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Consider a collision: What's m1/m2?

  • #1
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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?
 

Answers and Replies

  • #2
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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.
 
  • #3
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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.
 
  • #4
haruspex
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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.
 
  • #5
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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 seperately, i cannot think of a way to "multiply it out" unless you just mean m1(v1f-v1i)=m2(v2i-v2f)??
 
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  • #6
haruspex
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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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