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Momentum and Energy sliding blocks

  1. Oct 9, 2008 #1
    1. The problem statement, all variables and given/known data
    A cube of mass m slides down a circular path of radius r cut into a larger block of mass M. M rests on a table, and friction is ignored. Both blocks start at rest and m starts at the top of the circular path. What is the velocity v of the the small cube as it leaves the larger cube?

    2. Relevant equations
    U = mgr since the radius of the circular path is the same as the height the smaller cube starts
    K = 1/2mv^2
    pm = pM (momentum)

    3. The attempt at a solution
    At first glance, I just set U = K, and solved for v, but I realized the v the small cube leaves will affect the larger cube's momentum. The problem is I can't figure out how to setup the momentum portion of the problem.
  2. jcsd
  3. Oct 10, 2008 #2
    Here's a diagram for reference:
    http://img390.imageshack.us/img390/5575/blocksbi2.jpg [Broken]

    m slides along the circular slope and leaves the bottom with v, but I'm confused as to how to factor in the momentum between m and M since both blocks act without regard to friction. So the final v for m should also have the velocity that M is moving in the opposite direction, right?

    I tried setting up (M+m)vo = (M+m)v but vo is zero, and since the masses are constant, the final v can't be zero as well.

    I must be missing something.
    Last edited by a moderator: May 3, 2017
  4. Oct 10, 2008 #3
    What you've done with your momentum equation is say that initial momentum equals final momentum. That means that initial velocity (zero) equals final velocity (zero) which may be the relative case, but that's not what we're solving for. It becomes tricky because the force is down and the angle normal to the surface of the mass M is constantly changing so it could end up as an integral.

    However I believe the question is a lot more simple than that. We know that initially there is no kinetic energy and all potentially. At the end there is all kinetic and no potential.

    Therefore we can say that all potential energy is converted to kinetic through the course of this system:

    [tex] E_k = E_p [/tex]


    [tex] \frac{1}{2}mv^2 = mg\Delta h [/tex]
  5. Nov 24, 2011 #4
    I think you need to consider the kinetic energy of the large block aswell,

    1) mgR = (mv^2)/2 + (Mv1^2)/2

    then you can use momentum conservation to find v1 and finally replace in 1) to find v.
  6. Nov 27, 2011 #5
    Try using Lagrangian mechanics to solve this problem. I think it would be easiest if you used a linear generalized coordinate.
  7. Nov 27, 2011 #6


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