Conservation of momentum:

$\overline{){{\mathbf{m}}}_{{\mathbf{1}}}{{\mathbf{v}}}_{\mathbf{1}\mathbf{i}}{\mathbf{+}}{{\mathbf{m}}}_{{\mathbf{2}}}{{\mathbf{v}}}_{\mathbf{2}\mathbf{i}}{\mathbf{=}}{{\mathbf{m}}}_{{\mathbf{1}}}{{\mathbf{v}}}_{\mathbf{1}\mathbf{f}}{\mathbf{+}}{{\mathbf{m}}}_{{\mathbf{2}}}{{\mathbf{v}}}_{\mathbf{2}\mathbf{f}}}$

Condition for a perfectly elastic collision:

$\overline{){\mathbf{e}}{\mathbf{=}}\frac{{\mathbf{v}}_{\mathbf{2}\mathbf{f}}\mathbf{-}{\mathbf{v}}_{\mathbf{i}\mathbf{f}}}{{\mathbf{v}}_{\mathbf{1}\mathbf{i}}\mathbf{-}{\mathbf{v}}_{\mathbf{2}\mathbf{i}}}}$

Using conservation of momentum:

velocity of the first ball, v_{1i} = 2.00 m/s

velocity of the second ball, v_{2i} = -3.60 m/s

m_{1} = m_{2} = m

Two billiard balls of equal mass undergo a perfectly elastic head-on collision. If one ball’s initial speed was 2.00 m/s and the other’s was 3.60 m/s in the opposite direction, what will be their speeds and directions after the collision?

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