#### Answer

After the collision, bob A rises to a height of $0.57~m$ and bob B rises to a height of $2.3~m$

#### Work Step by Step

Let the masses be $m_A$ and $m_B$. Note that $m_B = 2m_A$.
Let $v_A$ be the initial velocity of bob A.
Let $v_B$ be the initial velocity of bob B.
Let $v_A'$ be the final velocity of bob A.
Let $v_B'$ be the final velocity of bob B.
We can use conservation of energy to find the velocity of bob A just before the collision:
$KE = U_g$
$\frac{1}{2}m_Av_A^2 = m_Agh$
$v_A = \sqrt{2gh}$
$v_A = \sqrt{(2)(9.80~m/s^2)(5.1~m)}$
$v_A = 10.0~m/s$
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 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.
$m_A~v_A + m_B~v_B = m_A~v_B' - m_Av_A + m_Av_B+ m_B~v_B'$
$v_B' = \frac{2m_A~v_A+m_B~v_B -m_Av_B}{m_A+m_B}$
$v_B' = \frac{2m_A~v_A+2m_A~v_B -m_Av_B}{m_A+2m_A}$
$v_B' = \frac{2~v_A+2~v_B -v_B}{3}$
$v_Bâ€™ = \frac{(2)(10.0~m/s)+(2)(0)- (0)}{3}$
$v_B' = 6.67~m/s$
We can use $v_B'$ to find $v_A'$.
$v_A' = v_B' - v_A + v_B$
$v_A' = 6.67~m/s - 10.0~m/s + 0$
$v_A' = -3.33~m/s$
After the collision, bob A is moving in the opposite direction at a speed of 3.33 m/s
After the collision, bob B is moving at a speed of 6.67 m/s
We can use conservation of energy to find the height that bob A reaches:
$U_g = KE$
$m_Agh = \frac{1}{2}m_A(v_A')^2$
$h = \frac{(v_A')^2}{2g}$
$h = \frac{(3.33~m/s)^2}{(2)(9.80~m/s^2)}$
$h = 0.57~m$
We can use conservation of energy to find the height that bob B reaches:
$U_g = KE$
$m_Bgh = \frac{1}{2}m_B(v_B')^2$
$h = \frac{(v_B')^2}{2g}$
$h = \frac{(6.67~m/s)^2}{(2)(9.80~m/s^2)}$
$h = 2.3~m$
After the collision, bob A rises to a height of $0.57~m$ and bob B rises to a height of $2.3~m$