Official 2011 AP Physics C Thread

Test Preparation AP Sciences
61

<p>^no, only teh constants sheet</p>

62

<p>Wonderer- refer to this page:<br>
<a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;

<p>It looks like you’re at a low 3. It all depends on your essays. </p>

<p>To whoever said that it looks like the MC is getting easier- well, the FRQs are getting exponentially more difficult (ever looked at 2004 and below?), so I think it’s fair if they are making the MC a bit easier.</p>

63

<p>Is there always a question when you have to create your own experiment? Or does it only happen often/sometimes?</p>

64

<p>Puggly - For whatever reason I did really well on that one. Usually I’m at -7 or 8. 24/35 still puts you well in the running for a 5.</p>

65

<p>Abrayo - I <em>think</em> there is a question on experimenting every time, but it might not always be about designing an experiment.</p>

66

<p>Help with E&M?

  1. <a href=“http://tinyimage.net/images/25790944428468568236.png[/url]”>http://tinyimage.net/images/25790944428468568236.png&lt;/a&gt;
    I don’t even… The answer is B… What should I review??</p>

<p>2)<a href=“http://tinyimage.net/images/81684598844478119651.png[/url]”>http://tinyimage.net/images/81684598844478119651.png&lt;/a&gt;
Why can’t I use the formula V^2/R for this one?? answer C btw ( I did get it right, but just because I didn’t see the answer choice I needed had I used V^2/R…)</p>

<p>3)<a href=“http://tinyimage.net/images/14929814429358178541.png[/url]”>http://tinyimage.net/images/14929814429358178541.png&lt;/a&gt;
How do I approach questions like 52? The answer is E… (I get 51 though)</p>

<p>4)<a href=“http://tinyimage.net/images/89661487184785817963.png[/url]”>http://tinyimage.net/images/89661487184785817963.png&lt;/a&gt;
Umm… I don’t remember learning this… What is this an application of?</p>

<p>5)<a href=“http://tinyimage.net/images/05849919813601766271.png[/url]”>http://tinyimage.net/images/05849919813601766271.png&lt;/a&gt;
57:B -> Why is R2 ignored? I put A originally…
58:A -> I thought there is no current after a significant amount of time b/c capacitor takes up all the charge? no?</p>

67

<p>Ahh, any predictions for the FR topics?
Is torque always on it? I fail at torques. . .</p>

68

<p>I think you posted the wrong one for number 2</p>

69

<p>I’ve had a bit of trouble understanding the concept of electrical potential as it relates to work.</p>

<p>So when a positive particle moves from a region that has a more negative potential to a region with a less negative potential (i.e., the potential difference is positive), it’s doing positive work, right? But then the electrical field is doing negative work on the particle?</p>

<p>Correct me if I’m wrong.</p>

70

<p>1998 #29) A particle moves in the xy-plane with the coordinates given by x=Asinwt and y=Asinwt, where A=1.5 meters and w=2.0 radians per second. What is the magnitude of the particles acceleration? A) 0 b) 1.3 c) 3 d) 4.5 e) 6</p>

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<p>^ are you sure that’s the right question? If you wrote it correctly, the acceleration would vary with time.</p>

72

<p>If you take the second derivative of the function, the magnitude of the acceleration ends up being Aw^2, which gives you E. Or something like that.</p>

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<p>^ I had trouble with that question as well. Taking the second derivatives didn’t really help me with that question. The only things that cancelled were -Aw^2, leaving the trig. </p>

<hr>

<p><a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;

<p>For 2.e), it says to use W = integral of F(x).
What’s wrong with using W=Fd? Everything was right, up until the (1/3) factor.
I thought I could do this and get: W = 1700(0.1)^2(0.1) = 1.7.</p>

74

<p>^ He must have meant for one of the sins to be “cos”, since otherwise the acceleration would vary between 0 and Aw^2. If this is the case, then the answer is simply Aw^2, or 6.</p>

<p>For 2e, it’s because the force varies with the distance, x. If this were not the case, you’d have a horizontal line on your graph. You could, however, take the area under the graph you plotted if you wish.</p>

75

<p>Could someone help me with these Mech questions?</p>

<p>[AP</a> Physics C Mechanics Practice Tests_°Ù¶ÈÎÄ¿â](<a href=“AP Physics C Mechanics Practice Tests - 百度文库”>AP Physics C Mechanics Practice Tests - 百度文库)</p>

<h1>34, doesn’t the escape velocity depend on the mass of the planet since Ve = sqrt(2GM/R)?</h1>

<p>why is the answer not B?</p>

<h1>32, I don’t know how to do this one.</h1>

<h1>28, if U =0.5kx^2, and k = F/x = 9, then 0.5<em>9</em>3^2 = 81/2? no? Why is the asnwer 81/4?</h1>

<ol>
<li>Again, I dont get this one</li>
</ol>

<p>25,26, not sure how angular acceleration varies with theta</p>

<p>Thanks!</p>

76

<p>Ahh I see.</p>

<p>Yes, one is supposed to be sin and one is supposed to be cos.
I still didn’t get an answer though.
The second derivative of both would be -Aw^2sin(wt) and -Aw^2cos(wt) or something (I forget exactly), and then dividing the d2y by d2x . . . but how would anything cancel out?</p>

77

<p>Abrayo, you know that SHM is related to Circular motion. Centripetal acceleration is v^2/r, and in this case, r is the amplitude, A. Vmax = Asqrt(k/m) (get this from 0.5mv^2 = 0.5kA^2, solve for V). </p>

<p>So, we get that a = v^2/r = A*(k/m)</p>

<p>You know that w= 2pi<em>f = 2pi</em>(1/2pi)*sqrt(k/m) = sqrt(k/m) and you already have A.</p>

<p>So k/m = w^2</p>

<p>and since you are looking for A(k/m), you need to do Aw^2 = acceleration, so the answer should be 6.</p>

<p>I think.</p>

78

<p>1984 # 19) A particle is moving in a circle of radius 2 meters according to the relation theta = 3t^2 +2t, where theta is measured in radians and t in seconds. The speed of the particle at t=4 seconds is a) 13 m/s b) 16 m/s c) 26 m/s d) 52 m/s </p>

<p>why is the answer d, i tried it and got c</p>

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<p>I got c as well.</p>

<p>@WongTongTong:
Too much work for one question! In my opinion.
Did it take you long to think about how to apply your knowledge?</p>

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<p>^No. It only took about a minute to do. I derived that last week so I didn’t have to think about it. The quickest way to do it is what Graviton said: a=Aw^2; I was simply explaining where that formula came from. </p>

<p>Also, I got D. You guys are forgetting to multiply by the radius. Remember, v=r*w. </p>

<p>Sent from my iPhone 4 using CC app</p>