Answer
The ball which originally had speed 2.00 m/s will have a speed of 3.60 m/s in the opposite direction to its original direction.
The ball which originally had speed 3.60 m/s will have a speed of 2.00 m/s in the opposite direction to its original direction.
Note that the two balls simply exchanged their velocities.
Work Step by Step
Let $v_A = 2.00~m/s$ and let $v_B = -3.60~m/s$. Let $m$ be the mass of each ball.
We can use conservation of momentum to set up an equation.
$m~v_A + m~v_B = m~v_A' + m~v_B'$
$v_A + v_B = v_A' + v_B'$
Since the collision is elastic, we can use Equation 7-7 to set up another equation.
$v_A - v_B = v_B' - v_A'$
$v_A' = v_B' - v_A + v_B$
We can use this expression for $v_A'$ in the first equation.
$v_A + v_B = v_B' - v_A + v_B+ v_B'$
$v_B' = \frac{2v_A}{2} = v_A$
$v_B’ = 2.00~m/s$
We can use $v_B'$ to find $v_A'$.
$v_A' = v_B' - v_A + v_B$
$v_A' = 2.00~m/s - 2.00~m/s - 3.60~m/s = -3.60~m/s$
Note that the two balls simply exchanged their velocities.
