<p>Answers!!!
Now that the 2 day period is over…
1 a) Impulse is force times time… J = F<em>t. t = J/F.
b) Impulse is change in momentum… J = m delta v. v initial = 0. J = m</em>v. m = J/v
c) W done in the neg. x direction. = delta KE. = 1/2<em>m</em>v^2 - 0.m = J/v; W = (1/2)<em>J</em>v
d) W = force * distance… W = Fb<em>d. Fb = (J</em>v)/(2d)</p>
<p>e) [Not 100% sure, didn’t have time to do in exam] Use cons. of energy minus work done by friction and the force. </p>
<ol>
<li><p>a) Fn perpendicular, mg straight down
b) centripetal = Fn perp – mg sin(theta) : note that theta was measured at the upper horizontal instead of the lower horizontal
c) Conservation of energy: mgh = mg(7R/4) = (1/2)<em>m</em>v^2. 7Rg/2 = v^2. V = sqrt(7Rg/2)
d) W = F<em>d = delta-KE = mg(7R/4) = mu</em>m<em>g</em>3R. = 7/12.
e) i. v’(t) = -(k/m)v.
ii. lambda = -k/m. v = e^[(-k/m)*t]
iii. y-intercept @ y = k/m.</p></li>
<li><p>a) torque = I<em>alpha. Alpha = 2nd derivative of theta. Theta’’ =( -b/I)</em>theta.
b) Using the analogy. Theta = A<em>sin(omega</em>t+phi). Phi is obviously the initial theta. Knowing that theta = theta-0 at t=0:::::::: A<em>sin(theta-0) -- By law of small numbers—> A</em>theta-0 = theta-0. Therefore, A =1.
Alpha is the second derivative of theta. Igitur, alpha = omega^2<em>sin(……). Omega^2 =B/I. Omega = sqrt(B/I). T = 2</em>pi/omega = 2<em>pi</em>sqrt(I/B)
c) Graphed
d) Eyeballed as T^2 = 150<em>I + 2.
e) t^2 = 4</em>pi^2<em>I/B. 4</em>pi^2/B = 150. B = 0.263
f) The intercept is the period where there is no rotational inertia of the pendulum, that is to say, there is nothing ON the torsional pendulum. The intercept is the period of the little rod alone.</p></li>
</ol>
<p>^^^
I intentionally set it so mine went through the origin…
Although in reflection that might be wrong.
but it made sense for f, because something that has no resistance to being turned will return to it’s starting point immediately.</p>
<p>I think that played a large role in it. MC was killer, and when you open up the FR and see part (e) (aka find __ in terms of 999 other variables), you definitely freak out.</p>
<p>That being said, it was still collectively harder than any past FR’s I’ve done.</p>
<p>^No. I drew one, but I’m a stupid idiot. I didn’t realize until last night that there wasn’t one and that I probably got 0/15 on the entire #2 FRQ.</p>
<p>Yeah I drew a centripetal force as well…but as long as we still have normal and gravity correct (hopefully one point per force…) I doubt they would dock points, and if anything, just 1 point. And unless I am mistaken, don’t the graders still give you credit if an answer is wrong because of a previous wrong answer but the methodology was still correct (i.e. using the forces in the diagram for part b).</p>
<p>^There really isn’t a centripetal force. It’s just the name for any force(s) pulling an object towards the center of a circle. So in this case, the normal force is the centripetal force, AFAIK.</p>
<p>It is always -1 point if you have any extra forces.
And yeah, they give you some leeway, but it has to make some sense. you can’t just make up a non-existent force that magically makes subsequent problems ridiculously simple.</p>
<p>Copied from another thread are my E&M answers: </p>
<p>Anyway, my answers. Some are definitely wrong:</p>
<p>E&M. 1.</p>
<p>(a) Surface is sphere with r<R and goes inside shell. Basically, the surface encloses no charge whatsoever. By Gauss’ Law, if no charge is enclosed, electric field must be 0.
(b) Yes. While the charge density on the outside of the sphere is not uniform, the Gaussian surface still resides within the shell, in which no charge is enclosed, so E is still 0.
(c) I said none, since I reasoned that no matter what, it would be impossible for all field lines to be parallel to the face. But I agree that it’s all faces with A
(d) Corner F
(e) E=(kQ)/(3L^2)
(f) I did some really random stuff, integrating E dA. I said A=L^2, so dA=2LdL. I subbed that in, said that E=(kQ/8L^2) and I integrated from L to L*sqrt(3). Definitely got something weird. Hopefully I get points.</p>
<p>E&M. 2.</p>
<p>(a) i. Q=CV. Make sure units of capacitance are farads and not mF.
ii. Q=0 from t=0 to t=t<em>1. Increases exponentially (concave down) with a horizontal asymptote of CV (or equivalent number)
iii. I=0 from t=0 to t=t</em>1. I jumps up to V/R (or equivalent number) at t=t_1 and decreases exponentially (concave up) with horz. asy. of I=0.
(b) i. U=.5QV. Modify so that you have Q and C and solve (I may have plugged in 9 for V without thinking…ugh)
ii. Conservation of energy. The energy from part (b)i equals .5LI^2, with L being the inductance. Solve for I (I got .297 A).
iii. Use V=Q/C to solve for voltage. V=-L(dI/dt). Solve for dI/dt.</p>
<p>E&M. 3.</p>
<p>(permittivity constant/2pi)=(2E-7)</p>
<p>(a) i. B=0, since dl is zero since no current flows in that section.
ii. B=(2E-7)(I)(r^2/b^2)
iii. B=(2E-7)(I/4b)
(b) I put that it went horizontally to the right, though it should be pointing right and down.
(c) No force. No electric field is present, so no electric force. For magnetic force to be felt, electron must be moving (F=qvB). However, v=0, so magnetic force=0.
(d) i. Graph
ii. Plug in pi, b, and I values from directions. Get B/r from the slope of line. Solve for permittivity constant (I got something like 1.46E-7; it was very close to the real value).</p>
<p>Ologos you really need to shut up. Youre not the only person who knows upper-level math, and while Im sure discussing how easy this test was and how easy the method of eigenvalues (stupid phrasing BTW) may make you feel self-important, you come off as a complete idiot to anyone who (shock!) may know as much if not more than you about physics/math.</p>
<p>For those who are interested
1E.
Work (scenario A) = Work (scenario B)
Fb<em>d = Fb</em>dn + fT*D
dn = d (fT/Fb) D</p>
<p>1F.
Work (scenario A) = change in KE
Fb d = 0.5m(vx)^2
Fb = m(vx)^2/(2d)</p>
<p>momentum is conserved just before/after the collision (instantaneous)
mvx = (M + m)v where v is the speed of the block/bullet
v = m/(M+m) vx</p>
<p>finally, the work/energy theorem can be used from when the bullet hits to when it stops after dn.
Ei + W = Ef
0.5m(vx)^2 Fbdn = 0.5(M+m)v^2
plugging in from above and simplifying, the answer should be:
dn = (1 m/(M+m))d</p>