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By Oovelknoovel (Oovelknoovel) on Friday, May 07, 2004 - 10:38 pm: Edit |

has anyone here taken the physics c exam or is anyone planning on taking it this monday? any advice/info would be helpful

By Relinquo1 (Relinquo1) on Saturday, May 08, 2004 - 04:11 am: Edit |

I'm taking it Monday, why the hell do they have to schedule it so damn early in the morning. I usually get to school later than the test is supposed to start. Basically I'm just going to try to cram, and I'm only taking the mechanics part I think.

By Oovelknoovel (Oovelknoovel) on Saturday, May 08, 2004 - 08:56 am: Edit |

bump

By Siavash8p (Siavash8p) on Saturday, May 08, 2004 - 11:43 am: Edit |

I am taking it too. I thought it is 12PM.

By Nelly (Nelly) on Saturday, May 08, 2004 - 06:03 pm: Edit |

do you guys know of any ap physics exams mc or free responses online?

please share

anything would be a big help

By Dartkid08 (Dartkid08) on Saturday, May 08, 2004 - 06:05 pm: Edit |

Anyone know what the curve is and whether or not the test is harder/easier than the Barron's book?!

By Siavash8p (Siavash8p) on Saturday, May 08, 2004 - 07:08 pm: Edit |

try the collegeboard for FR questions...I can't find any MC...

By Wordpad (Wordpad) on Sunday, May 09, 2004 - 09:19 am: Edit |

bump..lets help each other out!

By Jnatkins (Jnatkins) on Sunday, May 09, 2004 - 10:19 am: Edit |

Curve is typically around 64% for a 5, 56% for a 4, 45% for a 3...Dunno for 2's or 1's

By Wordpad (Wordpad) on Sunday, May 09, 2004 - 01:34 pm: Edit |

bump...

By Siavash8p (Siavash8p) on Sunday, May 09, 2004 - 01:51 pm: Edit |

that certainly makes me more hopeful...64% for a 5 is a good curve! were did u find that?

are u taking mech-elec or just mech?

I am just taking the mech

I never took the Physics C class though. I self-studied w/ Halliday's book and also my father helped me.

let's make predictions of our scores...

By Zzii (Zzii) on Sunday, May 09, 2004 - 05:59 pm: Edit |

My Prediction:

5 Mechanical

2 E&M

By Oovelknoovel (Oovelknoovel) on Sunday, May 09, 2004 - 09:33 pm: Edit |

hmm if your 64% = 5 statement is true, i'm feeling more confident than before . i am also taking it based on a barron's book and giancoli's physics for scientists and engineers... good luck tomorrow

By Guyute (Guyute) on Sunday, May 09, 2004 - 10:37 pm: Edit |

i am actaully pretty nervous.

By Jnatkins (Jnatkins) on Sunday, May 09, 2004 - 11:01 pm: Edit |

It's pretty accurate...Those are around the numbers my teacher goes for his B and C classes, and he's been teaching the course for quite a few years. Not to mention he's one of the best teacher's I've had in high school, and I trust his statements.

By Yellowrose (Yellowrose) on Monday, May 10, 2004 - 03:29 pm: Edit |

jon? is that you?

By Oovelknoovel (Oovelknoovel) on Monday, May 10, 2004 - 03:32 pm: Edit |

ouch that was a pretty hard test. i've been sitting in the library for 7 hrs . The only thing i was hoping wouldn't be on the free response, was. but electricity was ok.

By Wordpad (Wordpad) on Monday, May 10, 2004 - 04:01 pm: Edit |

Could you go over some general trends on the test? What general topics were covered on free response?

By Bigman82085 (Bigman82085) on Monday, May 10, 2004 - 04:04 pm: Edit |

so hard....

By Yellowrose (Yellowrose) on Wednesday, May 12, 2004 - 06:07 pm: Edit |

hey, time for discussion now...

By Yellowrose (Yellowrose) on Wednesday, May 12, 2004 - 06:24 pm: Edit |

http://www.collegeboard.com/prod_downloads/ap/students/physics/ap04_frq_physics_cm.pdf

heres the link to download this year's free response. ill get it started if someone else wants to help me out.

