What Are the Final Speeds of Billiard Balls After a Perfectly Elastic Collision?

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In a discussion about the final speeds of billiard balls after a perfectly elastic collision, two balls of equal mass collide with initial speeds of 2.0 m/s and 3.0 m/s in opposite directions. The conservation of momentum and kinetic energy equations are applied to find the final velocities. The simplified equations lead to two unknowns, which can be solved using algebraic methods. The final speeds after the collision are determined to be 3.0 m/s for the first ball and 2.0 m/s for the second ball. This demonstrates the principles of elastic collisions in physics.
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Homework Statement



Two Billiard balls of equal mass undergo a perfectly elastic head on collision
(Energy is conserved) . If the speed of one ball was initially at 2.0 m/s
and the other 3.0 m/s in the opposite direction, what will be their speeds after the
collision??

Homework Equations


m1v1 + m2v2 = m1v1' + m2v2'
1/2 m1v1^2 + 1/2 m2mv2^2 = 1/2m1v1 '^2 + 1/2m2v2 '^2


The Attempt at a Solution


so far i have simplified both formulas so that it's

v1' = v1 + v2 - v2'
v2' = v1^2 + v2^2 - v1 '^2

That's as far as a can go but I'm stuck what should my next step be...

The answer is v1 = 3 m/s & vs = 2 m/s if it helps anyone
 
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Well, you've done the bulk work of the problem already. You have two unknowns (v1 and v2 final) in two separate expressions. Use your algebraic skillz to solve (i usually substitute when there's only 2 unknowns, but you can eliminate too).
 
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