Question

-

Answer 1

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

Answer 2

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

Answer 3

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: