<p>Yeah, I made a similar one the other day. I wrote 6/4 = 1.6666 and was wondering how I got the question wrong.</p>
<p>Does anyone have any suggestions to work complex spring problems? For example, I saw one problem where a spring had two blocks connected to it, and the second block was wedged up against a wall. The first block was used to compress the spring, and then the second block left the wall and the spring was in SHM. So confusing…</p>
<p>
</p>
<p>11 & 12:</p>
<p>Friction opposes motion, so Figure A must be wrong because the box is not going up. Figure B has an incorrectly drawn normal force so it’s wrong. Figure E shows the thing accelerating up the wrong, so we don’t want it. Figure D is incorrectly drawn because either you show mg or it’s components, but not both. That’s why Figure C is right for both.</p>
<p>14: I didn’t know either. </p>
<p>17: A-D are wrong because they imply kinetic energy is conserved, which is clearly not the case in inelastic collisions.</p>
<p>19: Never learned this as well, but I think it’s an important one to know. Going to review that right now. </p>
<p>32: A&B are wrong because he’s in a lab so that’s impossible. C is nonsensical because air resistance decreases the downward acceleration. E is wrong because shape of the object is irrelevant - we care about mass. D is correct because air resistance makes the acceleration nonuniform. </p>
<p>33: You have to understand the concept of projectile motion. Although a zero theta maximizes the horizontal component, it also reduces the time of motion. The horizontal component decreases as theta increases, but the time of motion increases. Some value between 0 and 45 will produce the biggest product between time and horizontal component, creating the maximum value. </p>
<p>34: The rock enters uniform circular motion. Acceleration is via centripetal force, which is directed inward, so directly vertical in this case.</p>
<ol>
<li>How much net work must be done by an external force to move a -1 μC point charge from rest at point C to rest at point E ?
(A) -20 μJ (B) -10 μJ (C) 10 μJ (D) 20 μJ (E) 30 μJ</li>
</ol>
<p>Why is it B and not C?</p>
<p>-W = qV</p>
<p>-W(done from C > E) = (-1 μC)(Ve-Vc)</p>
<p>???</p>
<p>EDIT: it’s because it’s asking for the work done by the external force right? gah I need to learn how to read</p>
<p>We can use our calculator on free response to calculate a definite integral, right?</p>
<p>^ Sure, but I would set up the integral on paper and write the integrated form so they know you know.</p>
<p>Okay, good that’s what I thought. It looked like that’s what they wanted on one of their FRQs, but I just wanted to make sure.</p>
<p>Also, do you know if they would ask us to calculate the moment of inertia for an object (not point-masses, but something like a rod with uniform weight distribution) or the center of mass of something like an L-shape?</p>
<p>i think so because i saw them going over the integrals in my book…</p>
<p>The equation of motion of a simple harmonic oscillator is given by d^2x/dt^2 = -9x, where x is the displacement and t is the time. What is the period?
A) 6pi
B) 9/2pi
C) 3/2pi
D) 2pi/3
E) 2pi/9</p>
<p>EDIT: Nevermind.</p>
<p>a = -9x = -w^2*x, so w=3</p>
<p>T = 2pi/w = 2pi/3</p>
<p>Answer is (D)</p>
<p>I’m hoping for a very limited amount of SHM in the test tomorrow. I used to think I understood it well, but now I’m just completely confused! (Please CollegeBoard, no more than 1 free response with a spring!)</p>
<p>Unfortunately E is not the answer.
I would think it involves Calculus, but it could not.</p>
<p>since is SHM: x=Acos(wt) (omega is w)
so d^2/dt^2=-Aw^2cos(wt) = -w^2(x)
so 9=w^2
w=3
so T=2pi/w
so T=2pi/3</p>
<p>Could someone explain to me a few things:</p>
<p>1) Why is U_gravity negative (-Gmm/r), and what does a negative potential energy mean as opposed to a positive potential energy?</p>
<p>2) Can an object’s total energy be negative, and if so, what does that mean?</p>
<p>3) What is the significance of the direction of torque and angular momentum?</p>
<p>potential energy itself is arbitrary, only the change in potential energy matters. However if you really want to get into it, potential energy is 0 really at infinity; thus, when you do an integral of force over distance to get the energy (integral of Gmm/r^2 from infinity to the distance r), you get -Gmm/r from infinity to r which is -Gmm/r-0=-Gmm/r</p>
<ol>
<li><p>total energy can be negative if the potential energy is very negative. this usually means a potential well, meaning the object is trapped by the force and will never escape the pull of the force it is experiencing</p></li>
<li><p>without external forces, angular momentum is always conserved and angular momentum is p x r… torque causes a change in angular momentum like force does on momentum</p></li>
</ol>
<p>anyone know how barron’s compares to the real test? I know it is usually harder…</p>
<p>How’s a 25/35 on the Barron’s compare to the real one?</p>
<p>Yeah, Barron’s is always more difficult. I failed their free response on the first practice test, but didn’t think the second practice test was too bad.</p>
<p>I’m not sure how it scales over, but just remember there’s a big difference between missing x-amount and just leaving x-amount blank.</p>
<p>In other news, I actually calculated ‘I’ correctly in 2004 FRQ!!</p>
<p>whos got the physics c mech 04 answers?pm me. ill trade. email at : <a href="mailto:yearsofwisdom@gmail.com">yearsofwisdom@gmail.com</a></p>
<p>Here’s a good question, 2006 #1, bcd: <a href=“College Board - SAT, AP, College Search and Admission Tools”>College Board - SAT, AP, College Search and Admission Tools;
<p>Can anybody think of some good equations to store into our calculators for FR? (lol)</p>
<p>There’s really nothing you need besides what’s on the equations sheet.</p>
<p>Yeah, true, but some of the things you need to use are pretty annoying to derive.</p>