Conservation of momentum:

$\overline{){{\mathbf{P}}}_{{\mathbf{i}}}{\mathbf{=}}{{\mathbf{P}}}_{{\mathbf{f}}}}$

Therefore:

$\begin{array}{rcl}\mathbf{m}{\mathbf{v}}_{\mathbf{1}\mathbf{i}}\mathbf{-}\mathbf{m}{\mathbf{v}}_{\mathbf{2}\mathbf{i}}& \mathbf{=}& \mathbf{-}\mathbf{m}{\mathbf{v}}_{\mathbf{1}\mathbf{f}}\mathbf{+}\mathbf{m}{\mathbf{v}}_{\mathbf{2}\mathbf{f}}\\ {\mathbf{v}}_{\mathbf{1}\mathbf{i}}\mathbf{-}{\mathbf{v}}_{\mathbf{2}\mathbf{i}}& \mathbf{=}& \mathbf{-}{\mathbf{v}}_{\mathbf{1}\mathbf{f}}\mathbf{+}{\mathbf{v}}_{\mathbf{2}\mathbf{f}}\\ \mathbf{-}{\mathbf{v}}_{\mathbf{1}\mathbf{f}}\mathbf{+}{\mathbf{v}}_{\mathbf{2}\mathbf{f}}& \mathbf{=}& {\mathbf{v}}_{\mathbf{1}\mathbf{i}}\mathbf{-}{\mathbf{v}}_{\mathbf{2}\mathbf{i}}\\ \mathbf{-}{\mathbf{v}}_{\mathbf{1}\mathbf{f}}\mathbf{+}{\mathbf{v}}_{\mathbf{2}\mathbf{f}}& \mathbf{=}& \mathbf{2}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{0}\\ {\mathbf{v}}_{\mathbf{1}\mathbf{f}}& \mathbf{=}& {\mathbf{v}}_{\mathbf{2}\mathbf{f}}\mathbf{-}\mathbf{1}\end{array}$

Two billiard balls with the same mass m move in the direction of one another. Ball one travels in the positive x-direction with a speed of V_{1i}, and ball two travels in the negative x-direction with a speed of V_{2i}. The two balls collide elastically, and both balls change direction after collision. If the initial speeds of the balls were v_{1i}=2.0m/s and v_{2i} =1.0m/s, what would be the final speed and direction of ball two, v_{2f}, in m/s?

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