Using Special Case of Elastic Collisions in one dimension

  • Thread starter Seydlitz
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  • #1
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Is it to possible to use the special case of elastic collisions in one dimension with bodies that posses different mass. Ordinarily I know that if the body has same mass the velocity of the bodies will simply be exchanged but is the fact also hold for body with different masses?

[tex]v_1 - v_2 = v_2' - v_1'[/tex]
 

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  • #2
Doc Al
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Is it to possible to use the special case of elastic collisions in one dimension with bodies that posses different mass. Ordinarily I know that if the body has same mass the velocity of the bodies will simply be exchanged but is the fact also hold for body with different masses?
The equation you posted--which states that the relative velocity reverses in an elastic collision--applies regardless of the masses. But it does not allow you to conclude that velocities are simply exchanged. (That's even more of a special case.)
 
  • #3
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The equation you posted--which states that the relative velocity reverses in an elastic collision--applies regardless of the masses. But it does not allow you to conclude that velocities are simply exchanged. (That's even more of a special case.)
Ah okay, so what the above equation actually state? I'm confused with the meaning that the relative velocity is simply reversed.
 
  • #4
Doc Al
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Ah okay, so what the above equation actually state? I'm confused with the meaning that the relative velocity is simply reversed.
I'll give an example. Before they collide, say m1 is moving with speed 3 m/s to the east and m2 is moving with speed 4 m/s to the west. Taking east as positive, the relative velocity of m2 with respect to m1 before the collision is: V2 - V1 = -4 - 3 = -7 m/s.

So the relationship expressed in the equation above states that whatever happens during the elastic collision, they must end up such that the relative speed of m2 with respect to m1 after the collision is: V'2 - V'1 = +7 m/s. That's what reversed means. Of course this fact alone is not enough to determine those final speeds. You'll have to combine it with another relationship, such as conservation of momentum.
 
  • #5
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Ah now I get it, thank you for the help Doc. :D

This problem does not actually appear in my test today but it is really helpful to gain new insight from this.
 

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