Answer
The block swings through a maximum angle of $53.7^{\circ}$ after the collision.
Work Step by Step
Let $m_A = M$ and let $m_B = 3M$.
Let $v_A'$ be the final velocity of the ball.
Let $v_B'$ be the final velocity of the block.
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.
$m_A~v_A + 0 = m_A~v_B' - m_Av_A + m_B~v_B'$
$v_B' = \frac{2m_A~v_A}{m_A+m_B}$
$v_B' = \frac{(2)(M)(4.00~m/s)}{M+3M}$
$v_B' = 2.00~m/s$
The potential energy at the top of the swing will be equal to the kinetic energy at the bottom.
$P = KE$
$mgh = \frac{1}{2}m(v_B')^2$
$h = \frac{(v_B')^2}{2g}$
$h = \frac{(2.00~m/s)^2}{(2)(9.80~m/s^2)}$
$h = 0.204~m$
We can find the angle $\theta$ after the block swings up to a height of 0.204 meters.
$\cos(\theta) = \frac{0.500~m-0.204~m}{0.500~m}$
$\theta = \arccos(\frac{0.296~m}{0.500~m})$
$\theta = 53.7^{\circ}$
The block swings through a maximum angle of $53.7^{\circ}$ after the collision.
