Inelastic collision Two angles and final velocities

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Homework Help Overview

The problem involves an inelastic collision between two blocks on a horizontal plane, where one block is initially at rest. The scenario includes the loss of kinetic energy and the deflection of both blocks at specified angles after the collision. The objective is to find the angle and final velocities of the blocks.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply conservation of momentum and kinetic energy principles but expresses uncertainty about the setup of the equations, particularly in separating the momentum into x and y components.
  • Some participants question the interpretation of momentum as a vector and suggest the need for separate conservation equations for each direction.
  • There is a discussion about the initial conditions and how they affect the momentum equations, particularly noting the absence of initial momentum in the y direction.

Discussion Status

The discussion is ongoing, with participants exploring the correct formulation of momentum conservation equations. Some guidance has been offered regarding the need to treat momentum as a vector and to separate the equations for x and y components. However, there is no explicit consensus on the next steps or resolution of the problem.

Contextual Notes

The original poster mentions feeling overwhelmed by the complexity of the problem and the lack of resources that address intermediate inelastic collisions with differing masses. There is also a note about the homework being limited to a few problems over the week, adding to the pressure.

-PhysicsMajor-
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Homework Statement


Consider an inelastic collision between two blocks on a horizontal plane. Block M1 is moving with velocity Vo and collides with block M2 which is at rest. During the collision a fraction Q of the original kinetic energy is lost. It is observed that M1 is deflected by an angle theta (above the x axis), and M2 is deflected at an angle phi (below the x axis). After the collision M1 is moving to the right.

Find the angle phi, and the final velocities of the blocks.

M1=2kg
M2=4kg
Vo=10m/sec
Theta=30o
Q=.2

Homework Equations


Inital Momentum = Final Momentum (M1iVo+M2iVo =M1fVf+M2fVf)

KEf = (1-Q)KEi

*I'm sure there are more but this is what I have from class.

The Attempt at a Solution



Finding the final KE was easy enough using the above equation: KE inital =100 KE final =80

Then I tried to split them up in the x and y components. They both start with zero in the y direction and only M1 has initial momentum in the x direction. In the Y direction after the impact I got some sin and cos directions
So M1iVo = M1V1fCos(theta)+M2V2fcos(phi)+M1V1f-M2V2fsin(phi) (negative because it is below the x axis)
If that is right (big if) I have no clue where to go from here. I really don't know how to turn the final kinetic energy into two different objects and I REALLY don't know how to find that phi angle. I'm at a complete loss.

This is my first post here so hopefully formatted the question correctly. This stuff is really stressing me out. Every time I think that I've stumbled upon something useful online, it just seems way to simple, which is pretty much a guarantee that it can't help me. Our homework consists of just a few problems over the course of a week, and when he shows examples in class my professor fills several chalkboards full of calculations. So anything that is a quick answer must be wrong.

I don't want the answer, I just need a nudge in the right direction of how to even set up the problem. I have not found a single thing online that can help with these intermediate inelastic collisions with two objects of a different mass.

This forum is my last hope.

*Way too many words, not nearly enough physics. Sorry
 

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-PhysicsMajor- said:
So M1iVo = M1V1fCos(theta)+M2V2fcos(phi)+M1V1f-M2V2fsin(phi)
you seem to have a basic misapprehension regarding momentum.
Momentum is a vector, and your 'relevant equation' for its conservation should be interpreted in that light.
When resolving into separate X and Y directions, that gives you two momentum conservation equations, one for each direction. But you appear to have collapsed them into one scalar equation.
 
haruspex said:
you seem to have a basic misapprehension regarding momentum.
Momentum is a vector, and your 'relevant equation' for its conservation should be interpreted in that light.
When resolving into separate X and Y directions, that gives you two momentum conservation equations, one for each direction. But you appear to have collapsed them into one scalar equation.
Thank you for bringing this up, because it seemed wrong when I was doing it. I guess I just got confused because there is no inital momentum in the y direction.

So
Pix=Pfx = M1iVo = M1V1fCos(theta)+M2V2fcos(phi)
Piy = Pfy = 0 = M1V1fsin(theta)-M2V2fsin(phi)

??
 
-PhysicsMajor- said:
Thank you for bringing this up, because it seemed wrong when I was doing it. I guess I just got confused because there is no inital momentum in the y direction.

So
Pix=Pfx = M1iVo = M1V1fCos(theta)+M2V2fcos(phi)
Piy = Pfy = 0 = M1V1fsin(theta)-M2V2fsin(phi)

??
Yes.
 

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