AP Physics C - Mech HELP PLEASE!!

<p>I'm sorta desperate; I have atest soon and I don't understand how to do this. Can someone explain to me how to do 2002 Mech 3 (below)? Any help would be greatly appreciated!</p>

<p><a href="http://www.collegeboard.com/prod_downloads/ap/students/physics/physics_c_m_frq_02.pdf%5B/url%5D"&gt;http://www.collegeboard.com/prod_downloads/ap/students/physics/physics_c_m_frq_02.pdf&lt;/a&gt;&lt;/p>

<p>Well, have you looked at the solutions yet? If you go through AP Central, rather than just collegeboard, you can see the solution guide and some sample responses...if those don't clarify it for you, here's how I see it.</p>

<p>(a): Simple...just graph the function.</p>

<p>(b): F = -dU/dx. That is, the force equals the negative derivative of the potential energy function. Just take the derivative of U(x) and add a negative sign in front.</p>

<p>(c): If the object is at rest at the origin (x=0), its kinetic energy at x=0 is 0. Thus, the object's total mechanical energy (which will remain constant) is equal to its potential energy when x=0.</p>

<p>Total energy = U(0) = 2.0 J</p>

<p>Now, for any x the object's total energy (which will always be 2.0 J) is equal to the sum of its kinetic and potential energy at that point. At x=2, its potential energy is 1.0 J [U(2)=1]. Because the total energy will be 2.0 J, the object's kinetic energy at x=2 must be 1.0 J.</p>

<p>Since kinetic energy is K=(1/2)mv^2, and since K=1 and m=.5, v (the object's velocity when x=2) is 2 m/s. [1=(.5)(.5)v^2.....4=v^2....v=2]</p>

<p>(d) Take a meterstick, a spring, and a set of objects of different masses. There are quite a few ways to do this, but here's mine. For the following to make sense, you need to have done oscillations. If you haven't, let me know. Basically, all you need to do is devise an experiment to measure an object's velocity 2m away from its original position.</p>

<p>(e) First you have to find k for the spring. Let the spring hang freely in the air (vertically). Attach one mass, and record the mass and the distance the spring stretches from its rest position. Do this for several (5 is good) different masses. Take your data and make a F vs x graph (put the weight, ie w=mg, of the mass on the y-axis, and the distance stretched in meters on the x-axis). Because of Hooke's Law, you know F=kx, where k is the spring constant. Draw a best fit line for your data, and find the slope, which is equal to k.</p>

<p>Now rest the spring on its side on the air track, 2 meters away from the glider's initial position. Release the glider to be acted on by the force, F, and let it collide with and compress the spring. Measure how far (in meters) the spring in compressed. </p>

<p>Now use conservation of energy. Basically, the total energy of the glider-spring system just before the collision will be equal to the total energy at the point of maximum compression. That is, the kinetic energy of the glider at x=2 will be equal to the potential energy of the spring when it is compressed the most (and thus when the glider is stopped).</p>

<p>K = U
.5mv^2 = .5kA^2 , where A is the measured amplitude (distance of maximum compression), and m=.5</p>

<p>(.5)(.5)v^2 = .5kA^2
v^2 = 2kA^2
v = sqrt(2kA^2)</p>

<p>Compare this value of v with the calculated one in (c).</p>

<p>I hope that helps!</p>