<p>Yeah, V = IR works only if there’s a resistor between the two points.
If there’s a capacitor, use V = Q/C.
If you’re told how much power something dissipates inbetween the two points, use V = P/I.
For inductors, its -L(dI/dt).</p>
<p>Ok, does anyone know in what situation to use the E = rho(J) equation?
[where E = electric field, rho = resistivity and J = current density]
I honestly don’t remember doing any problem like that.</p>
<p>Any suggestion on how to review for the Mechanic part ? I am taking on Mechanic this year.</p>
<p>Also, in drag force problem, when they ask for terminal speed, how come the net force at that moment is 0 ??</p>
<p>From the review questions I have seen about Drag Forces, they are usually equal to a constant b times velocity, or some other form.</p>
<p>Terminal velocity occurs when the Drag force equals the Force due to gravity (or when it equals the net force, for example on an incline plane when Ffriction kinetc + Fdrag = Fg (the component parallel to the incline, mgsintheta).</p>
<p>So pretty much Drag force is proportional to velocity, so after a certain amount of time an object in free fall will be accelerated to a velocity such that bv = mg. At that point, there is no more acceleration and the object will continue at velocity v as their is no net force acting on it.</p>
<p>I have a question about Potentials differences:</p>
<p>If there is a sphere with radius R1 and charge +Q and a spherical shell of inner radius R2 and outer Radius R3 with charge +2Q, I can easily determine the Electric Fields using Gauss’s Law, and the charge on the inner surface of the spherical shell is -Q and on the outer it is 3Q. but what I have trouble with is determining potential.</p>
<p>So on the outside of the sphere the Electric Field is 3Q/4pi(e0)r^2 so the potential is the - integral of Edr. And V = 3Q/4pi(e0)r</p>
<p>How would I go about finding the potential on the inner surface of the spherical shell and on the surface of the inner sphere?</p>
<p>Do i use Vba = V(b) - V(a) to find the difference between the potential differences? do I use dV = dQ/4pi(e0)r, do I use V = the - integral of E dr? and if so with what limits.</p>
<p>Also if there is a semi circle with uniform charge and you are trying to find the potential at its center do you use dQ = lamba r dtheta? </p>
<p>Any help would be greatly appreciated. I learned all this stuff and just didn’t retain the potential stuff.</p>
<p>I see what you say, thanks. </p>
<p>In spring problems, is the max kinetic when the block passes the equilibrium point ? Where does elastic potential=kinetic ? Is it half of maximum displacement ?</p>
<p>Any suggestions on how to review and strategy to attack problems ? I am doing a lot of FRs right now and feel so uncertain about what approach to use.</p>
<p>Has anyone done the practice tests in the PR 2009 edition for Mechanics? For some reason, they seem unusually difficult (or at least much more intricate). Anyone else?</p>
<p>The PR ones do seem quite difficult. When I took the BC test last year the test was way easier than any test problems. With AP Physics C I expect them to be somewhat easier but not to the degree of BC calc.</p>
<p>As for the spring problem when U=(1/2)kx^2, then elastic potential equals kinetic when U=KE=(1/2)Umax.</p>
<p>U max is when X = A so (1/2)Umax = 1/4 kA^2 so at A/sqrt(2) or sqrt(2)/2 times displacement is when potential equals kinetic energy. (correct me if i am wrong though).</p>
<p>At equilibrium elastic potential = 0 and since there is no non conservative forces (assuming this is a frictionless problem) KE is equal to Umax and velocity is maximized.</p>
<p>sry to ask this again, but does anyone know the curves (general) for both Mech. and E & M for a score of 5? i took a practice test in MEch. and I got around 19/35 RAWWW, how do i improve this? can i still get a 5? plus, i have the bio test in the morning too =( .■■■</p>
<p>Phew! For a while, I kept thinking that I was an idiot for not getting those in about a minute which is how fast you need to work.</p>
<p>Also, just looked at the sample questions on CB and they seemed much more straight forward and <em>doable</em> in the allotted time. You think those are more realistic for the exam?</p>
<p>Defintely PR took it up several notches in their practice tests for Mechanics this year. Some of the questions are near impossible to solve in the allotted time limit, and its unrealistic to judge your performance against those tests. Unless of course you’re scoring perfect and finishing in 20 minutes, in which case you might pull off a 90/90 on the real test.</p>
<p>Question:</p>
<p>A ball is thrown straight up from a point 2 m above the ground. The ball reaches a maximum height of 3 m above its starting point and then falls 5 m to the ground. When the ball strikes the ground, what is its displacement from its starting point?</p>
<p>A) Zero
B) 8 m below
C) 5 m below
D) 2 m below
E) 3 m below</p>
<p>I couldnt get this, help??</p>
<p>I think its around 60 Mech 55 E&M, I think if you keep studying and can bring that up to at least a 28-30/35 you should be fine. </p>
<p>I am fairly confident about getting a 5. The only mistakes I make in mechanics are ones that are simple mistakes, that I notice (like having Work be the incorrect sign) and now I know them. </p>
<p>Keep doing practice problems. Free Response questions are important, if you don’t understand a problem, you should know that each part are worth somewhat equal points. So if you can’t figure out the first part don’t give up on the whole problem. Even if they are asking for specific answers giving them the formulas and solving it symbolically will give you a better chance at getting much needed points. </p>
<p>I think if you get 49 points on E&M and 55 on Mechanics (out of 90) you should be in the 5 range. Remember that the MC score is multiplied by 45/35 so every 4 problems you get wrong isn’t just a point off its a little bit more.</p>
<p>2m below. The ball was released 2m above the ground, so the ground is 2m below the starting point.</p>
<p>The potential on the inside of a shell is constant and equal to the potential on the surface because the electric field inside a shell is 0.</p>
<p>You just add up the potential differences from all the shells.</p>
<p>For a thick shell, you integrate kdQ/R from the inner radius to the outer radius.</p>
<p>Hmmm here’s a problem for you guys to try, I want to check my answers (haven’t done rotation stuff since first semester lol)</p>
<p>A student holds one end of a thread, which is wrapped around a cylindrical spool. The student then drops the spool from a height h above the floor, and the thread unwinds as it falls. The spool has a mass M and a radius R, with moment of inertia I = 1/2 M R^2.
