<p>deleted for reasons</p>
<p>On FR1, what do we need, energy or momentum?</p>
<p>1a and 1b were impulse-momentum. 1c and 1d were energy.</p>
<p>I’m mad because I didn’t label any of my asymptotes when they were approaching zero - I read the directions, I just guess I didn’t think anything of not labeling a graph that is obviously approaching zero.</p>
<p>anyone have any advice/tips? I’m taking both physics c tests next friday</p>
<p>^there are none. i just wish you the best of luck.</p>
<p>we can discuss now.</p>
<p>For 1 e did people get </p>
<p>Dn=FtD/fb</p>
<p>The questions aren’t online yet, it’ll be really hard to discuss without access to them (I didn’t get my green insert back yet).</p>
<p>I got mine green insert today, right it’;s safe to discuss answers now</p>
<p>for the differential equation in question 2, did you have M dv/dt = -kv? I think I had a lapse in judgement in my math when actually integrating it, I ended up with (ln|-kv|)/-k = t/M, and ended up as a final answer with v(t)=(e^-kt/M)/-k + v initial - 1/-k. I’m positive the final integral was wrong, but i’m wondering if I at least integrated it initially correctly. also, for 3a, would it have been I d^2 theta/ dt^2 = -B theta?</p>
<p>How I did it was:
M dv/dt = -kv
dv/v=-kdt/M
ln|v|=-kt/M + C
v=ce^(-kt/M)</p>
<p>initial value (0, vd)</p>
<p>v(t) = vde^(-kt/M)</p>
<p>^That’s right.</p>
<p>
</p>
<p>I believe so. Except I was rushing through it and wrote d^2 theta/ dtheta^2 = -B theta, lol. Woops :o</p>
<p>yeah, that’s the right answer. I should have only brought v to the side with dv, not sure why i took both. By that point I was so frustrated with the test, it must have worked on my focus. oh well. I figure the actual integration can’t be worth more than 3 points? I wonder how many points people reckon parts e and f will be worth on the first question. i figure no more than 10 as a maximum, and realistically perhaps 6.</p>
<p>!!! Did anyone else for the 2nd Mechanics FRQ, Part E, (iii.) try to graph starting from point A all the way to point E???</p>
<p>I just realized that that’s what I did because it seemed rather ambiguous in the wording from where to start and end… I still ended up graphing the correct acceleration function that I got from the velocity equation in the part above (ii.), but I added the free fall acceleration from point A to B, made something up for the acceleration from B-D, then graphed the “correct” acceleration for D to E.</p>
<p>Do you think they’ll still accept my answer or did I screw it up too bad? Also is it worth reporting this as an ambiguity, because although I might be like the only one who graphed from A to E, I really wasn’t sure where we were supposed to start.</p>
<p>What did you guys put for #3 part f? “What is the physical significance of the intercept of your line with the vertical axis?”</p>
<p>
</p>
<p>I did exactly the same thing haha. I agree that the wording was ambiguous. I think as long as we labeled our points on the x-axis it shouldn’t be that much of a problem.</p>
<p>^And for that one, I had like 10 seconds left at that point so I think I said something along the lines of at the y-intercept, the rotational inertia will be 0 so the string will not be twisting at that time, so the y-intercept is the initial period that the pendulum has, or something like that. I mean, the torsion pendulum will inevitably sway side to side, right?</p>
<p>did any of you wonder about the premise for 1 e and f? where it said assume that the collision of the bullet and the block are instantaneous and hence no frictional force was acting at that time. because if there was no frictional force acting and the block didn’t move until after the bullet had stopped moving in the block, then that would violate conservation of energy as either the bullet wouldn’t have entered the block and the block would have moved or the bullet would have gone in the block all the way and the block wouldn’t have moved. Correct me if I’m wrong.</p>
<p>And Takayu, I believe the problem expressly stated graph the acceleration from D to E.</p>
<p>^The problem did not explicitly state to graph the acceleration from D to E, I just checked. However, I believe it was implied.</p>
<p>^Did I need to have the acceleration hit 0, or would asymptote be OK? I know the cart stops at point E, but does it say that the graph has to end at point E?</p>