Answer
(a) $v_A' = (\frac{m_A - m_B}{m_a+m_B})~v_A$
$v_B' = (\frac{2m_A}{m_a+m_B})~v_A$
(b) $v_A' \approx -v_A$
$v_B' \approx 0$
(c) $v_A' \approx v_A$
$v_B' \approx 2v_A$
(d) $v_A' = 0$
$v_B' = v_A$
Work Step by Step
(a) We can use conservation of momentum to set up an equation.
$m_A~v_A + 0 = m_A~v_A' + m_B~v_B'$
Since the collision is elastic, we can use Equation 7-7 to set up another equation.
$v_A - 0 = v_B' - v_A'$
$v_A' = v_B' - v_A$
We can use this expression for $v_A'$ in the first equation.
$m_A~v_A + 0 = m_A~v_B' - m_Av_A + m_B~v_B'$
$v_B' = \frac{2m_A~v_A}{m_a+m_B} = (\frac{2m_A}{m_a+m_B})~v_A$
We can use $v_B'$ to find $v_A'$.
$v_A' = v_B' - v_A = (\frac{2m_A~v_A}{m_a+m_B}) - v_A$
$v_A' = (\frac{2m_A~v_A}{m_a+m_B}) - (\frac{m_A~v_A + m_B~v_A}{m_a+m_B})$
$v_A' = (\frac{m_A - m_B}{m_a+m_B})~v_A$
(b) When $m_A$ is much smaller than $m_B$
$v_A' = (\frac{m_A - m_B}{m_a+m_B})~v_A \approx -v_A$
$v_B' = (\frac{2m_A}{m_a+m_B})~v_A \approx 0$
A common example of this would be a tennis ball bouncing back off a wall.
(c) When $m_A$ is much larger than $m_B$
$v_A' = (\frac{m_A - m_B}{m_a+m_B})~v_A \approx v_A$
$v_B' = (\frac{2m_A}{m_a+m_B})~v_A \approx 2v_A$
A common example of this would be a person's foot kicking a tiny pebble on the sidewalk.
(d) When $m_A$ is equal to $m_B$
$v_A' = (\frac{m_A - m_B}{m_a+m_B})~v_A = 0$
$v_B' = (\frac{2m_A}{m_a+m_B})~v_A = v_A$
A common example of this would be a moving billiard ball hitting head on another billiard ball which is at rest.