<p>@Vigilante13 Correct.</p>
<p>@EZCaptainAmerica Same time the scores are released (early-mid July).</p>
<p>@EZCaptainAmerica they will post it only when we get our scores.</p>
<p>I put Q for that one also.</p>
<p>@Vigilante13 Oh my goodness, you are so right. I thought I nailed the first FRQ but I totally did not. I thought it was uniform circular motion since the problem specifically stated “circular arc.” I feel so freaking stupid and useless. All that studying for nothing.</p>
<p>Guys, I studied for about 6 days straight and nailed all FRQ over the last 4 years or so, maybe more. This test caught me off guard, probably because i spent so much time determining density on No. 2. It was pretty nasty for me, i just want to know if anyone is in the same predicament, and what you all will think the curve will be.</p>
<p>The following are my ■■■■ ups and what i think each is worth:</p>
<p>1) Easy kinematics, probably about a 13/15 i skipped D because my dislexia got the best of me and i thought it was asking something else
2) 7/10 maybe. Solved for a density of like 100,000 kg/m3, i noted i knew it was wrong, but plugged it into the next problem, and that should be fine. Everything else i know i got right, easy problem over all
3) 6/10 Skipped rankings, solved for it but didn’t rank lol, -1 or -2 they normally don’t cost much, could not determine Td in terms of p0 and v0, i left it in terms of Ta, Td = 4Ta, i didn’t know how to convert to proper terms. BC was the one that was having work done by the gas, because work done on the gas is negative, blah blah blah, we know the drill. Everything else i think was right
4)5/10 maybe, skipped d and e because of dislexic tendencies, and i was running low on time. Vector for the graph was to the top left because of components determined in part A
5)4/15, screwed this up bad, A was wrong, B i missed a force (+2 maybe), C didn’t try, D good (+2), E wrote F=BIL (+1 maybe, not including)
6)4/10 A and B were good, a was just v = f<em>lambda, b was phi = hf, nice and simple, but the rest i didn’t have enough time because i spent so much time on 2, stupid me
7)7/10, caught me by suprise during the test but easy enough, A use v = f</em>lambda, B is same as answer to A because frequency doesn’t change no mater what, c was just n = c/v then plug v into v = f*lambda with previous frequency, and you’re good. d was greater than, justification was probably wrong lol. e was a blind guess, but i said it had something to do with the angle of refraction changing, i doubt this is worth any big points because it’s the last question on the exam, they don’t expect you to get there.</p>
<p>Anyone else in the same boat?</p>
<p>@yoloswagins, I am sort of in the same boat as you are. However, I slightly did better on the frqs. Also, for number 7, I thought the index of refraction in the plate was less. I used the formula n1/n2 = lamda2/lamda1</p>
<p>If you plug in the values you can see the index of refraction decreases.</p>
<p>Can anyone post the solutions for questions 3-7???</p>
<p>@Vigilante13 You might have had your n’s backwards, all of us in my school got greater than (about 14 of us).
@huhululu My physics teacher is planning on going over them in my class tomorrow. I will take photos of each answer, and post them here.</p>
<p>@yoloswagins hmm you may be right, but I’m not sure. Could you explain how you arrived at your answer?</p>
<p>On the free response involving the magnetic field, the last question on the left page gave a bunch of variables you had to find your answer in in terms of (I can’t remember what you were finding though). But, how did m fit in there?</p>
<p>@WhySoStressful since they told you it was a delta y, you had to factor in gravitational force (mg), cuz the bar was hanging in the air</p>
<p>solutions anyone?</p>
<p>@sepehr3 Okay cool, that’s what I did. I was just concerned because my equation didn’t look as pretty once the minus sign was introduced 
Also, g was not one of the variables mentioned</p>
<p>Regarding question 5, I am fairly confident the correct expression for Δy does not include Fg because it is present before and after the battery is installed, meaning that it had the same effect on the rod before and after, so it shouldn’t have an effect on Δy.</p>
<p>@WhySoStressful while I agree with you that Fg should not be included, the question states that the variables k, m, I, L, B, and any constants are allowed, so g would fall under the last category.</p>
<p>My solution:</p>
<p>mAy = ΣF = Fs - Fb -Fg
0 = 2ky - ILB - mg
y=(ILB+mg)/(2k)</p>
<p>mAy0 = ΣF0 = Fs0 - Fg
0 = 2ky0 - mg
y0=mg/(2k)</p>
<p>Δy = y - y0 = ILB/(2k)</p>
<p>Notes:
Fs is 2ky since there are two springs in parallel
m is given as a variable that can be used in the answer, but it is not needed</p>
<p>@johnstucky Thank you! Your solution seems right, especially since in part e), using the straight line probably means using the slope value calculated as Δy/I to find B. College Board was probably trying to trick people into using m in their answer, but smart people like you don’t fall for that :). </p>
<p>@420Stalk3r Yeah for part e) B = 2kΔy/IL = 2(Δy/I)(k/L) = 2(0.0028)(25/0.35) = 0.4 T</p>
<p>@420Stalk3r @johnstucky I hate to break it to both of you, but I think your solutions are incorrect. Collegeboard wouldn’t put false information in saying that you have to include m in your solution. Also, y_final in your equation is actually delta y</p>
<p>@Vigilante13
According to the free response question:
“Derive an expression for Δy in terms of k, m, L, I, the magnetic field strength B, and fundamental constants, as appropriate.”
That clearly implies that you do not need to use all the variables.</p>
<p>“A conducting rod of mass m and length L hangs at rest from two identical conducting springs, each with spring constant k, as shown in the figure at left above.”
In the first figure, the rod is specifically at rest, meaning that gravity is already acting on the rod. We can solve for y0, the initial position of the rod in terms of the other variables using Newton’s second law. Since the rod is at rest, meaning it is not accelerating, so the net forces acting on the rod are equal to zero. The two parallel spring forces are acting up, and the gravitational force is acting down, with all other forces being nonexistent or negligible.
Therefore 2Fs = Fg, -2ky0 = mg
y0 = -mg/(2k).</p>
<p>“The rod is displaced downward, eventually reaching a new equilibrium position with the springs stretched an additional distance Δy.”
When the problem states that a new equilibrium is reached, it is again implying that the net forces equal zero. We can use the same logic to arrive at an expression for y final.</p>
<p>y = -(ILB + mg)/2k</p>
<p>Therefore Δy = y - y0
= -ILB/(2k)</p>
<p>Note:
I didn’t include the negative in my previous solution, since in my physics class, we just used Fs = kx (without the negative), and used logic to determine if the negative is appropriate. In this case, it would be, but I hadn’t remembered to use it. Also, since the Δy values are positive in part e), I’m assuming the College Board won’t be too picky about the sign.</p>
<p>@johnstucky my physics teacher used mg in his answer. Your method doesn’t seem false, but I’m just stating what my physics teacher got. I’ll show him this later and see if you are correct or not.</p>
<p>Sweet, I’d be interested to see what he says. For the first part, I got the same as you @johnstucky - But for the DeltaY I included the mg. We’ll see, but I do see your point for part e.</p>