<p>i got -.79% on the solenoid problem, you could also find the orbital speed of a planet by setting gravity equal to centripetal force and dividing that into the circumference of the orbit. the final velocity is when gravity equals the drag force so v=-mg/k. I had never done rotational collisions or induction so i sort of had to derive some induction stuff for that one. i think i got 3/5 on the collision one. does anyone remember what their graph looked like on the induction one. also, on the saturn one i derived everything correctly but i forgot to square 2 (in 2 pi) so my mass was half of what it should be. that error would only take like 1 point off right?</p>
<p>towerpumpkin, read what you just wrote, and you will realize that it would take longer to fall.</p>
<p>1.) haha, my % error on the solenoid was .798%.</p>
<p>2.) the saturn one was really easy, for me at least. It was the only one of the 3 mechanics questions that i fully completed. the mass was 5.6E24, right?</p>
<p>3.) im pretty sure it takes the ball longer to fall.</p>
<p>we went over all of the problems in class, the % error varies by graph, but we got around 5%. The true mass of saturn is around 6E26, and it definately takes longer to rise. Think about this, when the ball is going up, there are 2 forces creating a downward acceleration. When the ball is going down, 1 force is creating an upward acceleration (drag force) and another force is creating a downward acceleration (gravity). In addition, the ball can reach terminal velocity only when going down (when drag=gravity).</p>
<p>for the person who got 34% error...i think they forgot to divide by 3...it was in the equation like outside the slope that you graphed...that would have made the percent error small and like everyone elses. i thought the keplar's one was easy...like that one and miu/solenoid ones were my best friends. i messed up the inductor one though because i couldn't get straight when the current was max and when it was zero b/c i kept confusing it with RC circuits.</p>
<p>oh well. i think i got partial credit on most...i agree the mech rotation one was hard =/</p>
<p>The ball takes longer to fall since when it is falling the 2 forces are against each other. When the ball is rising, the friction force and gravitational force are in the same direction. Also the time it takes to reach the maximum height is shorter simply because the maximum height is lower due the friction force. However when the ball is falling it will fall slower because it reaches a terminal velocity</p>
<p>Yeah, I'm pretty positive that it takes longer to fall. lol... I was just afraid to post because I didn't take the class and you said so authoritatively that it takes longer to rise.</p>
<p>The problem with your reasoning (which megamike has already illustrated well - I'm just pointlessly adding my two cents) is that you assume that the fact that two forces are working "against" the velocity when the ball is going up means that it will take longer. However, the greater the force opposing the velocity, the faster the ball will reach its maximum height. In this case, the word "against" was just confusing.</p>
<p>ok, my AP physics teacher will be the ultimate authority on this:</p>
<p>"it takes longer to fall."</p>
<p>(opposing forces, instead of reinforcing forces, see above post)</p>
<p>but then it will fall that same max height in much shorter time....</p>
<p>Google it. It takes longer to rise; too bad I didn't think of any examples (the example I read used a whiffle ball - it takes much longer to come down to the forces opposing each other, and it's pretty obvious when you think about it) and put the wrong thing down.</p>
<p>I really thought Mech 2 was the easiest one, although I made plenty of stupid mistakes (like forgetting units and one of the points on the graph). Although I didn't really finish part A of the question, as I still had the mass of the moon in it (unless it's supposed to be in there). The first couple parts of Mech 3 was just screwing around with momentum and KE; I got 1/3 for the ratio, and didn't have time for the last part of 3.</p>
<p>edit: also, the mass of saturn is 5.7 x 10^26 kg. I got 5.8, so I don't know if the error was built in to the problem (doubt it) or I just punched in something wrong on the calculator (don't doubt it).</p>
<p>ohnoes, where did you find something online say that it takes longer to rise?</p>
<p>I've thought about it (instead of studying for macro/micro, which I still need to do), and I am pretty sure that it takes longer to fall. Consider a "proof" or sorts, using contradiction.</p>
<p>Assume that, at some height, the ball in the down trajectory is going faster than the ball did at the same height in the up trajectory.</p>
<p>Now, refer to a height that is smaller by a very small amount (I say "very small amount" to avoid the technicality of air resistance changing with changes in velocity, which doesn't affect the line of reasoning here) higher on both paths. Both balls will have less speed, because of the effect of gravity. </p>
<p>However, the ball going up will have speed at this new height that is <em>even less</em> than what could be accounted for by gravity, because of the air resistance. The ball going down will have speed that is <em>more</em> than what is accounted for by gravity, since by referring to a height that is further up you will be reversing the effects of air resistance (so, greater speed). As a result, if the speed of the falling ball is greater than the speed of the corresponding rising ball at any given height, the speed of the falling ball will have to be <em>even greater</em> than the speed of the corresponding rising ball at a higher altitude.</p>
<p>Basically, if the falling ball is faster than it was at the same height in its rising trajectory, it mathematically forces itself to also be faster at any higher height than it was in its rising trajectory at the same height (pardon the poor word choice).</p>
<p>But, if you continue with this, you'll reach a contradiction: at <em>every</em> height above the one initially assumed, the falling ball must have a higher speed than it did in the rising stage. However, at the top of its path, it can't have a higher speed, because the speed in both stages is zero (the stages merge). So, it would be impossible for the falling ball ever to have a higher speed in the first place.</p>
<p>If it never has a higher speed, it can't fall faster, and there we go.</p>
<p>I know that I phrased this terribly, but I'm pretty sure that it's the right logic...</p>
<p>um, randomperson, you are, um, wrong....didnt the question ask for justification? Clearly it was not seeking an answer requiring such an explanation as you just gave; proof by contradiction is not part of the AP Physics C curriculum. This is more reasonable and, I believe, correct:
The ball falls faster on the way down bc Fg works to accelerate in downward direction, wheras this decelerates motion in the upward direction.</p>
<p>Haha, I meant to say it takes longer to go down and couldn't edit it. Sorry about that</p>
<p>Evidentally the answer is going down. Oh well...its only going to be a 2 or 3 pt part.</p>
<p>"only going to be a 2 or 3 pt part"</p>
<p>yes, but you also had to show it on your graph!</p>
<p>yeah, but you might be able to get partial credit for the left hand side</p>
<p>wow. im in normal physics and did not take the Physics AP, but we did the same exact moon problem you guys are having difficulty with first semester. thats what a lot of problems every week in Giancoli's book does to you.</p>
<p>
[quote]
e) the graph I got was a 'v' shape that hit the x axis at the midpoint and went back up.
[/quote]
</p>
<p>Isn't v negative when its coming back down because of direction? </p>
<p>And, I got about .8% error. I think 34% is WAY too large. What was the mass for Saturn? I got something like 5E26 kg.</p>
<p>yeah. I got like 5.26E26 too...im pretty sure. did linear regression.</p>
<p>too bad I failed everything else. forgot to cancel out the d on 3, and totally skipped the last part (the find x so that rod completely stops), as well as forgetting totally about g in my diff. equ. arg.</p>
<p>what did everyone get for their diffeq?</p>
<p>i got something like
dv/dt = g/(1 + k/m), i dont remember exactly, might have been a minus sign.</p>