<p>If that thing is true, I am a little more optimistic…</p>
<p>Physics has a massive curve. But bear in mind, the cutoff for a 5 may be a bit higher now that there are no penalties for wrong answers.</p>
<p>If that thing is true, I am a little more optimistic…</p>
<p>Not to be a downer, but I’m pretty sure it will be higher since there are no points lost for guessing wrong.</p>
<p>Also, you can calculate the score there by just doing MC score * 45/35 + FR score</p>
<p>Mechanics, If I had 22/35 the MCQ and 2/3 of the FRQ how much would that equate to?</p>
<p>Also, if I had 29/35 on the E&M MCQ and 2/3 of the FRQ how much would that equate to?</p>
<p>It’s alright. I really don’t care anymore. </p>
<p>If I get a 5, I get a 5. If I don’t, I don’t. No use wasting my breath over it.</p>
<p>@PioneerJones: Bear in mind, that website has adjusted the scores for the new no penalty rule.</p>
<p>did anyone get something like µ(r^2-a^2)/(2πr(b^2-a^2)) for E/M FRQ 3 pt 2ii?</p>
<p>^I just multiplied (µI)/(2πb^2) by (r^2/b^2), but I forgot to take into account that there’s no current for when r<a, so I think you’re right (as long as you add in the I).</p>
<p>yeah sorry about that, forgot the I in there :P</p>
<p>you had to substitute the x into y</p>
<p>Mechanis MC wasn’t hard; it just had a ton of questions which were extremely time consuming, which made it seem impossible. FR…if anyone else noticed, there were part f’s for each question, or in other words, six parts! That’s eighteen parts total. If anyone like me was hoping for partial credit, he can kiss that goodbye. And from what I heard, the AP graders set the score range before they begin to grade the tests, so that ought to bring a record low…</p>
<p>E&M was actually relatively easy (keyword: relatively). I personally thought it was much easier than Mechanics, and usually I’m a lot better at Mechanics.</p>
<p>EDIT: Turns out my physics scores are gonna be cancelled because some people had an AP shirt that said “Flux this sheet” and put the equation for flux on it <_<</p>
<p>hahahahahaha.</p>
<p>but wow, that sucks</p>
<p>
</p>
<p>Same here. :P</p>
<p>
</p>
<p>Do you know the a–hole that called ETS/OTI about this?</p>
<p>Mech. 1</p>
<p>Parts (a) through (d) are fairly straightforward impulse - momentum questions.
Part (e) is rather difficult until you realize that the work is the same in both cases.
Part (f) is extremely difficult.</p>
<p>Mech. 2</p>
<p>Parts (a) through (d) are straightforward.
Part (e) is extremely difficult. It is a first order differential which has an exponential solution. The easiest way to solve this is by using an indefinite integral and evaluating the constant based on initial conditions. The acceleration is simply the derivative of the velocity.</p>
<p>Mech. 3</p>
<p>If you can write the differential equation and compare it to the differential equation for linear SHM, the rest of the problem is routine. Even if you cannot solve parts (a) and (b) you can still do parts (c) and (d) which are trivial.</p>
<p>Parts of this exam are extremely difficult while some parts are routine bordering on simple. I believe that a score of 4 is well within the reach of a competent student while a score of 5, while difficult is attainable.</p>
<p>^In your professional opinion, what is the answer to 2(b) (the magnitude of the centripetal force)?</p>
<p>The key is to avoid complicating the problem. The question asks for the magnitude of the centripetal acceleration in terms of the forces drawn in part (a). Don’t add anything. I can’t stress this enough to my students, who make problems more complicated than they need to be. </p>
<p>The equation will be:</p>
<p>F = ma
N - Wsin(theta) = Fc</p>
<p>Where: N = normal force, W = weight (mg), and Fc = centripetal force.</p>
<p>If you are uncertain about which function to use for the weight component, just remember that components of vectors always use sine or cosine. In this case consider the situation where theta = 0. Clearly, the weight acts vertically with no horizontal component (the direction of the normal), so sine is the correct function since it is equal to zero at zero degrees. You may confirm this by realizing that when theta = 90, N = mg, so the function must equal one at 90, which is the sine.</p>
<p>Newton’s law problems are all the same, the sum of the forces (net force) is on one side of the equation while ma is on the other. In this case, you do not need to use mv^2/r on the right side for centripetal force since the question asks for the equation for centripetal force, not m or a.</p>
<p>There is a typo in my previous post.</p>
<p>“The question asks for the magnitude of the centripetal acceleration…”</p>
<p>The line should be:</p>
<p>The question asks for the magnitude of the centripetal FORCE…"</p>