1) a)mg(2L)= 1/2(mv^2)+ mgL

mgL=1/2(mv^2)

v= square root of 2gL

b)T - mg = (mv^2)/L

insert v= square root of 2gL for v

T-mg = m(2gL)/L

T-mg=2mg

T= 3mg

c) conservation linear momentum

m1(square root of 2gL)=(m1+m2)(V)

V=m1(squareroot of 2gL)

---------------------

m1 + m2

d)initial KE =1/2 m(squareroot of 2gL)^2

= 1/2(m)2gL

=mgL

KE after =1/2 (m1+m2)(V)^2

insert V from answer c -> KE= (m1)^2 times gL

----------------

m1 + m2

ke before divided by Ke after

= (m1 + m2)

---------

m1

e) i suck at kinematics. someone else's turn!

anyone else feel free to point out any mistakes or to confirm my work

By Punjabiashiq (Punjabiashiq) on Wednesday, May 12, 2004 - 10:54 pm: Edit |

Lets keep the discussion goin... How the hell do you do #2 (The RC Circuit) from the E&M part because I really don't think I did that right.

By Calikid (Calikid) on Wednesday, May 12, 2004 - 11:00 pm: Edit |

I'd like to see #3 for Mechanics, please.

By Gabushida (Gabushida) on Wednesday, May 12, 2004 - 11:07 pm: Edit |

E was L + something I dont feel like figuring out right now. The first part of Mechanics was MAD easy.

By Skysurfer (Skysurfer) on Wednesday, May 12, 2004 - 11:09 pm: Edit |

no ur #1 is wrong as angular momentum is also conserved. so u add i/2 Iw2. So v comes root of gL.

By Skysurfer (Skysurfer) on Wednesday, May 12, 2004 - 11:11 pm: Edit |

come to chat room physics in aim.

By Punjabiashiq (Punjabiashiq) on Wednesday, May 12, 2004 - 11:52 pm: Edit |

Does anyone know if the answers for the E & M are posted somewhere on the web, or could someone please work out #2 for me. Thanks a lot.

By Theiceman (Theiceman) on Thursday, May 13, 2004 - 12:58 pm: Edit |

For E & M Question 1:

(c)i. E = L/(2(pi)r(eps))

ii. E = ((pi)(rho)(r - r1)^2 +L)/(2(pi)r(eps))

iii. E = ((pi)(rho)(r2 - r1)^2 + L)/(2(pi)r(eps))

where:

L = lambda

r = distance (the independent variable)

pi and rho are their respective greek letters

eps = epsilon

r1 = inner radius

r2 = outer radius

I did this problem just now MUCH slower than I did on the real test, and it seems to be right.

Anyone disagree?

By Calikid (Calikid) on Thursday, May 13, 2004 - 02:12 pm: Edit |

I got E = (lambda x r)/(2 x epsilon) for (c) part I...

By Andrew123s (Andrew123s) on Thursday, May 13, 2004 - 04:21 pm: Edit |

For i., I got the same answer as iceman.

I got different answers for part ii and iii. You did (r-r1)^2. I think it was supposed to be (r^2-r1^2) (which would be the entire outer circle minus the entire inner circle, as opposed to just one smaller circle).

I think ii is:

E*2*pi*r*h = (lamda)*h + (rho)2*(pi)(r^2-r1^2)*h

'''''''''''''''''''''''''''------------------------------------

''''''''''''''''''''''''''' epsilon

(replace the 's with spaces)

Therefore, the 2s, the Pis, and the hs all cancel

So I got the final answer as:

E = ((lamda) + (rho)*(r^2-r1^2))/(r*epsilon).

For iii, I got the same answer as I got in ii, except the r^2 in the NUMERATOR was replaced with r2^2 (the r in the denominator stayed the same).

Also, skysurfer, you said:

"no ur #1 is wrong as angular momentum is also conserved. so u add i/2 Iw2. So v comes root of gL. "

When is angular momentum conserved? The person is not rotating at all, he is just swinging. His initial potential energy is equal to his final translational kinetic energy. It is just centripetal force provided by the tension (I don't think any rotational variables are needed). I got root(2*g*L). For the momentum problems after, he isn't even swinging anymore; it is just a regular conservation of linear momentum problem from what I can see.

