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I have a question about the billiard game physics. I know that in a billiard game, balls collisions are elastic and therefore the angle of the collision of two balls can easily be calculated no matter how high the velocity of balls are.

But from what I have seen in a real game, I know that the power of the shot affects the angle of collision.

How could you explain this? Is this because of the collision is not elastic in the real world?

Thanks

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    $\begingroup$ No collision (except the ones happening at the atomic level ) is completely elastic. Especially in the case of billiard, the very fact that you hear the sound of balls colliding show you that there is some energy lost in creating that sound. $\endgroup$ – SagarM May 2 '19 at 13:31
  • $\begingroup$ Also, unless you are a very good cueist then you will be applying side to the white - even though you don't mean to, which explains why people miss easy shots. $\endgroup$ – user207455 May 2 '19 at 15:06
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How could you explain this? Is this because of the collision is not elastic in the real world?

Yes. At least in the macroscopic real world.

Microscopically, however, collisions can approach being completely elastic. An example is an ideal gas where the collisions of the molecules of an ideal gas are considered to occur between rigid point particles.

Hope this helps.

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