Answer
The final speed of the tennis ball is 2.50 m/s in the original direction.
The final speed of the other ball is 5.00 m/s in the original direction.
Work Step by Step
Let $m_A = 0.060~kg$ and let $m_B = 0.090~kg$.
We can use conservation of momentum to set up an equation:
$m_A~v_A + m_B~v_B = m_A~v_A' + m_B~v_B'$
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:
$m_A~v_A + m_B~v_B = m_A~v_B' - m_Av_A + m_Av_B+ m_B~v_B'$
$v_B' = \frac{2m_A~v_A+m_B~v_B -m_Av_B}{m_a+m_B}$
$v_B' = \frac{(2)(0.060~kg)(5.50~m/s)+(0.090~kg)(3.00~m/s)- (0.060~kg)(3.00~m/s)}{(0.060~kg)+(0.090~kg)}$
$v_B' = 5.00~m/s$
We can use $v_B'$ to find $v_A'$:
$v_A' = v_B' - v_A + v_B$
$v_A' = 5.00~m/s - 5.50~m/s + 3.00~m/s = 2.50~m/s$