By Theiceman (Theiceman) on Thursday, May 13, 2004 - 05:04 pm: Edit |

andrew is right...i wrote what he said on my paper and just copied it wrong. Thank you!

By Andrew123s (Andrew123s) on Thursday, May 13, 2004 - 05:12 pm: Edit |

What about the r-c circuit?

I got 20V, the 8V, R2=10,000 ohms, and energy = .0014J. I completely guessed on the graph, then said that reducing r2 would decrease capacitance due to an increase in voltage.

By Theiceman (Theiceman) on Thursday, May 13, 2004 - 05:14 pm: Edit |

wait, the 2s and the pi's don't factor completely in the numerator so they can't cancel! the h's do factor out of the numerator, so they do cancel...

i'm too lazy to reenter the answer so i'll just accept the fact that I understand the concept

By Andrew123s (Andrew123s) on Thursday, May 13, 2004 - 05:22 pm: Edit |

Yeah, I probabably cancelled wrong. It sucks because out of 3 points, that's probably -1 for each.

Wanna join AIM chat room physics to discuss more?

By Theiceman (Theiceman) on Thursday, May 13, 2004 - 05:39 pm: Edit |

nah, i'm okay.

The first 4 answers for R-C Circuit are right. The graph was an exponential decay graph from .002 A to .0008 A.

Your reasoning for the last one makes sense except that they were asking for the energy stored in the capacitor NOT the capacitance. And because E=.5CV^2, the energy stored in the capacitor is greater than as it was with R2 in the circuit.

By Theiceman (Theiceman) on Thursday, May 13, 2004 - 05:42 pm: Edit |

the increase in voltage increases energy...the last one I understand because I saw an IDENTICAL problem in my textbook which I should've probably studied better (oh well)...hopefully I might be able to pull off a five although I think I slept too much that day (I was thinking kind of slow and had just eaten a big breakfast -my other AP tests were the opposite)

By Andrew123s (Andrew123s) on Thursday, May 13, 2004 - 05:56 pm: Edit |

What about the order of potentials?

By Theiceman (Theiceman) on Thursday, May 13, 2004 - 06:27 pm: Edit |

Vb=Ve

Vd>Vc

Va is least. I think the order is:

Vd = 1

Vc = 2

Vb = 3

Ve = 3

Va = 4

By Andrew123s (Andrew123s) on Thursday, May 13, 2004 - 07:37 pm: Edit |

So what about the mechanics. On problem 1, did you have to use any rotational variables at all (other than centripetal force for the tension one)?

For number 2, was an average velocity for each interval vs. time graph acceptable (to find the slope)?

And for number 3, what was the answer to the tangential speed of the tip of the rod after it went from horizontal to vertical? I use the rotational kinematics equations and got v = root(2*alpha*pi), but how would you get alpha? I don't think A/r (or 9.8/r) works, cuz that would mean a different angular accelleration for each point on the rod, which isn't possible. Did you use conservation of energy for that one?

By Theiceman (Theiceman) on Thursday, May 13, 2004 - 08:12 pm: Edit |

On the first one it's kind of hard to find the rotational inertia of a rope (NOT a rod). So yeah, rotational energy is not considered.

On the second one, using the average velocities in the interval to compute the acceleration was exactly what I did. It works.

You have to use conservation of energy on the third one. HERE rotational energy is considered and you have to use the rotational inertia calculated in part (a).

The potential energy relative to the lowest position of the center of mass is MgL/2. When the rod reaches its lowest postion, the energy is entirely rotational energy (from Textbook).

.5MGL = .5(1/9)(M)(L^2)(w^2)

w= sqrt(9g/L)

v=rw

v=(2L/3)sqrt(9g/L)

For the period:

T = 2(pi)/w = 2(pi)/sqrt(9g/L) = 2(pi)sqrt(L/9g)

By Andrew123s (Andrew123s) on Thursday, May 13, 2004 - 08:18 pm: Edit |

TheIceMan, why couldn't you use rotational kinematics?

