<p>can we discuss the 2014 test now?</p>
<p>Sometimes when I’m alone I like to quote myself
</p>
<p>I thought this year was pretty easy. I’ve spent one year self studying physics before this test and I project about 24 right and 1 wrong for 23.75 (after extensive comparisons after school with some of my classmates). I still hope the cutoff is low though just because :D.</p>
<p>I agree. This year’s test seems a little easier than last year’s test. I just hope the cut off doesn’t jump into the 17-20 range.</p>
<p>I feel that this test was the easiest of all years. Thirteen of my friends took it, and including me, we scored from 19-25. I would be surprised if the cutoff was anything lower than 20. </p>
<p>The cut off doesn’t change that drastically from year to year, does it? </p>
<p>I mean, ~15 to 12.25 was a big jump. And now you’re saying it’ll go from 12.25 to 20? </p>
<p>Do you go to Stuy or something? </p>
<p>I seriously doubt that the cut off will be 20. The test was easier than last year’s, but it’s not that easy. I would guess a cut off somewhere in the range 14-16. Last year’s cut off just seems unusually low. </p>
<p>Let’d post physics problems to take up time until we can discuss! Here’s one:</p>
<p>A particle of mass m is constrained to move in one dimension. A force F acts on the particle. F always points toward the origin. For example, when the particle is to the left of the origin, F points to the right. The magnitude of F is a constant F except at the origin where it is zero.</p>
<p>a. The system is horizontal. F is the net force acting on the particle. The particle is
displaced a distance A from the origin and released from rest. What is the period of the motion?</p>
<p>b. The system is vertical. Both the force F described above and gravity act on the system. The particle is displaced a distance A above the origin and released from rest. F > mg. What is the period of the motion?</p>
<p>Two more interesting problems:</p>
<p>1) Find the time required for a spherical cloud of matter with uniform density and a given radius to collapse due to gravitation. Assume that there is no pressure or other particle-particle interactions.</p>
<p>2) A long straight wire carries a current I. At distances a and b from it there are two other wires, parallel to the former one, which are interconnected by a resistance R. A connector slides without friction along the wires with a constant velocity v. Assuming the resistances of the wires, the connector, the sliding contacts, and the self-inductance of the frame to be negligible, find:
(a) the magnitude and the direction of the current induced in the connector;
(b) the force required to keep the connector’s velocity constant.</p>
<p>1.</p>
<p>Consider a point on the edge of the cloud of radius R. The inside portion of the cloud attracts the point as if all the mass were concentrated at the center. Thus, no matter how small the cloud is, as long as it is spherical, the outside point will experience a force equivalent to the situation in which the entire mass of the cloud were concentrated at the center. The situation is therefore equivalent to a point mass falling directly into a mass due to gravity.
The cloud has mass M = 4/3<em>pi</em>R^3*p. </p>
<p>By Kepler’s Third Law,
T^2 / R^3 = (4 * pi^2) / (G * M)
T = sqrt(3<em>pi/(G</em>p)).</p>
<p>This is the period of orbit for a full elliptical revolution around a parent body, and we can assume that falling into a body is equivalent to traversing the semi-major axis of a very flat (degenerate) ellipse. Therefore the time it takes to fall in is T/4, or t = sqrt(3<em>pi/(16</em>G*p)). Somebody else can do #2.</p>
<p>^
That’s actually incorrect. For an ellipse, the foci are off-center, so the falling time is not T/4. </p>
<p>Hmm why does that affect the falling time?</p>
<p>Does anyone remember approximately when the results come out?</p>
<p>Around six days.</p>
<p>
</p>
<p>It means that you can’t simply say that it’s equivalent to traversing the semi-major axis of the ellipse. The problem is quite beautiful (in my opinion), but I would recommend trying a different approach ~ ;)</p>
<p>Is the semi final like the f=ma exam where we can choose a certain day to take it or do we all have to take it on the same day?</p>
<p>@mapletree, I’m afraid I’m not quite sure what’s wrong. The semimajor axis would be the distance from the center of the cloud to the initial edge of the cloud, yes? And by Kepler’s Third Law that is the distance that matters. What is the solution supposed to be?</p>
<p>@fooiey, the semifinal can be taken in a span of a week or so, but it depends on when your teacher decides to administer it.</p>
<p>I can’t wait to get the results back! But if I do get to the next round, my answer sheet will probably be completely blank because I looked at the previous tests and had no idea how to do them. Do you guys think it’s possible to learn everything in the rest of my physics b textbook and a physics c textbook by exam day? I really don’t want to sit in a room for 3 hours and not have anything written down by the end of the exam.</p>
<p>probably not…but one can try. and physC is a lot easier than semifinals btw.</p>
<p>don’t worry, that’s basically what I did last year. Passed F=ma on a guessing fluke, took one glance at the semifinal problems, handed the paper back to the teacher, and was like “can’t do any, can I just not waste both of our time :3” and he was like “ok”</p>
<p>i shoulda prepped better this year. maybe I can do 1 semis A problem now…lol…</p>
<p>Isn’t a great honor just to make it to semifinals? Doesn’t it look great on apps and stuff? </p>
<p>Fingers crossed yo…</p>