Post-Impact Speed of Colliding Billiard Balls

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SUMMARY

The discussion focuses on a perfectly elastic collision between two identical billiard balls, one moving north at 15.0 m/s and the other south at 10.0 m/s. The key principles involved are the conservation of momentum and the conservation of kinetic energy. Since the masses of the billiard balls are identical, the post-impact speeds can be calculated directly using these conservation laws without needing the mass values. The post-impact speed of the first ball is 10.0 m/s north, and the second ball's speed is 15.0 m/s south.

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Homework Statement



Two billiard balls, one heading north at 15.0 m/s and a second heading south at 10.0 m/s, collide head-on. Take the collision to be perfectly elastic and choose the positive direction north.

What is the post-impact speed of the first ball
Answer: m/s

What is the post-impact speed of the second ball?
Answer: m/s


Homework Equations





The Attempt at a Solution



Don't even know where to start.
 
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You start by applying conservation of momentum and conservation of kinetic energy.
 
But don't i need the mass in order to do that?
 
They are identical billiard balls I assume, in mass and otherwise.
 

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