<p>You could solve for Acceleration of the first block, and the acceleration for the second block would be the same, so you could find it that way.</p>
<p>Rwchong I agree with you on E&M 3 D and E</p>
<p>I think it stayed the same at first and then got dimmer…</p>
<p>Also for 1c how is there charge on the surface of the sphere?
Its just a cloud of charge…not a conducting sphere</p>
<p>For those using conservation of energy, you must consider the energy of the system.</p>
<p>Take the reference line / datum to be the tabletop and the positive direction to be down.</p>
<p>By conservation of energy,
1/2 Mv^2 = (M/2)gd
v^2 = gd
v = sqrt(gd)</p>
<p>Why the first equation? Only the lowering mass exhibits a change in gravitational potential hence (M/2)gd. However, the entire system (with the mass on the tabletop) exhibits a change in speed and therefore, kinetic energy, hence 1/2 Mv^2.</p>
<p>Why not
1/2 (M/2)v^2 = (M/2)gd? In that case, we are considering the system to be the lowering block only. However, work has been done by the system on the mass on the table top because it has moved from rest (work-energy theorem), so really, the equation would be
1/2 (M/2)v^2 = (M/2)gd - W
1/2 (M/2)v^2 + 1/2 (M/2)v^2 = (M/2)gd because W = 1/2 (M/2)v^2 (∆K of tabletop mass).</p>
<p>Hope this clears things up for those using energy.</p>
<p>For those using force, tension must be considered and it exists independent of friction. Tension is the force responsible for accelerating the mass on the tabletop. As kmoore notes, it does not need to be calculated, simply considered. The final bit is that each block’s acceleration is equal.</p>
<p>For Mech 3 (?–the one with the blocks and the rope)
I got v=sqrt(2gd) and v=sqrt(2gL)…
I used v^2=2ad for both. And if you do it this way, it agrees with the energy people’s 1/2mv^2=mgh => v=sqrt(2gh).
Bah, I never know when and when not to do energy.
And I think the tension force in the string (on the table, masses) is equal to the force of gravity on the hanging block, and the frictionless (read: nonexistent) conditions make it so that the acceleration of the whole system is g.</p>
<p>For E+M 1 (a) + (c): The negative derivative of the equation for the potential, r<r, is=“” negative,=“” so=“” the=“” field=“” here=“” radially=“” inward.=“” for=“” r=“”>R, the negative derivative of the potential equation is positive, so the field is radially outward. A radially inward field signifies a net negative charge distribution, so there must be a net negative charge for r<r. but=“” a=“” radially=“” outward=“” field=“” signifies=“” net=“” positive=“” charge=“” distribution,=“” so=“” there=“” must=“” be=“” for=“” r=“”>r. Therefore, in (c) there must be charge on the surface of the sphere because there must be enough positive charge here to equal and then overcome the net negative charge on the interior of the distribution, and in doing so produce an outward electric field for R>r.</r.></r,></p>
<p>For E+M (e): Though we would describe the current as going clockwise here, the actual electrons are moving counterclockwise and therefore from right to left within the bar. So if we use the right hand rule for v, pointing left, and B, pointing into the page, the resultant vector points upward. But since electrons are negatively charged they will move opposite this direction, and there will be an accumulation of negative charge on the bottom of the bar. With a relative positive charge at the top of the bar, the fact that electric fields point from positive to negative charges tells us that the resultant electric field is downward within the bar.</p>
<p>Ok, well I decided to take a break from the physics free response and come back. FRQ #3 a for Mech is definitely sqrt(gh). Want to know why? Conservation of mechanical energy. (Kinematics sucks because you have to do twice as many calculations, plus it only works for constant accelerations. Lucky you had one this time.) So right…to begin with the energy.
(M/2)<em>g</em>h=(1/2<em>(M/2)</em>v^2)+(1/2<em>(M/2)</em>v^2)
=> (M/2)<em>g</em>h=(1/4<em>M</em>v^2)+(1/4<em>M</em>v^2)
=> (M/2)<em>g</em>h=1/2<em>M</em>v^2
=> Mgh=Mv^2
=> v^2=gh
=> v=sqrt(gh).
I was wrong…forgot to account for the energy in the other block. Final speeds are the same due to my realization of this.</p>
<p>P.S. Just read orangebase’s explanation. He’s absolutely right. I forgot to consider the kinetic energy of the 2nd block in my rush. Definitely didn’t get a 5. Kinematics still suck though. :p</p>
<p>nowell91: doesn’t the right hand rule show that the electrons are forced to the top, not the bottom?</p>
<p>bco09: Well if the moving charges were protons, then the right hand rule would show that they were moved to the top, but since they are electrons and negatively charged they move opposite this direction, towards the bottom. For negatively charged particles you can actually use the left-hand rule with F=qv x B, and then your thumb points in the right direction.</p>
<p>hmm… i guess the difference is in how we do the right hand rule. Here is what I did:</p>
<p>-put right hand in front of you
-turn your hand to point your four fingers in the direction that the electrons are moving (from b to a)
-flip your hand such that your fingers will curl in the direction of the B field. (Since the B-field goes into the page, your hand will be flipped such that you see the back of it, not the palm).
-Your thumb will now extend to point in the direction of the force on a positive charge. In this case, your thumb is pointing down. Since we are talking about electrons, we can conclude that the force on the electrons will be up.</p>
<p>this could very well be the wrong way to do the right-hand rule, but it is what i used.</p>
<p>Because you chose the direction the the electrons are flowing, the downward direction is where the electrons will move.</p>
<p>Here’s what I did:</p>
<p>-place your index finger in the direction of conventional current, from a to b
-point your other fingers in the direction of the magnetic field, into the page
-extend thumb perpendicular to both I and B and this is the direction of force on positive charge, meaning that electrons go to the bottom.</p>
<p>I’m not sure if this helps, but here’s a nice explanation of the right hand rule for Electricity and Magnetism. [Electromagnetic</a> Fields](<a href=“http://www.ndt-ed.org/EducationResources/CommunityCollege/MagParticle/Physics/ElectromagneticFields.htm]Electromagnetic”>http://www.ndt-ed.org/EducationResources/CommunityCollege/MagParticle/Physics/ElectromagneticFields.htm)</p>
<p>the electrons are moving from right to left though, (b to a)… so shouldn’t you point your index finger in that direction initially?</p>
<p>It doesn’t matter, but if you do that, then you have already accounted for the negative charge and there is no need to flip your hand.</p>
<p>umm… no pointing from b to a accounts for the charged particle’s velocity being in the direction b to a. flipping my hand later is what accounts for it being negative</p>
<p>Actually, you know what I think you’re right bco09. I forgot to envision the field vector as if it were tail to tail with the velocity one, in which case it would produce the opposite effect than what is true. So i think you’ve shown me a mistake I made.</p>
<p>I don’t know, we will see once they post the scoring guidelines on collegboard.</p>
<p>When do they do that?
How long does it take for them to post the scoring guidelines?</p>
<p>^ Hopefully soon =)</p>
<p>What did you guys get for 2 (a) on C:Mech?</p>
<p>^ scoring guidelines are posted after July</p>
<p>what you get for Mech 1 e)? Somehow my equation for position x vesus time t is very weird…it has x^3 and everything.</p>