Momentum and Energy sliding blocks

AI Thread Summary
A cube of mass m slides down a circular path, converting potential energy into kinetic energy as it descends. The initial potential energy, U = mgr, transforms into kinetic energy, K = 1/2mv^2, as the cube exits the larger block of mass M. The momentum conservation equation must account for both blocks, as the motion of m affects M's momentum. The relationship between the velocities of both blocks is crucial for solving the problem, indicating that the final velocity of m will also depend on the velocity of M. The discussion suggests using Lagrangian mechanics for a more straightforward solution.
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Homework Statement


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?

Homework 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)

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.
 
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Here's a diagram for reference:
http://img390.imageshack.us/img390/5575/blocksbi2.jpg

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.
 
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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:

E_k = E_p

therefore:

\frac{1}{2}mv^2 = mg\Delta h
 
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.
 
Try using Lagrangian mechanics to solve this problem. I think it would be easiest if you used a linear generalized coordinate.
 
Please note: this question was asked 3 years ago.
 
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