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

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SUMMARY

The discussion centers on calculating the final speeds of two billiard balls after a perfectly elastic collision. Given initial speeds of 2.0 m/s and 3.0 m/s in opposite directions, the final speeds are determined to be 3.0 m/s and 2.0 m/s, respectively. The conservation of momentum and kinetic energy equations are utilized, specifically m1v1 + m2v2 = m1v1' + m2v2' and 1/2 m1v1^2 + 1/2 m2v2^2 = 1/2 m1v1'^2 + 1/2 m2v2'^2. The solution involves algebraic manipulation to isolate the final velocities.

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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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