<p>it seems that i ruined the free-response section of mech (barely half right)…to be safe, i i get 80% in mcq and about 50% on free-response, will it guarantee a 5?</p>
<p>btw, i hv some doubts:</p>
<p>1.for mech 1, do i get any credit if i just plotted vt against time instead of vt^2 against time? (i just used the data given as it is) i included the best-fit line as well. </p>
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<li>for mech 2, i got the 3rd q wrong (i equated mgh = 1/2 Iw^2 instead of mgh = 1/2 mv^2 + 1/2 Iw^2…apparently forgot to include translational KE…), consequently leading to a wrong answer for the last q (finding the total distance). If my working was all correct except for the incorrect value of velocity from the 3rd q, will i be able to get credit, if any, for the last q?</li>
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<p>You may lose 1 - 2 points for not vlotting v_t^2 versus time. I’m basing this estimate on 2006 #2c.</p>
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<p>Well, first off, the last part is asking for final velocity of the ball plus box plus wagon. Assuming you did find the final velocity, you will get 100% credit if everything else is right for part d. You could end up getting no credit for part c because by not including KE_translational you’re basically saying the center of mass is not moving.</p>
<p>For E&M 1E I was interested in how you all were coming up with odd answers as opposed to the general equation itself. In terms of the negative derivative of potential you would have kQ/r. If you take the negative derivative of that you would have something like kQ<em>deriv(1/r) or kQ</em>deriv(r^-1) which would end up being kQ/r^2. Either the general equation for the electric point charge field is correct or the potential solution is incorrect. Interestingly enough each dQ is a distance r from point P. Just some general thoughts as I’ll be taking the form B this friday. I hope I’m ready.</p>
<p>JDong can you please elaborate on some of your E&M solutions. Especially E&M 3. Thank you.</p>
<p>You have to integrate (k*dQ/R) rather than assume that all the charge is concentrated at the center. Since it is uniformly distributed over a flat quarter circle, it’s uniform relative to length. So dQ = (σ) dL. The σ value is equal to Q/(π /2 * R), because it’s total charge over total length. </p>
<p>And if you observe a small piece of length dL, you know the relation is dL = R * dΘ. So all in all, your total expression is dQ = [Q/(π /2 * R)] * R * dΘ, which ultimately cancels out into dQ = 2Q/π dΘ. When you integrate k*dQ/R, you substitute this into the expression, getting :</p>
<p>2kQ/πR dΘ. When you integrate theta from -π/4 to π/4 (quarter circle), you get π/2. This multiplied towards 2kQ/πR cancels out into kQ/R.</p>
<p>For part (e), instead you integrate k*dQ/r^2 * cos Θ (Because all the vertical components cancel out due to symmetric property). You still have the same dQ value, so you integrate (2kQ/πR^2 * cosΘ dΘ). The first half is constant and you pull it out, and the integral of cos Θ dΘ is sin Θ. When you subtract, sin(π/4) - sin (-π/4 ), you get sqrt(2). You multiply this to the constant you have on the outside, and get 2kQ *sqrt 2/πR. Now just substitute k = 1/4πe0, and you get the same answer that they provided. Hope that helped!</p>
<p>you can’t just do -grad V to get E, you dont have E as a function of x, you only have a value at one point. i was tempted to do that at first also.</p>
<p>also i dearly hope these solutions are correct, cause that implies that i got a 40/45 on the mech and 45 on em</p>
<p>Yeah, I got a 2 in mechanics which I was really shocked by since I studied so hard for the test. I am for sure going to get my test rescored because it sounds like the curve was super generous so it would be really hard for me to have gotten a 2. I think maybe they added my score wrong or something…</p>
<p>I reviewed my free response answers after the exam, and I felt like I should’ve gotten at least 60% right. And the multiple choice, I left 5 blank and felt confident in all the rest but 1 or 2. If my instincts are close to right on this, I should’ve had at least a 65%. Usually the curve is 50% or better for a 5-either I should get it rescored or the curve wasn’t so generous.</p>
<p>It sounds like from what I’ve been reading the curve was really good, I think they must have messed up my score, a 2 is just really low for how I thought I did, and I thought I did like a 1 in E and M but I got a 3.</p>
<p>for the second problem in e%m, i took the circuit to be in parallel instead of series. then for the calculations in part d and e, would i get error carried forward or both parts are wrong?</p>