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## Homework Statement

The problem is pretty simple, however I don't understand which value to use after using quad formula to solve. See below.

V1(initial) = 2.5 m/s

V2(Initial) = -5.9 m/s

A pool ball moving with a speed of 2.5 m/s makes an elastic head-on collision with an identical ball traveling in the opposite direction with a speed of 5.9 m/s. Find the velocities of the balls after the collision.

## Homework Equations

ΣP(initial) = ΣP(final)

M

_{1}V

_{1}+ M

_{2}V

_{2}= M

_{1}V

_{1}' + M

_{2}V

_{2}'

M

_{1}= M

_{2}

2.5 + (-5.9) = V

_{1}' + V

_{2}'

V

_{2}' = -V

_{1}' - 3.4

Elastic collision: KE(initial) = KE(final)

(Mass cancels out)

V

_{1}

^{2}+V

_{2}

^{2}= V

_{1}'

^{2}+V

_{2}'

^{2}

## The Attempt at a Solution

From above, I got:

0 = 2V

_{1}'

^{2}- 6.8V

_{1}' - 29.5

by substituting "V

_{2}' = -V

_{1}' - 3.4" for V

_{2}'

After applying quadratic equation I got:

V

_{1}' = -2.5 m/s, 5.9 m/s

My problem is I do not know which one is the correct answer. My answer key says V

_{1}' is equal to -2.5 m/s, but I have no idea why.

Any help is appreciated!