Consider a collision: What's m1/m2?

In summary, the conversation discusses a collision between two masses, with one initially at rest and the other with an initial velocity of 1i-1j. After the collision, mass 1 moves with a velocity of 2i-3j and mass 2 moves with a velocity of -1.5i+3j. The question is asked about the ratio of m1/m2, and the discussion leads to the suggestion of applying conservation of momentum to obtain a general equation that can be used for both elastic and non-elastic collisions. The equation m1(v1i^2)+m2(v2i^2)=m1(v1f^2)+m2(v2f^2) is mentioned, but
  • #1
Charlene
23
0

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

Related to Consider a collision: What's m1/m2?

1. What does m1/m2 represent in a collision?

m1/m2 represents the ratio between the mass of the first object involved in the collision and the mass of the second object involved. It is used to calculate the final velocities of the objects after the collision.

2. How is the value of m1/m2 determined?

The value of m1/m2 is determined by dividing the mass of the first object by the mass of the second object. This can be done by using a scale or by looking up the mass of the objects in a reference table.

3. Why is m1/m2 important in a collision?

m1/m2 is important in a collision because it helps to determine the amount of momentum and energy that is transferred between the objects. It also affects the final velocities of the objects after the collision.

4. Does the value of m1/m2 change during a collision?

No, the value of m1/m2 does not change during a collision. However, the mass of each individual object may change if there is a transfer of mass between them. In this case, the value of m1/m2 would change accordingly.

5. How does m1/m2 affect the outcome of a collision?

The value of m1/m2 affects the outcome of a collision by determining the amount of kinetic energy and momentum that is transferred between the objects. It also affects the final velocities of the objects after the collision, with larger values of m1/m2 resulting in smaller final velocities for the first object and larger final velocities for the second object.

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