Conservation of Linear Momentum problem help

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Homework Help Overview

The problem involves a block resting on a wedge, both of which are on a frictionless surface. The task is to determine the velocity of the wedge when the block touches the table after falling a distance h. The context is rooted in the principles of conservation of momentum and energy.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of the center of mass (COM) approach and its limitations, particularly regarding the conservation of momentum in different directions. Questions arise about the validity of using COM when one component of momentum is not conserved.

Discussion Status

The discussion is ongoing, with participants exploring the implications of momentum conservation in this scenario. Some guidance has been offered regarding the conservation of horizontal momentum, while others express confusion about the professor's statements on the matter.

Contextual Notes

There are references to previous similar problems and the constraints of the problem setup, including the frictionless nature of the surfaces and the initial conditions of the system.

wsender
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Homework Statement


Having a little trouble with this problem. I've tried a few different manipulations using COM & COE and wasn't able to get it to fit the form.

"A block of mass m rests on a wedge of mass M which, in turn, rest on a horizontal table as shown in the figure. All surfaces are frictionless. If the system starts at rest with point P of the block a distance h above the table, find the velocity of the wedge the instant point P touches the table."



Homework Equations



KE+PE+W=KE+PE

P1=P2

See attached image file for solution and diagram.


The Attempt at a Solution



Tried using COM to solve this but ran into issues in the y direction regarding preservation.
 

Attachments

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wsender said:
Tried using COM to solve this but ran into issues in the y direction regarding preservation.
If your y direction is vertical, momentum is not conserved in the y direction. (The floor exerts a force.) But the horizontal component of momentum is conserved.
 
Doc,

I worked that out as well and my Professor said that you can't use COM for any parts of the equation if either the X or Y component isn't conserved, is it true?
 
wsender said:
Doc,

I worked that out as well and my Professor said that you can't use COM for any parts of the equation if either the X or Y component isn't conserved, is it true?
No, that's not true. (Perhaps you misheard him.) Momentum is a vector and it can be conserved in one direction but not another, which is the case here. There are no external horizontal forces on the system, so the horizontal component of momentum will be conserved.
 
Doc Al said:
No, that's not true. (Perhaps you misheard him.) Momentum is a vector and it can be conserved in one direction but not another, which is the case here. There are no external horizontal forces on the system, so the horizontal component of momentum will be conserved.

No I definitely didn't mishear him. I brought up this exact point in class and he dispelled is validity. Thank you for confirming my suspicions.
 
You don't have to assume conservation of momentum, it will automatically fall out of the equations from Newton's laws. Give it a shot.
 

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