<p>m1v1 + m2v2 = (m1+m2)vf => (3kg)(2m/s)-(5kg)(2m/s) = (8kg)vf => vf = -.5 m/s. Speed = |-.5| = .5 m/s. Answer: B?</p>
<p><em>EDIT</em> damn, dude I need to study up on rotational</p>
<p>I had this on another thread, but in case you missed it:</p>
<p>A charge -1 C is stuck at x=0. A charge +4 C is stuck at x=2 m. You need to place a positive charge Q somewhere along the x-axis so that it doesn’t move. No glue, staples, nails etc. are available.</p>
<p>It should be placed at:</p>
<p>a) x = 2/3 m
b) x = -1 m
c) x = -2 m
d) x = 5 m
e) x = -1/2 m</p>
<p>C, first notice that it must be placed on the negative x-axis. Now let x be the distance between the origin and the position of the particle on the negative x-axis.</p>
<p>Now the force on Q due to the -1C charge is denoted by kQ/x^2 and directed to the right. The force on Q due to the 4C charge is denoted by 4kQ/(x+2)^2 and is directed to the left. Fnet = 0, so equate the opposite forces, solving the quadratic yields x = 2, and thus Q’s position is -2.</p>
<p>does anyone know any good tricks to memorizing certain formulas, if so, please share
:)</p>
<p>vhoidz, I don’t know any specific tricks but dimensional analysis may often point you in the right direction when it comes to recalling formulas. </p>
<p>Anyways new question:</p>
<p>Suppose that an electron (charge -e) could orbit a proton (charge +e) in a circular orbit of constant radius R. Assuming that the proton is stationary and only electrostatic forces act on the particles, which of the following represents the kinetic energy of the two-particle system?</p>
<p>let k = 1/(4pi(e0))
a) k(e/R)
b) (k/2)(e^2/R)
c) -(k/2)(e^2/R)
d) k(e^2 / R^2)
e) -k(e^2 / R^2)</p>
<p>oops, i’m only taking the B exam, so that’d be why i answered C (that’s the only way i know how to do it lol)</p>
<p>No problems, I didn’t want to confuse anybody. Just ignore rotational KE for the Physics B test, they never test you on it to my knowledge.</p>
<p>Snipe: Is the answer b)?</p>
<p>yep, I’ll post some more later. Anyways, can someone explain the formulas relating magnetic flux, time-varying magnetic fields, induced emf, current, and area of loop? In particular I would like to know which quantities need to be integrated/differentiated to get the others.</p>
<p>The following problem is for Physics C students only:
Snipez, this question integrates (pardon the pun) all of the concepts you noted into one problem:</p>
<p>A wire has a time dependent current: I = 3t where t is measured in seconds. The current carrying wire is placed next to a square, conducting loop shown in the picture: <a href=“http://i30.■■■■■■■.com/2r7a749.jpg[/url]”>http://i30.■■■■■■■.com/2r7a749.jpg</a></p>
![http://i30.■■■■■■■.com/2r7a749.jpg[/url]](http://i30.tinypic.com/2r7a749.jpg%5B/url%5D){kind=link}
<p>The square loop has dimensions 3x5 meters as shown by the picture and the wire is 4 meters away from the loop.</p>
<p>Calculate the magnitude of the force on the wire at t = 20 seconds and give the direction.</p>
<p>This problem is very tedious and cannot be done in 1 step. If any of you need help, just ask me.</p>
<p>Yatta. </p>
<p>Can you explain how the answer to snipez’s question is B?</p>
<p>And also let me see if I can figure out the answer to your question.
At t = 20, the current is 60Amps. Right hand rule says magnetic field is directed into the page. So essentially a wire is placed in a changing magnetic field which is why there is a force on the wire? Now heres where i am confused. what equations do i use? F=BIL? or some variation of ampere’s/biot-savart law?</p>
<p>As you can see, i am not very good at E&M at all.</p>
<p>Snipez’s question asked what the kinetic energy of the system of an electron orbiting a proton was.</p>
<p>When anything is in orbit (let it be planets, electrons, etc…), there is some centripetal force acting on the orbiting object, in this case, the electrostatic force between the proton and the electron. Because F = k<em>q1q2/R^2 and both charges have a magnitude e, the electrostatic force is k</em>e^2/R^2</p>
<p>Now we set this force equal to the centripetal force for reasons explained above, so k<em>e^2/R^2 = m</em>v^2/R. where m = mass of the electron, v = velocity of the electron, and R = radius of orbit.</p>
<p>Note that KE = 1/2<em>m</em>v^2 so we need to manipulate the above equation to get the KE in terms of the given quantities.</p>
<p>Starting with k<em>e^2/R^2 = m</em>v^2/R, multiply both sides by R to get k<em>e^2/R = m</em>v^2. Now simply divide both sides by 2 to get k<em>e^2/2R = m</em>v^2/2, therefore the answer is k*e^2/(2R).</p>
<p>Now note one important thing that I did not include a negative sign with the force expression. I did this because the negative simply represents that the direction of the force on the electron is towards the center of orbit, which is the same direction of centripetal force.</p>
<p>As for the problem I gave, I made a slight mistake in the wording. The question should ask what the force on the loop is, not the force on the wire. Sorry for the confusion.</p>
<p>Heh, I was just about to ask you that Yatta. Hmm ok I just skimmed Halliday-Resnick and I know how to find the force on one of the four sides of the loop, say for instance, the top segment (3 m)… will this answer the question?</p>
<p>do we get an equation sheet for both MC and free response? i’m not talking about that s*itty table of symbols and constants, but the sheets with the lists of equations?</p>
<p>NVM answered my own question.</p>
<p>no, only for the free response.</p>
<p>I heard that the special focus for the exam this year is electrostatics. Anyone else hear this?</p>
<p>Keep scoring 3s on practice tests :(</p>
<p>^^ where did u hear that???</p>
<p>is the single slit interference/diffraction grating stuff on the test??
cuz princeton review only covers double slit</p>
<p>yatta, that was a great explaanation. ty so much. you are like a physics genius.</p>
<p>Here’s something to ponder: that orbiting electron from the problem is accelerating. An accelerating charge will radiate away energy. So, why doesn’t the electron fall into a death spiral toward the proton?</p>