<p>Would it be alright if I posted my answers and solutions to the free response section on Mechanics and Electricity and Magnetism now that the 48 hours are up? Do you know anyone who may have a problem with that?</p>
<p>by all means, yes.
FYI, whether you are correct or incorrect, some ppl will disagree so be prepared to back them up if you care enough?? haha</p>
<p>So for the past 48 hours, I’ve done little but worry about the Physics C Mechanics and E&M test we took Monday. Overall I felt both multiple choice tests went well, but free response gave me quite a stumble, especially in Mechanics. The official free response questions are here </p>
<p>Mechanics
<a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
<p>Electricity and Magnetism
<a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
<p>I’m going to do my best to project my answers/solutions. I’m too lazy to put myself through the hell of taking the exam again; so most likely I will just describe my approach to the FR. Here we go.</p>
<p>Mech 1.</p>
<p>a) 7 joules. I used the potential energy formula to find PE and used 1/2mv2 for KE. Added the two together and voila.</p>
<p>b) +sqrt (7/4), -sqrt (7/4). Set the total energy to PE and solve for x.</p>
<p>c) Find KE first by plugging in the x in the PE equation. Then solve for velocity by setting it to 1/2mv2
p = mv.</p>
<p>d) F= -dU/dx. So I took the derivative and divided by 3 kg. a=F/m. So I plugged in .60m for my answer.</p>
<p>e) At this point, I didn’t know what the hell I was doing so I believe I started taking integrals and punching in functions into my calculator. </p>
<p>I’m pretty sure this is wrong–WAIT! I know this is wrong… but I took the integral of acceleration function twice to get displacement. I just graphed that in my calculator and copied that to my paper.</p>
<p>For the KE graph, I graphed 7 - U(x) in my calculator and copied that down.</p>
<p>Pretty ■■■■■■■■, I know.</p>
<p>Mech 2. (■■■)</p>
<p>a) </p>
<p>i)Again, I’m 99.9% sure I’m wrong. But…
I started with the cosine function for SHM. x = A cos (wt). Took the derivative. dx/dt = -Aw sin (wt)
Ngl, I felt kinda smart writing that one out Monday. I feel like an effin’ ■■■■■■ now.</p>
<p>ii) T = 2piR/w = 2(pi)R / -Aw sin (wt) / R = 2 pi/ -Aw sin (wt) ???
Okay, missed that one too.</p>
<p>(b) Wrote random equations involving I.</p>
<p>(c) Totally gave up and cried deeply.</p>
<p>Mech 3. (Made me feel stupid. It’s easy. But it’s not.)</p>
<p>(a) mgh = 1/2mv2 ??? I think it’s just v = sqrt (2gh). I’m a little suspicious b/c it seems too easy. I just assumed that the hanging block was in free fall since there was no friction on the table.</p>
<p>(b) F = mg. m = My/L. F = Myg/L. This also seems too deceptively simple.</p>
<p>(c) + (d) OMG! These ones ****ed me off. W = Fd. I had trouble getting y as a function of L. So I kinda BSed these.</p>
<p>(e) Speeds are equal b/c Galileo said so.</p>
<p>Time for a break…I’ll be back.</p>
<p>E&M 1.</p>
<p>(a) </p>
<p>(i) I hope you didn’t do what I did and waste 6 minutes deriving the electric field using Gauss’s Law. Good. Now you made me feel like an idiot. E = -dV/dr. Just focus on the part with r and neglect the rest since most of it are constants, trying to trip you up. I put “Radially inward” b/c idk.</p>
<p>(ii) Do the same thing w/ the derivative. I said radially outward by Gauss’s Law.</p>
<p>(b) Now would probably be the time to use Gauss’s Law. I don’t think I did this right either, but I believe I multiplied E by the Area and set it to q/e0, and solved for q.</p>
<p>(c) I said yes. Charge can reside on the surface b/c they move freely on the surface of conductor. I think that’s right.</p>
<p>E&M 2.</p>
<p>(a) P = V2/R. I calculated resistance by dividing resistivity by length. It seemed like the logical thing to do at the time. THIS IS WRONG! I’m such an idiot. My cousin later told me to use the equation sheet. R = p*L/A. That’s how resistance should be calculated. </p>
<p>(b) ----> I justified this by saying that it would be in the same direction of positive charge flow. Yeah, I know I’m stupid.</p>
<p>(c) E = V/d; that’s all I did. Someone agree w/ me please.</p>
<p>(d) F = I l B; that’s all.</p>
<p>(e) The magnetic force in the bar is up; therefore the opposing electric force must act downward to oppose this force and create an opposing magnetic field out of the page. So just like my AP free response score the electric field is going downward.</p>
<p>(f) OOOH!!! I think I got this one!!! F= qvB = Eq. Charges cancel out and you’re left with E= vB. I hope I didn’t eff that one up. </p>
<p>E&M 3. I’m really curious about how I did on this one.</p>
<p>(a) EMF = change in flux/ change in time = change in B * A / change in time. Change in B (using the formula) should = a (change in time) since the b’s cancel out. Time should cancel out leaving. a*A or aL2, which I found a little too simplistic for something like induced EMF.</p>
