<p>@crazybobl
There is velocity, you set ke + gpe equal to the gpe in the first part of the problem (conservation of energy) so you just solve for v. </p>
<p>What didu guys say for the questions about the photoelectric effect? Choices were increase/decrease/remain the same </p>
<p>If it was all gpe at the original height, it would have to all be gpe at the same height on the other side. There’s no KE left to solve for the velocity, because there is no velocity. Total mechanical energy has to be the same all the way through. </p>
<p>What did you guys get for part d of that problem? The one that asked for how magnitude of momentum changed?</p>
<p>Guys, I think its just best if we don’t constantly fret over what you got for what questions: Relax, your test is done. If you want to calculate worst case scenario, use this: <a href=“http://www.appass.com/calculators/physicsb”>http://www.appass.com/calculators/physicsb</a></p>
<p>@yullari27 It was all GPE at the initial height, but at the point it was asking, there was a smaller amount of GPE, the rest of the energy had to go to KE, which is in the form of velocity. The GPE of the original amount is equal to the GPE of the point near the bottom (Let’s say it’s B because I can’t remember) plus the KE. therefore, KE=GPE(A)-GPE(B)
Because KE=1/2MV^2, 1/2MV^2=GPE(A)-GPE(B). The only unknown in that equation is velocity, therefore we can solve for it. At least, that’s what I did.</p>
<p>@1998golfer On my form, it specifically said the point was the same height as where the person began. I’m positive on this one, at least for my form. </p>
<p>@yullari27 I’m talking about velocity when it was at the lowest point (2.4m from surface of the water)</p>
<p>Free Response is out, someone want to do it with the right answers? </p>
<p>Ok, here is the solution to question 1: </p>
<p><a href=“Physics Solutions | PDF”>Physics Solutions | PDF;
<p>I’m not positive that it’s all correct, but I’m almost certain it is. </p>
<p>@adidinescu You must be kidding me. Well I’m done.</p>
<p>For number 1, did anyone put a parabola to describe the trajectory of the man? I’m pretty sure it was a straight line.</p>
<p>Solution to questions 1 & 2:</p>
<p><a href=“Physics Solutions | PDF”>Physics Solutions | PDF;
<p>@Vigilante13 I think it’s a parabola. He is accelerating downwards while his horizontal velocity stays constant. That means that his trajectory is parabolic in nature.</p>
<p>For number 1 the last part, shouldn’t momentum be conserved. so Pc=Pb???</p>
<p>@kingofxbox99 , did u see the fact that the frq said DRAW A SOLID LINE? lol</p>
<p>And also, just b/c his trajectory is a parabola, doesn’t mean you can draw it</p>
<p>you don’t know whether it is big, small, wide, or thin?</p>
<p>You have not information other than the FACT THAT IT A PARABOLA.</p>
<p>on the FRQs it stated “DRAW A SOLID LINE”</p>
<p>just b/c some guys posts solutions doesn’t mean he’s right.</p>
<p>don’t mean to brag but I probably got a high 5 on that exam. Trust me, I know what I’m talking about.</p>
<p>Oh ya when I said straight line I meant solid. Also, you are not supposed to give an exact trajectory, just an approximate path. This is what it should look like:</p>
<p><a href=“Screenshot - 41b131d12b995250e2427f8be05845f7 - Gyazo”>http://gyazo.com/41b131d12b995250e2427f8be05845f7</a></p>
<p>The solid line refers to the shading of the line to distinguish it from part ii</p>
<p>@kingofxbox99 Sry for getting a little angry there. I was just trying to make a point. I understand you can argue it is a parabola, but AP said to draw a line, and I’m pretty sure a line isn’t a parabola.</p>
<p>@huhululu If you look at the formula for momentum: p = mv, and you look at point D on his path where his velocity is 0, his momentum will be 0 which is different than when his velocity is a maximum at point C. I agree momentum is conserved, but that also involves vectors and directions, and we were only asked about magnitude.</p>
<p>A straight line is different than a line. A line can be curved or bent, as in this case. I don’t understand how it wouldn’t be a parabola because his trajectory is a projectile. His vertical velocity is increasing faster than his horizontal velocity is.</p>
<p>I think you’re just getting caught up on the semantics of the question. The question didn’t say “draw a straight line”, it said “draw a solid line”.</p>
<p>@kingofxbox99, AP stated that to draw a solid line, meaning a parabola isn’t a solid line. Also, in the guy’s solution, the line from D shouldn’t be pointing down. That’s completely wrong. It should be TANGENT. There will ALWAYS be a horizontal component. Again, not sure why you are listening to these solutions, but I’m trying to point out that they are wrong.</p>