<p>I was angry that the parallel axis theorm wasn't on the sheet.</p>
<p>S was thankful he took MV/DiffEq last year. Felt he did very well on both parts.</p>
<p>
[quote]
and, don't be hard on yourself the frq's were in fact "challenging"...i'm not going to say anything specific, but lets just say I had trouble finding "C--the integral constant" in the calculus questions
[/quote]
u don't have the constant of integration. just put the limits on the integral when u integrate both sides after separating the variables.</p>
<p>
[quote]
Omg, good thing i am taking multivar calc and diff eq this year! Those frq's (1o f them )were challenging if you didnt know advanced calculus.
[/quote]
not really... it was just separating the variables and u-substitution</p>
<p>I missed half the FRQ questions because I didn't have time.</p>
<p>And I was freaking how on how to solve a differential equation with mg sin theta in it ...</p>
<p>Oh J.C., I'm pretty sure I got a one on that. So did everyone else in my class.</p>
<p>The MC was pretty conceptual, which kinda sucked for me, cause I'm good with more mathematic MC's. The FRQ's were living hell: 1 I sucked up really badly, 2 I managed half of it before sucking up really badly, and 3, I got until the end when I sucked up really badly.</p>
<p>The reason could've been because I just finished the AP Bio/HELL exam 15 minutes prior to this one.</p>
<p>the component vector (ie mg something something) = constant. It also freaked me out</p>
<p>My prediction:
MC: Correct: 25; Wrong: 6
FR: #1: 8-12, #2: 9-12, #3: 8-12</p>
<p>I should get a 4 or 5 :)</p>
<p>^^^^^^^^^^ Those scores should easily give you a 5 by a good margin, if your predictions are correct.</p>
<p>For MC, I'm predicting 28 right and 4 wrong, . I guess that my extensive studying, going over all the Princeton Review MC questions at the ends of chapters helped me out there. That's translate to a score of about 35/45 on that half of the test.</p>
<p>For FR, the questions are worth 15 each, unless they changed how they've done the scoring in past exams.</p>
<p>I'm not going to mention specifics here, but people who took the mechanics should know what I'm talking about.</p>
<p>For one question, I'm hoping that I got 3/15. All I managed to do was get the free-body diagram for easy points. I blanked out and didn't even attempt the rest of it when they started asking me to derive equations.</p>
<p>I'm expecting 12/15 on the second problem and 8/15 on the last one. That gives me a total of 23/45 on free response. I can't explain why until this Free-Response embargo is lifted, but I think I'm pretty accurate there.</p>
<p>So, 35/45 on MC and 23/45 on Free-Response should technically get me a 58/90, which should be enough for the 5. But then again, maybe I'm being a bit generous with my scoring on MC, so, for the sake of being pessimistic and completely realistic, I'll say that I got a 55/90. Based on how people are responding to this test specifically, I'm thinking that the cut-off is going to be lower than the 55/90 reported in 1998 and hopefully more like the 49/90 in 1998.</p>
<p>I guess that I have a chance at the 5. I hope they mention the specific cut-offs before scores are mailed home, so I don't get a huge surprise and end up shooting myself when I open the envelope to a big fat 4.</p>
<p>Okay, 48 hours have passed, so we can now discuss the FRQs. </p>
<p>here is the first mechanics FRQ:
1. A skier of mass M is skiing down a frictionless hill that makes an angle θ with the horizontal, as shown in the diagram. The skier starts from rest at time t = 0 and is subject to a velocity-dependent drag force due to air resistance of the form F= -bv, where v is the velocity of the skier and b is a positive constant. Express all algebraic answers in terms of M, b, θ , and fundamental constants.
(a) On the dot below that represents the skier, draw a free-body diagram indicating and labeling all of the forces that act on the skier while the skier descends the hill.
(b) Write a differential equation that can be used to solve for the velocity of the skier as a function of time.
(c) Determine an expression for the terminal velocity V(sub T) of the skier.
(d) Solve the differential equation in part (b) to determine the velocity of the skier as a function of time, showing all your steps.</p>
<p>I have a couple of questions. For the free-body diagram, should the drag force point back horizontally or back up the slope? Also, what's the differential equation in part (b)? How do you solve it?</p>
<p>Here's what I got, but I think it's wrong.
(b) m(d^2x/dt^2)=mgsinθ-b(dx/dt)
(c) bv=mgsinθ ===> v=(mgsinθ)/b
(d) ma=mgsinθ-bv ===> mv=mgt<em>sinθ-bx ===> v=(mgt</em>sinθ -bx)/m</p>
<p>The horizontal uniform rod shown above has length 0.60 m and mass 2.0 kg. The left end of the rod is attached to a vertical support by a frictionless hinge that allows the rod to swing up or down. The right end of the rod is supported by a cord that makes an angle of 30 with the rod. A spring scale of negligible mass measures the tension in the cord. A 0.50 kg block is also attached to the right end of the rod.
(a) On the diagram below, draw and label vectors to represent all the forces acting on the rod. Show each force vector originating at its point of application.
(b) Calculate the reading on the spring scale.
(c) The rotational inertia of a rod about its center is (1/12)ML^2 , where M is the mass of the rod and L is its length. Calculate the rotational inertia of the rod-block system about the hinge.
(d) If the cord that supports the rod is cut near the end of the rod, calculate the initial angular acceleration of the rod-block system about the hinge.</p>
<p>Here's what I got:
(a) If I made a mistake, I think I made it here. I only drew vectors for F(sub T), W(sub rod), and W(sub block). Here's my question: Does the contact point at the hinge provide a force? If so, everything else I did was wrong.</p>
<p>(b) ma=0=F(sub T) - W(sub rod) - W(sub block) ===> F(sub T) = W(sub rod) + W(sub block) = 25N.</p>
<p>(c) I = I(sub rod) + I(sub block) = I(sub rod, center) + M(1/2L)^2 + M(sub block)L^2 = 1/3ML^2+M(sub block)L^2=.42 kg*m^2</p>
<p>(d) torque=I<em>alpha ===> alpha=torque/.42 kg</em>m^2
torque = L(F(sub T) - 1/2W(sub rod) - W(sub block))=(.6m)(25N - 10 N - 5 N) = 6 N*m ===> alpha = 6/.42=14.29rad/s^2</p>
<p>Here's the third FRQ:</p>
<p>In an experiment to determine the spring constant of an elastic cord of length 0.60 m, a student hangs the cord
from a rod as represented above and then attaches a variety of weights to the cord. For each weight, the student
allows the weight to hang in equilibrium and then measures the entire length of the cord. The data are recorded
in the table below:
Weight (N) 0 10 15 20 25
Length (m) 0.60 0.97 1.24 1.37 1.64</p>
<p>(a) Use the data to plot a graph of weight versus length on the axes below. Sketch a best-fit straight line through the data.
(b) Use the best-fit line you sketched in part (a) to determine an experimental value for the spring constant k of the cord.
The student now attaches an object of unknown mass m to the cord and holds the object adjacent to the point at
which the top of the cord is tied to the rod, as represented above. When the object is released from rest, it falls
1.5 m before stopping and turning around. Assume that air resistance is negligible.
(c) Calculate the value of the unknown mass m of the object.
(d) i. Calculate how far down the object has fallen at the moment it attains its maximum speed.
ii. Explain why this is the point at which the object has its maximum speed.
iii. Calculate the maximum speed of the object.</p>