Wf^2 = Wi^2 + 2*alpha*theta

theta = pi/2

Wi=0

Wf = root(2*alpha*pi/2) = root(alpha*pi)

v = w*r = 2/3L*root(alpha*pi)

My only problem was finding alpha. Out of curiousity, how would you find alpha using this approach? Do you think they would give some credit for this incomplete alternate solution?

By Theiceman (Theiceman) on Thursday, May 13, 2004 - 08:19 pm: Edit |

As I said before, I should've studied the example problems in the textbook

Oh yeah, the rotational inertia of the rod was calculated using the parallel axis theorem:

I = (1/12)(M)(L^2) + (M)(L/6)^2 = (1/9)(M)(L^2)

By Theiceman (Theiceman) on Thursday, May 13, 2004 - 08:21 pm: Edit |

You would have to use:

Sum of the torques = Rotational Inertia * Angular Acceleration

By Andrew123s (Andrew123s) on Thursday, May 13, 2004 - 09:40 pm: Edit |

But why doesn't my solution in my above post work, except for not being able to find alpha?

By Justice (Justice) on Thursday, May 13, 2004 - 09:55 pm: Edit |

You can't use it because alpha is not constant. That's what I did wrong. You need to use conservation of energy ideas whenever acceleration isn't constant.

By Andrew123s (Andrew123s) on Thursday, May 13, 2004 - 10:14 pm: Edit |

Why isn't alpha constant? You are saying alpha increases with time?

By Justice (Justice) on Thursday, May 13, 2004 - 10:35 pm: Edit |

Think about the torque on it. It's caused by MG, which is obviously not going to remain perpendicular to R as it goes to the vertical, so the applied force is not going to be the same. It follows that acceleration will not be constant. It sucks to hear but you'll definitely get zero points on that part like I will. It's supposed to be a huge no-no in physics and I don't know what I was thinking.

Don't worry the curve is nice.

By Andrew123s (Andrew123s) on Thursday, May 13, 2004 - 10:52 pm: Edit |

Well you never know. On the rubrics, they have alternate solutions, and they also sometimes have partial credit alternate solutions.

Also, what was the answer to the pendulum question. I thought it was just 2pi*root(l/g) with l substituted in as 2/3L, but then I've heard you have to use the formula for a physical pendulum, 2pi*root(l/Mgd) with d being the distance from the center of mass. Why would they ask a question requiring knowledge of a specific formula that wasn't given? Is it possible to derive it?

By Muawan (Muawan) on Thursday, May 13, 2004 - 11:27 pm: Edit |

It is in fact possible to derive it using the concept of simple harmonic motion. You set up the differential equation: I alpha = Torque net. Torque net you can find by subtracting the torque on one half of the lever from the torque on the other using force of gravity cross R. Then since at small angles Sin Theta = Theta plug that in you get T = k Theta where k is omse constant. Using the F = kx analog Period = 2 Pi Root (m/k) Imples Period = 2 Pi Root ( I / k) where K was the constant in front. I can prove it if anybody would like to see it. Also you must include rotational motion in 1 A because the tension holds the person in. The Veolicty is Root (2gh) and the tension is in fact 2mg.

By Muawan (Muawan) on Thursday, May 13, 2004 - 11:28 pm: Edit |

If anyone has any questions on the physics free response ask me, I got every single question on the free response. Just email me.

By Subtrunks (Subtrunks) on Thursday, May 13, 2004 - 11:34 pm: Edit |

"What about the r-c circuit?

I got 20V, the 8V, R2=10,000 ohms, and energy = .0014J. I completely guessed on the graph, then said that reducing r2 would decrease capacitance due to an increase in voltage"...i did tau=rc to find r2 ...the graph was nto scaled for that but thats CB's fault not mine ..so yea I win

By Andrew123s (Andrew123s) on Thursday, May 13, 2004 - 11:54 pm: Edit |

Wouldn't tau=rc find the resistance of the capacitor?

By Justice (Justice) on Thursday, May 13, 2004 - 11:57 pm: Edit |

Wait--the last part asked energy stored in the capacitor, not the capicitance! So I got that it increased. What do you guys think?

What did your field lines look like for 1a for E and M?

