get partial credit for stating conservation of energy for part D1. I also put "when potential energy is least" for part D2.. credit?</p>
<p>OMG. D was so deceptively simple!!!!!!</p>