<p>(b)
(i) current = V/R. Just divide by the resistance</p>
<p>(ii) Up (or CCW) through bulb 2 b/c the current must create a magnetic field through the square out of the page.</p>
<p>(c) P = V2/R; but remember the voltage is half of 9 b/c some of it is dropped by the other bulb. </p>
<p>(d) I thought bulb 1 would be the same since any current from the parallel curcuits would add up (to the earlier amount) by Kirchoff’s Junction rule and conservation of charge. </p>
<p>(e) Dimmer, b/c the EMF induced is halved by the smaller area. Power dissipated is 1/4 what is was before.</p>
<p>Once again. DO NOT use this as a scoring guide. That’s the last thing you want to do. Also, don’t leave this note thinking you failed FR because most likely, you aced this section and I failed. Please, let me know if these answers match your responses.</p>
<p>mech3a is wrong i think, because you forgot to take into account the normal force acting on the second block. here’s how i did it:
F=Mg/2=ma=Ma, A=g/2, v^2=v0^2+2a(d-d0)=2ad, v=sqrt(gd)
I’ll leave the rest to other people</p>
<p>Part D and E, E+M 3.</p>
<p>D reduced the overall resistance in the circuit, which increased the current, which in turn made the current through the parallel branch equal to the current in the single Bulb 1 branch, meaning it got brighter.</p>
<p>The second one was no change, becuase the region of flux got smaller, but the wire inbetween them created two seperate circuits with half the resistance. It was a dead short that made it two different circuits.</p>
<p>Thank you. That was very insightful and made a lot of sense. I’m hoping my multiple choice section made up for these errors.</p>
<p>Oopsies!!! The part (b) of question 2 on Mechanics requires you to use the small angle approximation. Now, correct me if I’m wrong, but I believe the assumption is sin (x) = x.
For that problem, the answer should read 2pi / (-A(w^2)t). Or at least that’s what I put on my FR section.</p>
<p>E & M 2e: i said E field is up. I agree that the force due to the B field is up, but that is because those are negative charges. So what you end up with is more electrons on top and a positive charge on bottom. E field goes from positive to negative… so i thought it had to be going up.</p>
<p>I applied right-hand rule because I thought circuit analysis was always in terms of positive charge (even though we know electrons flow). The battery source current is from the positive end. But I understand your reasoning, and you may still be correct.</p>
<p>I did F=qv x B but neglected that they asked for the force on an electron… :(</p>
<p>Oh but wait! They asked for the direction of the electric field. Score!</p>
<p>Well, as far as (c)+(d) on Mech 3 go you had your function right. Essentially (y/L)<em>M</em>g. Take the integral of that with respect to the y and you get your work. Set work equal to the change in kinetic energy and realize that it had 0 kinetic energy at the beginning.</p>
<p>Also, for FRQ #2 (b) Why do you have an R in there? It’s T=2pi/w</p>
<p>Oh!!! That makes soooo much sense in hindsight. Thanks for letting me know. That problem messed with me because y changed as a function of time, and you couldn’t use time in the equation. It was weird and I could tell some serious algebra or calculus was involved.</p>
<p>Yeah, I agree. There was a lot of calc on this exam. Like a strange amount of it.</p>
<p>mech #2 part a:</p>
<p>equate the two equations for torque: r x F = Ia (a= angular acc, and i’ll use y=theta because i can’t do greek symbols lol)</p>
<p>you get xMgsiny = I(y’') <= that’s your d’ble equation for part i</p>
<p>you’re right, small angle app. means siny=y so now you have xMgy=I(y’')</p>
<p>by definition, y’’ = (-k/I)y, and you should be able to get the period from that (w=sqrt(k/I))</p>
<p>and you completely left out the tension forces on mech 3 which is why it seemed so easy</p>
<p>i did the same exact thing until there was about 5 minutes left and i realized i forgot the tension so i had to rush and redo the whole problem…</p>
<p>What causes the tension if there is no friction on the table?</p>
<p>"What causes the tension if there is no friction on the table? "</p>
<p>Umm, there is no relation between friction and tension in this problem. So why would it matter to the tension if there is no friction?</p>
<p>Good point! So how is tension calculated?</p>
<p>tension really only plays a part in part a, your part b is right after i tried it again</p>
<p>tension is the only force acting on the mass on the table, so T=(m/2)a, and the net force on the hanging block is (m/2)g-T=(m/2)a</p>
<p>plug in T and solve for a and you get g/2. you don’t need to actually compute the tension, just know that it’s the force acting on the object on the table.</p>