## chris215 one year ago -

1. chris215

2 m/s, -3m/s -3 m/s, 2 m/s 2 m/s, 3m/s -2 m/s, -3 m/s

2. Michele_Laino

here you have to apply the total momentum conservation law, furthermore, since the collision is elastic, you can apply the kinetic energy conservation law

3. Michele_Laino

so we can write the subsequent equations: $\Large \left\{ \begin{gathered} m{\mathbf{u}} + m{\mathbf{v}} = m{\mathbf{u'}} + m{\mathbf{v'}} \hfill \\ \frac{1}{2}m{u^2} + \frac{1}{2}m{v^2} = \frac{1}{2}mu{'^2} + \frac{1}{2}mv{'^2} \hfill \\ \end{gathered} \right.$ where u, v are the initial velocities, namely before collision of the two balls, and u', v' are the corresponding velocities after collision. Furthermore m is the common mass of the two balls. Please note that the first equation is a vector equation. you can use this subsequent reference system: |dw:1435934610110:dw|