<p>What’s the curve going to look like for Form O E&M? I need to get a 5, but at best I got 48-49ish out of 90.</p>
<p>That’s probably a 4 unless the curve is really nice.</p>
<p>Understood, I will be sending my cancellation form. </p>
<p>I wouldn’t, you might’ve done better than expected.</p>
<p>Fixed google doc: Someone deleted everything.</p>
<p>Here is the backup…
<a href=“Backup - Google Docs”>https://docs.google.com/document/d/1rXPLJxXdiKqKE-XjVU1-GCPEV8ROq0htZBokgH4JqvM/edit</a></p>
<p>Combine the current version with the missing stuff from the backup =]</p>
<p>I can’t edit for some reason without signing in!?</p>
<p>I’ll post here:</p>
<p>For the polar bear question I put md/(m+M) with the smaller weight on top. I was pretty sure that was right.</p>
<p>On the v escape comparison: I put sqrt(2)/2. the first v was v = sqrt(2MG/R) ==> if R is doubled the new v is sqrt(MG/R) which is equal to 1/2 * sqrt(2) * sqrt(2MG/R)</p>
<p>There was another question about how to affect the period of a pendulum by changing certain characteristics. </p>
<p>I put (A) length only. I’m nearly 100% sure that’s right. </p>
<p>@SuperN0va What it a simple pendulum or a physical one? Because if it was physical it would be (B) which is length and mass</p>
<p>Simple pendulum. </p>
<p>Hello,</p>
<p>Intuitively, the bear thing is md/(m+M) because think about it: if you pull a heavy object, the heavy object has more of a “tendency” to stick in its place; this also satisfies the idea that heavier objects have greater inertia, so naturally, they’ll move less distance. </p>
<p>For the weird ball thing, my thought process (I was really unsure of this one though): “bigger surface area = the thing hits more air, so it’ll get resisted more. Air resistance is greater for the bigger ball. Also, if I drop a piano and a ping pong ball from the same ridiculous height of, I feel like the piano (point of this is that the piano is MUCH heavier than the ping pong ball) would hit the ground first.” Hence, I selected the appropriate answer choice.</p>
<p>However, now that I looked up the terminal velocity formula (which is \sqrt{\frac{2mg}{\rho(AC)}}), I notice that C (drag coefficient: dependent on shape: both balls obviously had the same shape), g, and \rho (density of air) would all be the same, leaving the only considerations to be mass and area. I don’t quite remember the givens, but I think we can figure what the actual answer is from here.</p>
<p>The simple pendulum period formula is 2\pi\sqrt{\frac{L}{g}}. The only thing that is an intrinsic property of the pendulum that apparently affects is period is length.</p>
<p>I personally feel like the EM curve will be more generous this year given that the FRQs were… nonstandard, in my opinion (don’t take my word for it though; I say this with very little confidence). However, the mech curve should be alright since this test was slightly easier than some of the practices I took (again, don’t take my word for it).</p>
<p>-DVA</p>
<p>Did anyone really get the whole test finished in time?</p>
<p>I barely finished all the MC in the last minute and I leave like 1/3 of the FRQ blank (Mechanics Form O). I really feel bad that I was too nervous to think of how to do the (b),© part of the 1st FRQ while I am taking the test, but I got the answers in my head right at the moment I stopped writing. So I just lost all these points even though I actually know how to do it.</p>
<p>And for E&M (Form G), I got all the MC answered also although 3 or 4 of them I just I guess and I got 3/4 of the FRQ part answered but I already know I made some mistakes on the last one (got a - written +).</p>
<p>Not confident at all, worried if I still have a chance to get a 5.</p>
<p>Guys, I’m almost sure that the answer to the very first MC for Mech form O was 5 N. Please tell me if any of my work sounds flawed: the coeff of fric was .25 between the two blocks and the bottom block and ground. weights were 4 and 10 N for top and bottom block, respectively. So the total Max static friction was 4(.25)+14(.25)=4.5. Therefore 5N would’ve been the least force needed of the choices to start motion</p>
<p>Not impossible for you to pull out the 5! </p>
<p>Do you remember any hard MC from Mech form O?</p>
<p>@SuperN0va i just posted a minute ago about a hard MC. Tell me what u think</p>
<p><a href=“http://media.collegeboard.com/digitalServices/pdf/ap/ap14_frq_Physics_C-M.pdf”>http://media.collegeboard.com/digitalServices/pdf/ap/ap14_frq_Physics_C-M.pdf</a></p>
<p><a href=“http://media.collegeboard.com/digitalServices/pdf/ap/ap14_frq_Physics_C-E-M.pdf”>http://media.collegeboard.com/digitalServices/pdf/ap/ap14_frq_Physics_C-E-M.pdf</a></p>
<p>Form O FRQS for both exams. Anyone want to solve them?</p>
<p>@dskate Your reasoning is correct. The only problem is that I believe the problem asked for the force needed to move Block B which was on top of Block A. Since A = 10 N and B = 4 N, the normal force on B was 4 N and thus friction = 0.25(4N) = 1 N. 3 N was the first answer choice > 1 N. </p>
<p>For E&M FRQ #2, did anybody put the acceleration was upwards? And why is that wrong?</p>
<p>@shutterstock Because there was no vertical acceleration. Any net force would cause vertical acceleration (Newtons 2nd law) so therefore it is 0</p>
<p>@Repede When are the scoring guidelines released? </p>