<p>I just took that lol (btw what did you get for the FRQ #3? I just want to check my answers)</p>
<p>The other mass has to move west because of conservation of linear momentum.
Since the only velocity resulting from the explosion is horizontal, both fall at a=g m/s and hit the ground at the same time.
Since m1 * v1 = m2 * v2 and m1 < m2, v1 > v2 must be true. In which case the second block would fall shorter than the first.</p>
<p>@ Ramblin man, how did u know they took the same amount of time?</p>
<p>@ feuxfollets, i dont think i got the Free Response set from my teacher, sry =(</p>
<p>Well when there is no air resistance all objects with the same vertical velocity will hit at the same time no matter there mass since there acceleration is determined by gravity.</p>
<p>They all fall with equal acceleration, and both blocks have 0 y components.</p>
<p>I just want to say thank you to everyone who is helping me and others. I hope everyone does well. I still have one more practice test from this other book called physics C by Mooney. I have taken the Barrons ones and have done pretty well on them.</p>
<p>Also the potential difference question is from the 5 steps to a 5 book if that helps. It is a Free response. I am going to study Potentials a lot today and hopefully will do well.</p>
<p>oh, ok i got it now…its these kinds of mistakes that give me no hope in getting a 5 =(</p>
<p>Well like I said with any kinematics problem, look at what you are solving for. See if it has y components or x components. then split everything up to x’s and y’s.</p>
<p>if there is a collision or explosion, think conservation of momentum. (must not be a net force, like there is no net horizontal force in the example you gave).</p>
<p>So always split it up into the basics, look at the question, then say well this is symmetric, or this doesn’t move so it is only vertical, etc…</p>
<p>Edit: Does anyone know if we need anything other than pencils, rulers, Graphing Calculators? Do we need pens for the FRQ’s?</p>
<p>allright, gotcha</p>
<p>I posted that question in this thread, 3rd on this page ^_^</p>
<p>I think the problem with potential I am having is somewhat conceptual.</p>
<p>Potential is the change in Potential energy divided by q right? So when calculating it for the sphere situation, do you calculate potential from the center of the sphere, from infinity? and if so what does the potential of the outer surface is? Is it in relation to anything else?</p>
<p>I think I just need to remember Electric Field is the negative change in potential and that potential. I understand what potential between two spots means, but what does the potential of one spot refer to.</p>
<p>another question:</p>
<p>A specially designed spring is stretched from equilibrium to the distances x given blow, and the restoring force F is measured in each case. What is the potential energy of the spring when it is stretched 3 m from equilibrium?</p>
<p>Values:</p>
<p>X(m) F(N)
0 0
1 1
2 8
3 27
4 64</p>
<p>A) 9/2 Joules
B) 9 Joules
C) 81/4 Joules
D) 27 Joules
E) 81/2 Joules</p>
<p>help?</p>
<p>btw, i answered E, which was wrong</p>
<p>another question:</p>
<p>An ideal spring with spring constant K is cut in half. What is the spring constant of either one of the two springs?</p>
<p>A) k/2
B) K^(1/2)
C) K
D) k^(2)
E) 2k</p>
<p>shouldnt the spring constant remain the same? i put C but it was wrong, why? and wat is the right answer? thnx</p>
<p>Well at 3 meters the force is 27 N. It doesn’t say it is in Simpe Harmonic Motion so it seems that the force = x^x or something like that, just note that k = 1 where k = 1 so U=(1/2)(1)(9) so A.</p>
<p>Im not sure I am doing this right, but it does not say SHM.</p>
<p>Ramblinman, the correct answer is C, but i dont know how they get that?</p>
<p>no remember two springs with the same spring constant are equivalent to k1k2/k1 +k2. So together they would equal k^2/2 so if the original is K = k^2/2 then k = sqrt (2k)</p>
<p>wait that isn’t an answer. Hm. I am not sure anymore, it makes sense that it would stay the same but it also doesn’t.</p>
<p>oops for the other one I change my answer from x^3 to X^x for no apparent reason if F = x^3 then U= - integral of F dx so U = x^4/4 for x = 3 thats 81/4</p>
<p>for the spring question, the answer is E. any clue on how they got that?</p>
<p>Yeah, it does NOT say SHM…which means you can’t use that, can you?
If you look at the values given, you can tell that the function is something like F = -kx^3.
Now, your potential energy function would be the integral of that force, i.e., -k(x^4)/4.
Now plug in 3m, and you get: -k(81/4).</p>
<p>Is it C?
Edit: Oh yes it is C. Nice lol.</p>
<p>Axeback your right I was doing something else when I did that and totally messed it up It works with integrating the force though.</p>
<p>I wrote down the first part then totally did something else for the last part.</p>
<p>I messed up again wow. its k1k2/K1 +k2 so thats k/2 = k. so the new ones are 2k? maybe, don’t quote me on that lol.</p>
<p>for the spring stuff it now seems to me like a better idea to always use u = - integral of F. (as long as you note that the negative is referring to how its a restoring force) as it won’t mess me up like the 1/2 kx^2 did.</p>
<p>So it is C, right? yea and that method works well. lol cool.</p>
<p>
</p>
<p>Lol, I totally forgot how to do part a). Do you have to take into account the rotational motion to find angular acceleration, and then convert that to linear acceleration?
I know how to get linear and angular velocity, but not the acceleration.
This was more than 6 months ago! I forgot everything!!1</p>
<p>Btw, all parts can be solved using conservation of energy, right? Or do you need conservation of angular momentum in there?</p>
<p>I’ve just been looking over the PR. I’m not too worried / pretty sure I’ll get a 5. JW, how much gravitational force type stuff will be on the AP?</p>
<p>The MC questions I’ve seen so far on orbits have mainly been like kepler’s 3rd law plug and chug and conservation of angular momentum type stuff. Is there anything else I’m missing?
o.o.</p>
<p>oh, and of course potential energy / calculations of escape velocity.</p>
<p>Free response Gravitation is what I’m most worried about. Likely hood of one showing up?</p>