Question about two elastic collision formulas

In summary, the two equations are saying that the momentum and kinetic energies of the objects after the collision are the same, but the momentum of the system before the collision is different.
  • #1
as2528
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Homework Statement
Explain what is the difference between:
1/2m1v1i^2+1/2m2v2i^2=1/2m1v1f^2+1/2m2v2f^2=>1
m1v1i+mvv2i=m1v1f+m2v2f=>2
Relevant Equations
1/2m1v1i^2+1/2m2v2i^2=1/2m1v1f^2+1/2m2v2f^2
m1v1i+mvv2i=m1v1f+m2v2f
Equation 1 is equating the kinetic energies of the objects before and after the elastic collision. Equation 2 is equating the momentums of the objects after the elastic collision. They can be used interchangeably as long as the collision is elastic.

Am I right in my conclusion?
 
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  • #2
as2528 said:
Equation 2 is equating the momentums of the objects after the elastic collision.
No, it does not say the objects have same momentum as each other after the collision. Is that what you meant?
as2528 said:
They can be used interchangeably
No, they are not interchangeable. They say different things.
 
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  • #3
haruspex said:
No, it does not say the objects have same momentum as each other after the collision. Is that what you meant?

No, they are not interchangeable. They say different things.
No what I mean was equation one says the momentum of the system was the same. I should have worded it better. I also believed that they were interchangeable, but I learned that they give different answers unless it is a perfectly elastic collision, so I will redact that.
 
  • #4
as2528 said:
I learned that they give different answers unless it is a perfectly elastic collision
Still not quite right. Neither by itself is enough to figure out what happens.
You can in general assume momentum is conserved - just check there are no impulses you have overlooked. But to be able to determine the subsequent motions you need more information. One possibility is that you have some reason to believe mechanical work is conserved; another is you may be told the objects coalesce, so you know the final velocities are the same; a third is you are told the maximum possible KE is lost; a fourth, you are given the coefficient of restitution.

In a one dimensional case in which mechanical work is conserved, there is a useful equation that can be obtained by combining momentum and work conservation laws: ##v_{1i}-v_{2i}=v_{2f}-v_{1f}##. Note that there is no mention of mass and no quadratic term.
As an exercise, derive that equation.
 
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  • #5
haruspex said:
Still not quite right. Neither by itself is enough to figure out what happens.
You can in general assume momentum is conserved - just check there are no impulses you have overlooked. But to be able to determine the subsequent motions you need more information. One possibility is that you have some reason to believe mechanical work is conserved; another is you may be told the objects coalesce, so you know the final velocities are the same; a third is you are told the maximum possible KE is lost; a fourth, you are given the coefficient of restitution.

In a one dimensional case in which mechanical work is conserved, there is a useful equation that can be obtained by combining momentum and work conservation laws: ##v_{1i}-v_{2i}=v_{2f}-v_{1f}##. Note that there is no mention of mass and no quadratic term.
As an exercise, derive that equation.
Thanks! I do believe that equation was derived in lecture and in my textbook, I will try to read both more closely since I've clearly not fully understood what was being discussed. I have not heard the term coefficient of restitution, but I have heard of the coalescing of objects which I believe means perfectly inelastic collision. Inelastic collision I do feel comfortable with, but this elastic collision phenomenon is new to me as we covered it in lecture just today.
 
  • #6
haruspex said:
mechanical work is conserved;
"Conserved" means same value before and after a process (or just a at two different times). It applies to state parameters like energy, momentum, angular momentum. Work is a process parameter and "conservation" does not apply to it. You don't have a work before the collision and another after the collision. The work is done during the collision.
Same as there is no conservation of heat (but it may be conservation of internal energy or thermal energy). Not that the work are heat are not conserved, it simply does not make sense to apply the term "conservation" to these quantities.
Maybe you mean conservation of kinetic energy.
 
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  • #7
as2528 said:
the coalescing of objects which I believe means perfectly inelastic collision
It's an example of a perfectly inelastic collision. You can also have an oblique impact in which the bodies do not coalesce but the impact is as inelastic as it can be. In that case, the objects end up with the same velocity component in a particular direction, namely, the normal to the contact plane. You can see that in a head-on impact that does imply coalescence.
 
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