By Andrew123s (Andrew123s) on Friday, May 14, 2004 - 12:11 am: Edit |

Yes, I made a mistake with that.

By Xdebugger (Xdebugger) on Friday, May 14, 2004 - 06:13 pm: Edit |

I typed up my solution for the mechanics test. Let me know if there are any errors. If you need to see the E&M part, I can type it up too.

http://xdebugger.tripod.com/AP_Physics_C_2004_Mech.pdf

By Wharton2008 (Wharton2008) on Friday, May 14, 2004 - 10:37 pm: Edit |

please put electricity if you get a chance

By Lemmethink (Lemmethink) on Friday, May 14, 2004 - 11:11 pm: Edit |

crap...i forgot it was a cylinder...i thought it was a sphere...do you think i'll get major points off?

By Justice (Justice) on Friday, May 14, 2004 - 11:15 pm: Edit |

yes pretty major considering that you definitely used gauss's law incorrectly....

xdebugger please post the answers to e & m!

I think you messed up on 2.c. Why is it T-mg=ma and not mg-T-ma?

By Korinfox (Korinfox) on Saturday, May 15, 2004 - 09:35 am: Edit |

can you pleassssse type up e&m for us? pleeeease thanks!

By Xdebugger (Xdebugger) on Saturday, May 15, 2004 - 01:34 pm: Edit |

Here goes the E&M part. Let me know if I did anything wrong. Enjoy!

http://xdebugger.tripod.com/AP_Physics_C_2004_EM.pdf

By Xdebugger (Xdebugger) on Saturday, May 15, 2004 - 01:56 pm: Edit |

Thx Jusice, I left out the negative sign there. Hope I didn't do that on the test...

By Andrew123s (Andrew123s) on Saturday, May 15, 2004 - 02:06 pm: Edit |

How many points do you think I would lose for algebraic errors on Gauss' law, such as incorrectly cancelling out 2s and pis or having an extra 2 in the numerator? If I did this for both II and III, would they just take off the points for II, or would they take off the points for both?

By Punjabiashiq (Punjabiashiq) on Sunday, May 16, 2004 - 03:40 pm: Edit |

Hey Xdebugger, I think you messed up on the bounds of integration, it should be from l to 4l, since the outer bound is 3l + l.

By Mehere (Mehere) on Sunday, May 16, 2004 - 05:03 pm: Edit |

Xdebugger,

for mec num 2, can i draw a V vs T graph? i can find V by inegrating a = 2D / t squred right? and then the integral of that graph from 0 to 1.38.

i find the way u did it, graphing 2D vs t2, the slope is not constant and its hard to find the exact a. But my way, i doono if my estimate of the V is right. i just used the D / t for every point. what do u guys think?

By Xdebugger (Xdebugger) on Sunday, May 16, 2004 - 06:12 pm: Edit |

Thx Punjabiashiq, I definitely messed that one up. It turns out to be Ln(4) instead and it should be ln(4) throughout the problem. Do you think I would get major points off on that one? Damn, I always make mistakes like that! Please let me know if any other parts don't look right.

To Mehere, I think your method is a little redundant because it's hard to estimate V. If you graph D against T^{2}, you should get a pretty straight line.

By Punjabiashiq (Punjabiashiq) on Sunday, May 16, 2004 - 06:23 pm: Edit |

I think they will only take points off the first time and since you were consistent with that mistake throughout you should only lose 1-2 points total. I made a stupid mistake on that one as well. For some reason I forgot to square the (e^(-k*t)) for the last part.

By Xdebugger (Xdebugger) on Sunday, May 16, 2004 - 06:29 pm: Edit |

That makes me feel a lot better now. Hopefully they only deduct 1 or 2 points on that.