(a) Find the linear acceleration of the spool.
(b) Calculate the angular velocity as the spool hits the ground.
(c) Assume the spool doesn’t bounce and rolls with slipping on a ground with coefficient friction u, find the angular velocity and linear velocity as functions of time. The initial linear velocity is 0.
(d) If t=0 at the moment the spool strikes the ground, when does the spool stop slipping?</p>
<p>Yeah with that kind of problem always think, is there an x-component, no there isn’t, is there a y-component, obviously. What are you solving for (well think this first lol), displacement that is a vector quantity, don’t get it mixed with distance.</p>
<p>Then just think displacement is the change in position.</p>
<p>Try to come up with your own method for each type of problem.</p>
<p>@feuxfollets The potential on the inside of the shell is equal to the potential of what surface? the outer surface?</p>
<p>and you add up the potential differences from all the shells to get what? (I can’t tell if you are referring to my post in terms of the questions I asked, if so I will have to take a look at my question).</p>
<p>The problem had an inner sphere surrounded by a shell with thickness r3 - r2.</p>
<p>
</p>
<p>Yes, thats how you get potential from the E-field.
To solve for the potential on the surface of the sphere, all you do is integrate is E . dr, but in your electric field equation, replace the ‘r’ with the given radius of the sphere (say that ‘a’ is the radius of the inner sphere and ‘b’ is the radius of the outer sphere; since you’re on the surface, the Gaussian surface is your actual radius).</p>
<p>And yep, thats how you find it between the spheres. Your limits of integration would be ‘a’ to ‘b’, in this case.
Yeah, thats also how you’d do it for the semi-circle with linear charge density. Just don’t mix up point charge physics and continuous charge distribution physics.</p>
<p>oh woow, thx for that solution. i guess i was thinking too hardd!</p>
<p>So the potential on the inner surface of the shell is v= - integral of E dr from R2 to R3?</p>
<p>so that would be 3Q/4pi(e0) times (1/r3 - 1/r2) ?</p>
<p>The question asks for the potential on the outside of the outer spherical shell the inside of the outer spherical shell and the outer edge of the inner sphere.</p>
<p>For the inner sphere would the stuff on the outside affect the potential difference? or would it not since the Electric Field is determined by the enclosed charge?</p>
<p>@Ramblinman
For a thick shell you treat it as an infinite number of thin shells and integrate to find the total potential. Integral of kdQ/R from the inner radius to the outer radius</p>
<p>how bout this question: btw, all these questions are from an actual COllege board Physics Mechanics AP MC exam that my teacher gave out to us!</p>
<p>A toy spacecraft is launched directly upward. When the toy reaches its highest point, a spring is released and the toy splits into two parts with masses of 0.02 kg and 0.08 kg respectively. Immediately after the seperation, the 0.02 kg part moves horizontally due east. Air resistance is negligible. True statements about the 0.08 kg part include which of the following?</p>
<p>I. It could move north immediately after the spring is released.
II. It takes longer to reach the ground than does the 0.02 kg part.
III. It strikes the ground farther from the launch point than does the 0.02 kg part.</p>
<p>A) None
B) I only
C) III only
D) I and II only
E) II and III only</p>
<p>I couldnt get this one either, help?</p>
<p>Well that is a conservation of Horizontal Momentum I think so I think it is none since its velocity would be less than the other, it can’t go up, and they take the same time</p>
<p>@ feuxfollets</p>
<p>so the potential on the outer spherical shell doesn’t have anything to do with the inner one. And are there different potentials for the inside and outside of the shell.</p>
<p>so if the outerspherical shell has net charge 2Q and I have determined the Electric Field outside the potential from inside to outside is determined by V= integral of kdQ/r from r2 to r3 and how would I split up dQ 2Qdr? then the potential on the inner side of the Spherical Shell (when I say shell I am talking about the surface that is surrounding the sphere like it would in a spherical capacitor) is the potential derived from V = - integral of E dr minus the difference in potential gotten from kdQ/r.</p>
<p>Thanks for your help, this is the one thing I don’t understand (I am used to double integrals and I hate using this integral building stuff).</p>