Elastic Collision is kinetic energy conserved?

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SUMMARY

The discussion centers on the principles of elastic collisions, specifically involving two identical billiard balls, each with a mass of 0.2 kg. The initial kinetic energy of the moving ball is calculated to be 0.625 Joules, which must be conserved post-collision. It is established that in elastic collisions, both momentum and kinetic energy are conserved, meaning the total values remain constant, although their distribution between the two balls may change. The final velocities can be determined using the conservation equations for both momentum and kinetic energy.

PREREQUISITES
  • Understanding of elastic and inelastic collisions
  • Familiarity with the conservation of momentum
  • Knowledge of kinetic energy calculations (E = 0.5mv²)
  • Basic algebra for solving simultaneous equations
NEXT STEPS
  • Study the principles of elastic collisions in detail
  • Learn to derive equations for conservation of momentum and kinetic energy
  • Practice solving problems involving two-body collisions
  • Explore real-world applications of elastic collisions in sports and physics
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Physics students, educators, and anyone interested in understanding the mechanics of collisions, particularly in the context of billiard games or similar scenarios.

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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.
 
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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.
 


The concept of elastic collisions can be a bit confusing, so it's understandable that you're having trouble with it. To answer your question, yes, kinetic energy is conserved in elastic collisions. In fact, this is one of the defining characteristics of an elastic collision.

In an elastic collision, both momentum and kinetic energy are conserved. This means that the total momentum and total kinetic energy of the system before and after the collision are the same. This does not mean that the individual values for momentum and kinetic energy have to be the same, but rather the total value.

In your example of the billiard balls, the initial kinetic energy is indeed 0.625 J. After the collision, this same amount of kinetic energy will be present, but it may be distributed differently between the two balls. This means that the final velocity of the two balls may not be the same, but the total kinetic energy will still be 0.625 J.

To find the final velocity of the two balls, you can use the conservation of momentum equation, which states that the total momentum before the collision is equal to the total momentum after the collision. In this case, the total momentum before the collision is 0, since one ball is stationary and the other is moving towards it. After the collision, the total momentum will still be 0, but it will be distributed between the two balls. You can set up an equation using the masses and velocities of the two balls to solve for the final velocity.

Remember, the key concept in elastic collisions is that both momentum and kinetic energy are conserved. So, while the individual values for momentum and kinetic energy may change, the total values will always remain the same. I hope this helps clear up any confusion you may have had about elastic collisions.
 

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