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Elastic Collision is kinetic energy conserved?

  • Thread starter Cummings
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Here is a little problem i am having with elastic colisions.

With inelastic colisions, momentum is conserved but kinetic energy is not.

And so from my point of view it would be impossible to have an elastic colision where both momentum and kinetic energy would be conserved as if kinetic energy was conserved, momentum would rise.

One solution i thought up was that conserved did not mean "the same" That the momentum could rise. I also have been thinking that the two balls will move off with the same speed as each other.

The question i am tackling involves two identical billard balls. Both of mass .2 kg

One is stationary, and the other is traveling at 2.5 ms towards the stationary

From E = .5mv^2 i found the initial kinetic energy to be .625 j

and so this amount of kinetic energy must be present in the final velocity of the balls once they colide.

I am asked to find the final velocity of the two billard balls but am stuck.
 

Answers and Replies

HallsofIvy
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And so from my point of view it would be impossible to have an elastic colision where both momentum and kinetic energy would be conserved as if kinetic energy was conserved, momentum would rise.
That's a very unusual point of view! Do you have any reason for thinking that "if kinetic energy was conserved, momentum would rise"?

One solution i thought up was that conserved did not mean "the same" That the momentum could rise. I also have been thinking that the two balls will move off with the same speed as each other.
No, "conserved" does, in fact, mean "doesn't change"- "is the same".

The question i am tackling involves two identical billard balls. Both of mass .2 kg

One is stationary, and the other is traveling at 2.5 ms towards the stationary

From E = .5mv^2 i found the initial kinetic energy to be .625 j

and so this amount of kinetic energy must be present in the final velocity of the balls once they colide.
Yes, the moving billiard ball has kinetic energy (1/2)(.2)(2.5)^2= 0.625 Joules. Since the second billiard ball is not moving, it has kinetic energy 0. The total kinetic energy is 0.625 j.

Let v1 and v2 be the speeds of the two balls after the collision. The first ball (the one that was moving) has kinetic energy (1/2)(.2)(v12) and the second ball (the one that was stationary) has kinetic energy (1/2)(.2)(v22) and so, conserving energy we have
(1/2)(.2)(v12)+ (1/2)(.2)(v22)= 0.625 j.

That is one equation for two unknowns. Fortunately for us, momentum does NOT increase but, just as your teacher told you, is conserved. That gives us a second equation.

The first ball, with mass 0.2 kg and speed 2.5 m/s has momentum
(0.2)(2.5)= 0.5 kg m/s. The second ball has speed 0 and so momentum 0. The total momentum is 0.5 kg m/s. (Strictly speaking, momentum, unlike kinetic energy, is a vector quantity. I am assuming that everything is moving on one line.)

After the collision, the first ball has momentum 0.2 v1 and the second ball has momentum 0.2 v2. Conservation of momentum tells us 0.2 v1+ 0.2 v2= 0.5.

Now solve the two equations (1/2)(.2)(v12)+ (1/2)(.2)(v22)= 0.625 j (which is the same as
v12+ v22= 6.25) and
0.2 v1+ 0.2 v2= 0.5 (which is the same as
v1+ v2= 2.5.
 

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