Momentum and Energy sliding blocks

Click For Summary

Homework Help Overview

The problem involves a cube of mass m sliding down a circular path on a larger block of mass M, which is at rest on a table. The scenario is set in a frictionless environment, and the objective is to determine the velocity of the smaller cube as it exits the larger block after starting from rest at the top of the circular path.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss energy conservation principles, equating potential energy to kinetic energy. There is an exploration of momentum conservation and how the motion of both blocks affects the final velocity of the smaller cube. Some participants express confusion about the relationship between the velocities of the two blocks and the implications of momentum conservation.

Discussion Status

The discussion is active, with various approaches being suggested, including energy conservation and momentum conservation. Some participants are questioning the setup of the momentum equations and the implications of the initial conditions. There is no explicit consensus yet, as different interpretations and methods are being explored.

Contextual Notes

Participants note the absence of friction and the initial rest state of both blocks. There is also mention of the complexity introduced by the changing angles during the motion of the smaller cube.

diablo2121
Messages
10
Reaction score
0

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.
 
Physics news on Phys.org
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.
 
Last edited by a moderator:
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]

therefore:

[tex]\frac{1}{2}mv^2 = mg\Delta h[/tex]
 
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.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
Replies
13
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K