<p>How can you determine whether angular momentum is conserved?</p>
<p>how many percentages do we need to get for a 5?</p>
<p>For the second one of mine, the answer is that the force is toward the magnet.</p>
<p>I kind did the right hand rule like mooble said, but if the current is clockwise around the loop, I get the north pole of the loop to be to the left. Therefore, the south pole is to the right, creating an attractive force with the north pole of the magnet</p>
<p>@ishanz:</p>
<p>when there’s no net external torque</p>
<p>Does anyone have any predictions for the FRQs for the Mechanics exam?</p>
<ol>
<li><p>Mass M1 is moving with speed v toward stationary mass M2. The speed of the center of mass of the system is?</p></li>
<li><p>A system consists of two objects having masses ml and m2 (m<em>1 < m</em>2). The objects are connected by a massless string, hung over a pulley, and then released. When the speed of each object is v, the magnitude of the total linear momentum of the system is?</p></li>
</ol>
<p>v_com = [ M1(v) + M2(0) ] / (M1 + M2) = M1*v / (M1+M2)</p>
<p>@ChemE14: Could you elaborate more? Thanks. Also, it would be great if you could answer the other problem too. :)</p>
<p>x<em>com = SUM(M</em>i * x<em>i) / M</em>total</p>
<p>If we differentiate this, we get</p>
<p>v<em>com = SUM(M</em>i * v<em>i) / M</em>total</p>
<p>So</p>
<p>v_com = [ M1(v) + M2(0) ] / (M1 + M2) = M1*v / (M1+M2)</p>
<p>Oh wow thanks ChemE14! Based on your posts in this thread, you seem like a 5’er for sure!</p>
<p>For second question:</p>
<p>They have the same speed, but M2 is going up while M1 is going down, so</p>
<p>p<em>system = p</em>1 + p_2 = M1(-v) + M2(v) = (M2 - M1)v</p>
<p>Can anybody make some predictions about the topics that will be on the Mechanics FRQ based on the past few years trends? </p>
<p>I’m thinking Universal Gravitation(with Angular Momentum), Newton’s Second Law(with 1 block on top of another), and a hard Moment of Inertia problem.</p>
<p>If you saw 09 and 08 FRQ, the questions are getting much more difficult each year.</p>
<p>
</p>
<p>Hopefully! Unless the CollegeBoard’s out to fail me - it’s their last chance to get me this year! I just know at least one of the FRQs will be something I have no idea how to do. And if that does happen, I’m not even going to take the time to attempt to work it out - I’m writing them a note and walking out!</p>
<p>if M1<M2, the M2 is going down and M1 is going up, thus p=(M1-M2)v</p>
<ol>
<li>The sum of all the external forces on a system of particles is zero. Which of the
following must be true of the system?
(A) The total mechanical energy is constant.
(B) The total potential energy is constant.
(C) The total kinetic energy is constant.
(D) The total linear momentum is constant.
(E) It is in static equilibrium.</li>
</ol>
<p>It’s obviously A or D, but I’m confused why the answer is D. Is it because the sum of forces could be zero, but if one of those forces is friction or something, then energy is dissipated?</p>
<p>A is not right…think of the inelastic collison</p>
<p>If F=0, a=0, v=constant, mv=constant, linear momentum=constant. Thats all.</p>
<p>^^
whoops, yeah, aznjunior has it right. I read the problem wrong (thought m1 > m2)</p>
<p>ChemE14, you can miss an entire FR (0/15), and get a 5, can’t you? The curve is HUGE!</p>
<p>will they give us a list of formulas or will we have to memorize them??</p>