Consider a collision: What's m1/m2?

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Homework Help Overview

The discussion revolves around a collision problem involving two masses, where one mass is initially in motion and the other is at rest. Participants are exploring the relationship between the masses based on their velocities before and after the collision, with a focus on whether the collision is elastic or inelastic.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are considering the nature of the collision, questioning whether it can be classified as elastic based on the given velocities. There is discussion about applying conservation of momentum and the validity of various equations related to kinetic energy and momentum.

Discussion Status

Some participants have offered guidance on applying conservation of momentum, while others are questioning the assumptions made regarding the equations used. There is an ongoing exploration of how to correctly apply these principles to the problem at hand.

Contextual Notes

Participants are grappling with the implications of using vector quantities in their equations and the potential need to separate components for analysis. There is also a recognition that the initial equation presented may not be suitable due to the vector nature of the velocities involved.

Charlene
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Homework Statement


Consider a collision: If mass 1 has initial velocity 1i-1j of and mass 2 is initially at rest. After collision mass one moves with a velocity of 2i-3j and mass 2 moves with velocity of -1.5i+3j. What's m1/m2?

Homework Equations


m1/m2-=(v2f-v2i)/(v1i-v2f)

The Attempt at a Solution


Reading this question i would think its an elastic collision because they don't end up sticking together because mass 1 moves in a positive x direction and negative y direction and for mass 2 it moves in the opposite direction of mass 1. Would this be a correct assumption?
 
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Apply conservation of momentum, which works for both elastic and nonelastic collisions. The relevant equation does not make sense. How can a scalar equal to division of vectors which is not allowed. Derive your own formula by conserving total momentum. Remember when two vectors are equal their x and y components are also equal.
 
okay, would the equation m1(v1i^2)+m2(v2i^2)=m1(v1f^2)+m2(v2f^2) be accurate? I'm pretty sure the right hand side is correct but i may have messed up on the left hand side.
 
Charlene said:
okay, would the equation m1(v1i^2)+m2(v2i^2)=m1(v1f^2)+m2(v2f^2) be accurate? I'm pretty sure the right hand side is correct but i may have messed up on the left hand side.
You are now corectly quoting the equation for conservation of KE, but you are wrong to assume you can apply that here.

Conservation of KE is one extreme possibility, where no KE is lost. Coalescing (sticking together) is the opposite extreme, where as much KE is lost as can be. In between there is a whole range of possibilities in which some KE is lost.
As Let'sthink recommends, apply conservation of momentum. The equation you originally quoted would be right for that if the velocties were all in the same direction, but here they are not. The more general form can be obtained by multiplying it out so that you are not dividing by vectors anywhere. See if you can quote it correctly.
 
haruspex said:
You are now corectly quoting the equation for conservation of KE, but you are wrong to assume you can apply that here.

Conservation of KE is one extreme possibility, where no KE is lost. Coalescing (sticking together) is the opposite extreme, where as much KE is lost as can be. In between there is a whole range of possibilities in which some KE is lost.
As Let'sthink recommends, apply conservation of momentum. The equation you originally quoted would be right for that if the velocties were all in the same direction, but here they are not. The more general form can be obtained by multiplying it out so that you are not dividing by vectors anywhere. See if you can quote it correctly.

So would the first equation i listed be able to be used if i used it for the x component and y component separately, i cannot think of a way to "multiply it out" unless you just mean m1(v1f-v1i)=m2(v2i-v2f)??
 
Last edited:
Charlene said:
m1(v1f-v1i)=m2(v2i-v2f)??
Yes. That equation is still valid when the velocities are vectors. Plug in the vectors you have.
 

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