<p>This year, I am trying out for the US Physics team with several friends. I was wondering, could the applications be mailed individually by the students, or should they be mailed together as a group by the teacher who nominated us?</p>
<p>mailed by the teacher, who has to sign each form saying that they agree to administer the exam.</p>
<p>when is the first thing, comparable to AMC for phsyics and chem?</p>
<p>The postmak deadline for registration for the physics olympiad is in 2 days, Jan 21st. Prelim tests to be given the first week in Feb.</p>
<p>i was wondering how that worked...my teacher said he can give it any time that week...but then can't people compare answers with others who took it earlier?</p>
<p>i know there are 30 multiple choice questions and the free response portion is worth 100 points. so is the total possible out of 130? because i thoguht i read somewhere that both portions were weighted equally. or that we needed half of the points? (80 ish, so 2 points each for multipe choice?) thanks.</p>
<p>Spyder - everyone at the same school is supposed to take it at the same time. Yes, there's the possibility of students from different schools sharing info. But that would be cheating, wouldn't it? Also, the teacher submits students' written work on the free response, so the judges can see how they answered the problem as well as how the teacher graded it.</p>
<p>hmmm, also, does anyone know any good physics forums to ask for help? (i forget if youre allowed to post it here, if not please PM it to me)</p>
<p>I also had a qeustion about this problem (1999 USPhO creative response #4):
A car accelerates uniformly from rest. Initially, its door is slightly ajar. Calculate how
far the car travels before the door slams shut. Assume the door has a frictionless hinge, a uniform mass distribution, and a length L from front to back.</p>
<p>it seems like you would need a specific angle or coefficient of friction or something. thanks.</p>
<p>I took this last year. The multiple choice counts much less than the free-response. The total is over 100 points and I think you have to get a 75 or above to qualify for the 2nd round. Feel free to PM me if you have more specific questions.</p>
<p>ssshafted: I was having trouble with the same problem. If you manage to figure it out sometime, I'd really be happy it if you could post the solution here. :)</p>
<p>Yeah it does seem that way. If it were open .1 radians compared to pi radians, it shouldn't take nearly as long to close.</p>
<p>pi radians is 180 degrees, i doubt it would be open that far :o</p>
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<p>"I also had a question about this problem (1999 USPhO creative response #4): A car accelerates uniformly from rest. Initially, its door is slightly ajar. Calculate how far the car travels before the door slams shut. Assume the door has a frictionless hinge, a uniform mass distribution, and a length L from front to back"</p>
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<p>Does anyone else get <3/4 pi-squared L>?</p>
<p>The angle doesn't matter very much. It's a pendulum, so the angle has little bearing on it, but it says "slightly ajar" so you can easily use pendulum/spring equations.</p>
<p>o, i asked my dad for some help. he said the asnwer is instantly because you take the moment about a moving point (which is a really obscure equation found in some graduate level dynamics course). i wonder how most people would do it though (wihtout the equation) and if anyone else got instnatly.</p>
<p>now that i think about it, the pendelum way does make more sense. ill look into that. thats probably how texas also did it (i think b/c of the pi^2 in his answer)</p>
<p>(1999 USPhO creative response #3): Two masses, m<em>1 and m</em>2, attached to equal length massless strings, are hanging side-by-side just in contact with each other. Mass m<em>1 is swung out to the side to a point having a vertical displacement 0.20 m above mass m</em>2. It is then released from rest and collides elastically with the stationary hanging mass m_2. Each of the masses is observed to rise to the same height following the collision.</p>
<p>The problem becomes really simple if you can assume that you know the two balls are travelling in oppisite directions (and along the same plane as m<em>1 swung). What is the reasoning for showing that this is the case? and also, is there an argument to show that both the balls do not contain any velocity components in the direction perpendicular of the way m</em>1 was swung (so you wont have to break it down into v_1,x etc. etc.) Sorry if these are really simple or obvious questions, my friend said it was "obvious" but i couldn't relaly see why (maybe it's because of the string).</p>
<p>If m1=m2, then you will have a Newton's cradle, meaning the observation won't be the same. So is there any restriction on the mass ratio between m1 and m2?</p>
<p>Regarding the above question, I see no reasoning to initially exclude the case the two masses cannot travel in the same direction, if they are mass-points. But really, a couple of simple equations taken on that assumption will contradict with that assumption itself.</p>
<p>Yeah....for the 1999 Free Response...</p>
<h1>3 basically...because the 2 balls rise to the same height...they must have the same speeds at their start of the rising (directly after the collusion)...this is because mgh = 1/2 mv^2...and as you can see...the masses cancel out.</h1>
<p>BTW, in this case, the balls MUST travel in the opposite direction for both momentum and KE to be conserved.</p>
<p>Basically, utalizing this fact...+ conservation of momentum and the fact that the collusion is elastic, you should be able to derive the height. (I got 0.05 as my height..just wondering if anyone else got the same answer.)</p>
<p>Basically, for #4, I used the pendulum rule:</p>
<p>T = 2(pi)(I/2mgL)^0.5....with g = a.</p>
<p>I basically worked from there...found the time, and plugged it in for distance.</p>
<p>forgot to add that time would be half the period....</p>