# A ball with a mass of  2 kg is rolling at 2 m/s and elastically collides with a resting ball with a mass of 9 kg. What are the post-collision velocities of the balls?

The initially moving ball would move $1.273 \frac{m}{s}$ to the left while the ball initially at rest would move $0.72 \frac{m}{s}$ to the right.

#### Explanation:

(NOTE: velocities with apostrophes denotes the arbitrary object's final velocity )

Consider the total momentum of the system.

$m {v}_{1} + m {v}_{2} = m v {'}_{1} + m v {'}_{2}$
Note that final velocities are positive because I assumed them to move to the right after the impact.

$m {v}_{1} + m \cancel{{v}_{2}} = m v {'}_{1} + m v {'}_{2}$
$\left(2\right) \left(2\right) = \left(2\right) \left(v {'}_{1}\right) + \left(9\right) \left(v {'}_{2}\right)$

Moreover, the problem states that the impact involves an elastic collision. Therefore, the coefficient of restitution of the system, $e$, is 1.
$e = 1 = \frac{v {'}_{2} - v {'}_{1}}{{v}_{1} - {v}_{2}} = \frac{v {'}_{2} - v {'}_{1}}{2 - 0}$

$\therefore , 2 = v {'}_{2} - v {'}_{1}$

To solve the post-velocities solve the system of equations:
$\left(1\right) : 9 v {'}_{2} + 2 v {'}_{1} = 4$
$\left(2\right) : v {'}_{2} - v {'}_{1} = 2$

Where $v {'}_{1} = - 1.273$
and $v {'}_{2} = + 0.72$

Since $v {'}_{1}$ is negative, the assumed direction of its final velocity is the opposite.