<p>The 48 hours is up....discuss away...</p>
<p>What did y'all find to be the hardest one? I thought the mech 3 rotational one was killer, um, yeah.....no idea how to do it...</p>
<p>I forgot which one was Mech 3...someone please refresh my memory.</p>
<p>All I remember is that the Mass of Saturn problem was unbelievably difficult. </p>
<p>E&M 1 was also pretty difficult. I think i got A and B right, but I don't know if I provided a correct explanation of everything. For E&M3, did anyone else get M1/M2 = 3?</p>
<p>Let's discuss!!</p>
<p>I was talking to my cousin who is damn good at physics (he got a 5 on Physics C like 5 years ago) about the mass of Saturn problem and I told him that all we were given was radius and period and that I didn't think it was possible and he explained to me pretty simply.</p>
<p>T= 2pi (r/v) [I had never seen this before in my life, but he told me that's what he would use]</p>
<p>Plug in period and radius and solve for Velocity.</p>
<p>v²/r = GM/r² and solve for M.</p>
<p>He also said that M = 4pi² r³ / G T²</p>
<p>Bah...seems easy enough if you knew all these formulas...many of which I had never seen.</p>
<p>no you just made fg=Gm1m2/r^2=fcentripetal=mv^2/r
yah, and then Mmoon cancelled, so you solve for v, and then to get T, do 2piR/v to get T...</p>
<p>and it all worked from there, sans any esoteric equations...:D</p>
<p>no but they gave you T i think..</p>
<p>i think the variables were mass of moon and velocity of moon
but stupid me..i forgot T=2pir/v.....</p>
<p>New Message because edit time for the previous message expired:</p>
<p>All I remember is that the Mass of Saturn problem was unbelievably difficult. </p>
<p>E&M 1 was also pretty difficult. I think i got A and B right, but I don't know if I provided a correct explanation of everything. </p>
<p>Mech 3: a) (1/3)(M1d^2)(omega) b) Forgot, but something relatively close to part a). c) M1/M2 = 3 and that the distance x in part d) was equal to d/3. </p>
<p>Mech 2 was just evil.</p>
<p>Mech 1: a) decreases because the intial upward acceleration upon release dissipates due to the constant acceleration down from gravity, b) m(dv/dt) = -kv (which I just realized was wrong) c) sqrt (2gh) d) longer up (not sure if this is right or wrong) and e) the graph I got was a 'v' shape that hit the x axis at the midpoint and went back up.</p>
<p>E&M 1: ai) Point C aii) Point A (not clear on explanations for either, just took a stab at how I got the answers) bi) electron moves along y = .06 with a constant acceleration and an constantly increasing speed bii) Something like 1.28 x 10^7 (or 8. Not fully sure) c) Estimate of 10 because V = -integral of E . dl and its halfway in between the two points d) top of my line was through y = .08 and bottom of the line went through point d and ended up at x = .08</p>
<p>E&M 2: a) I = 0 because the inductor does not change anything in the circuit immediately. If there is no current running through the inductor, there is no current through R2 and therefore the circuit shorts and there is no current at all. b) dI/dt = (emf - IR1 - IR2)/L c) emf/(R1 + R2) because inductor acts as a wire d) graph was curved from left to right and tapers off at value emf/(R1 +R2) e) Inductor has the total current as it was in part C because nothing changes immediately after a change in the overall circuit. V = (emf/(R1 + R2))R2</p>
<p>E&M 3: a) I had my turns starting at 25 and ending near 106 b) Was almost a straight line c) u0 = 3.33 x 10 ^ -6 d) 34% error (way too large I believe, I may have misread my decimal. Not sure)</p>
<p>Anyways, let's discuss!! Best of luck to all of us with the upcoming grades.</p>
<p>Problem two was just the derivation of Kepler's Third Law....The third one, you set K=.5Iw^2=.5mv^2 doing various things and you set m1v1=m2v2. Since the bar stops when it hits the ball, all its kinetic energy is transferred to the ball and momentum is conserved (you find the translational velocity of the rod right before it strikes using v=rw).</p>
<p>Jaug1, that answer for Mech 3 looks familiar...I think its right because the Ms and the Vs cancel out so you get something like 1/(1/3) which is the coefficient of the moment of inertia.</p>
<p>Yeah, that's exactly what happened to me tower. Yeah! Maybe we got something right!</p>
<p>what percent error did you get for the solenoid problem....? (i think EnM 3)</p>
<p>I got like .2% or something....really small...I was surprised that it was less than 1, though I may be wrong. All that was was B=(muo)nI, right?</p>
<p>I rocked the third problems for e/m and mechanics, not so sure about the first 2 though. The e/m third problem was quite possibly the easiest ap question ever. Just draw a graph (it gives you what the values should be around) plug in for buni, and do (e-o)/o*100.</p>
<p>the % error all depended on the point(s) you used. also, mech #2 was one of keplers laws. all you needed to do was derive it ;-) </p>
<p>2+2=5, d = vt .... look familiar? thats where the eqn for T came from ;-)</p>
<p>i got like 1.03 % error.....:/</p>
<p>the equation was 2(10^-4)=mu0(50)3 I think</p>
<p>Everything went well for me but the second problem for e/m... lol. The problem is that to answer that problem, you have to know how an inductor works, which is fine if you've taken the class, but... lol, for us "self-study" people who managed to put off reading the book until the night before, sometimes these things can be tough.</p>
<p>I did, however, just realize something: even with an 80% on the multiple choice and a 50% on the free response (both conservative estimates), my overall score is still a 5 under all the curves posted on the other thread from previous exams. Go low standards! :)</p>
<p>"T= 2pi (r/v) [I had never seen this before in my life, but he told me that's what he would use]</p>
<p>Plug in period and radius and solve for Velocity.</p>
<p>v²/r = GM/r² and solve for M.</p>
<p>He also said that M = 4pi² r³ / G T²"</p>
<p>Wow, that's EXACTLY what I did. I got the equation to find the mass, but couldn't put two and two together. I hope I got at least some points on that question for using Kepler's third law. </p>
<p>On FR #1, I couldn't do the differential equation (because I have almost no calculus background) and I wasn't sure if it took longer for the ball to go up or longer to fall back down (frankly, I thought at first it would take the same time both ways). As for the graph, I thought the graph had to cross the y axis since velocity would be negative roughly half of the time. Also, the graph has to be curved, since accleration is not constant. I'm hoping for some points here. </p>
<p>I seem to have gotten the same answers you guys got on FR #3. Didn't get the last part on that one though.</p>
<p>All in all, a very difficult FR section.</p>
<p>For the first one, it would take longer to reach the top than for it to fall. On the particles ascent, there are two forces that are in the opposite direction of its velocity--air resistance and gravity. When it is falling, however, gravity and air resistance are in opposite directions--gravity is "helping" the particle fall even though air resistances is retarding its movement. For the shape of the graph, since its a first order differential, I think the branches of the graph should look somewhat exponential/logistic.</p>