By Punjabiashiq (Punjabiashiq) on Sunday, May 16, 2004 - 06:30 pm: Edit |

Oh I had a question for you, Xdebugger, for Mech 3, part C. I got the same answer except I just used the general equation T = 2*pi*(L/G)^(1/2) and I plugged in 2L/3 for L since 2L/3 is how much that is below the pivot oscillating. Did I just get lucky or is this a correct method? Also do you think I would get any partial credit for the same question part A if for some reason I made the pivot the center of mass and calculated the rotational inertia that way? (It came out to 5M*L^2/9)

By Xdebugger (Xdebugger) on Sunday, May 16, 2004 - 06:37 pm: Edit |

Hm... that's interesting. I need to think more about that although it somehow seems that you were lucky on that one because T = 2*pi*(L/G)^(1/2) perhaps is only good for simple pendulum. But I may be wrong.

In part A, if you made the pivot the center of mass, I should've come out 1/12.

By Punjabiashiq (Punjabiashiq) on Sunday, May 16, 2004 - 06:37 pm: Edit |

Well the final answer was wrong, but you carried your original wrong answer through correctly so you shouldn't lose too many points. I am not sure if you will get the correct answer point for all of them but you may, I doubt they would penalize you for one mistake 3 times over.

By Punjabiashiq (Punjabiashiq) on Sunday, May 16, 2004 - 06:44 pm: Edit |

How did you get 1/12. If the pivot is the center of mass then isn't the rotational inertia M*(L/3)^2 + M*(2L/3)^2 = (M*L^2)/9+(4*M*L^2)/9 = (5M*L)/9?

By Punjabiashiq (Punjabiashiq) on Sunday, May 16, 2004 - 06:44 pm: Edit |

The last L should be L^2, sorry for the typo.

By Xdebugger (Xdebugger) on Sunday, May 16, 2004 - 06:49 pm: Edit |

Your method on part C was not correct. If the pivot were at the center of mass, the rod should not swing at all and if the pivot were at the end of the rod the period also turns out to be 2*pi*(2L/3G)^(1/2).

By Punjabiashiq (Punjabiashiq) on Sunday, May 16, 2004 - 06:53 pm: Edit |

Thats what I was thinking... I hope I will get some partial credit for that.

By Xdebugger (Xdebugger) on Sunday, May 16, 2004 - 06:56 pm: Edit |

To calculate the rotational inertia of a rod you need to integrate using I=ò_{ }^{ }R^{2}dm. I got around that because I knew the inertia of a rod at the CM is 1/12xxx.

Here is a link to a moments of inertia table.

http://scienceworld.wolfram.com/physics/MomentofInertia.html

By Punjabiashiq (Punjabiashiq) on Sunday, May 16, 2004 - 07:01 pm: Edit |

Yea I used that formula...(Hopefully I'll get parial credit for attempting to use that formula since I did write it down) I don't know what I was thinking when I made the pivot the center of mass though.

By Andrew123s (Andrew123s) on Sunday, May 16, 2004 - 07:56 pm: Edit |

Xdebugger, you are saying using the formula for a simple pendulum is incorrect, yet it yields the correct answer by chance? Would we get credit then, since we got the same answer? I don't think we are supposed to know how to solve second-order differential equations on the test.

By Mehere (Mehere) on Sunday, May 16, 2004 - 08:22 pm: Edit |

Xdebugger, for number 3 part b, the angular acceleration is constant right? so i found angular acc and then used kinematics to find w, Wf squared = Wi scored + 2 a d. however, my answer had an extra pie in it, do you know why is that?

By Xdebugger (Xdebugger) on Sunday, May 16, 2004 - 08:37 pm: Edit |

no, angular acceleration is not constant. It's greatest when the rod is horizontal and zero when it's vertical.

By Mehere (Mehere) on Sunday, May 16, 2004 - 10:54 pm: Edit |

.. hrmm how would u able to identify to see which situation the angular acceleration is constant and wen its not? cuz this case looks as tho its constant...

By Justice (Justice) on Monday, May 17, 2004 - 12:13 am: Edit |

thanks for the e&m xdebugger! i'm feeling a lot better about the FR now. Mehere, whenever you have a non-constant torque causing the acceleration, then you know it's not constant. the torque is called by gravity, which will not always be the same direction with the rod.

By Poiz0n (Poiz0n) on Friday, May 21, 2004 - 03:46 pm: Edit |

is it worth is to call July 1st to find out what i got on physics c and calculus...i could care less about the other 3 ap exams i took, but is it worth the money to find out 2 weeks earlier?

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