<p>Ok, makers of the that test didn’t make it THAT complicated. </p>
<p>The acceleration is without question (g/2)</p>
<p>yeah, i agree that it’s g/2
there needs to be a force that contributes to the acceleration of the horizontal block, since the acceleration of the two blocks are the same
this force pulls upwards on the vertical block, so that’s why its acceleration is only g/2</p>
<p>Exactly. Hyper222 Sulthernao and sat2500 are right.</p>
<p>By the way does anyone know how to do 1d?</p>
<p>Yeah, do -dU/dx=F or in words, the negative derivative of potential energy with respect to position is equal to the force. Force is equal to ma so set the -8.0<em>x=3.0kg</em>a. a=-1.6m/s^2</p>
<p>omg…i omitted the negative sign in the f=ma equationl…</p>
<p>It only asked for the magnitude. So don’t worry.</p>
<p>oh…yeah…that’s awesome…</p>
<p>r we gonna continue the g/2 and g thing today?</p>
<p>I think we’re each pretty firm in where we stand so we’ll just have to see who gets points and who doesn’t…</p>
<p>…yeah, but will we ever get to see our specific scores?</p>
<p>Well, we won’t get to see our exact scores, but I’m sure they’ll release the answers eventually. They give us the MC score, FR score and the composite score.</p>
<p>Hey! I have a thread started discussing the actual answers we got for the Free Response section, in case anyone was curious about getting similar answers/solutions.<br>
<a href=“http://talk.collegeconfidential.com/ap-tests-preparation/714142-ap-physics-c-my-answers-performance-evaluations.html[/url]”>http://talk.collegeconfidential.com/ap-tests-preparation/714142-ap-physics-c-my-answers-performance-evaluations.html</a></p>
<p>ok…so the colleges will see this too, huh?</p>
<p>Yes sir. The colleges will see your scores. I’m satisfied because I got a 4 or a 5…probably a 4. Hopefully a 5. It doesn’t really matter to me though because I’m getting out of first year physics no matter 4 or 5. :)</p>
<p>what did the graphs look like for 1-e? i wasn’t sure how to express either in terms of t.</p>
<p>also, the acceleration is definitely g/2 for 3-a.
the kinematic equations still apply.</p>
<p>For Mech 3</p>
<p>a) F(net)=Ma=Mg/2
a=g/2 and then plug that into v^2=2ad
b)F(net)=Myg/L
c)for a variable force W=antideriv(F(net))=My^2g/(2L)
d)W= change in K and solve for v</p>
<p>Mech:
MC-OK
FRQ_!!!*****
(although I loved #1)
E/M:
MC-nice
FRQ-OK
!!! Please let the curve be low…
I might even have got 4’s or 5’s.
Did anyone understand Mech FRQ 3a? Something about diff eqs for theta???</p>
<p>I think you mean 2a b/c thats the diff equation one, I think get it, I’m pretty sure I wrote the right thing down on the test</p>
<p>T=F x R= -mgxsin(theta)
T=I<em>alpha= I
so -mgxsin(theta)=I</em>d2(theta)/dt2
(I think has a negative sign b/c Keffective has to be positive and Keff=-dT/d(theta) at equilibrium angle)</p>
<p>In response to brahms, check out the page before this one with bobwrit’s response to number two. I wrote it down and did the math…it works. -_- Which means I got a point for T=2pi/w :D</p>
<p>I meant T for torque, i had to differentiate it from time t somehow, I got the same answer for the second part of 2a, period T=2pi*sqrt(I/mgx)</p>