Problem source

A uniform smooth wedge of mass $m$ has congruent triangular end faces $A_{1}B_{1}C_{1}$ and $A_{2}B_{2}C_{2},$ and $A_{1}A_{2},B_{1}B_{2}$ and $C_{1}C_{2}$ are perpendicular to these faces. The points $A,B$
and $C$ are the midpoints of $A_{1}A_{2},B_{1}B_{2}$ and $C_{1}C_{2}$ respectively. The sides of the triangle $ABC$ have lengths $AB=AC=5a$ and $BC=6a.$ The wedge is placed with $BC$ on a smooth horizontal table, a particle of mass $2m$ is placed at $A$ on $AC,$ and the system is released from rest. The particle slides down $AC,$ strikes the table, bounces perfectly elastically and lands again on the table at $D$. At this time the point $C$ of the wedge has reached the point $E$. 
Show that $DE=\frac{192}{19}a.$

Solution source

Conservation of energy, tells us that $2m \cdot g \cdot 4a = 8amg$ is equal to $\frac12 m v_{wedge}^2 + \frac12(2m)v_{particle}^2$.

Conservation of momentum (horizontally) tells us that $m v_{wedge}+2mv_{particle, \rightarrow} = 0 \Rightarrow v_{particle, \rightarrow} = -\frac12 v_{wedge}$.

\begin{center}
        \begin{tikzpicture}[scale = 2]

        \coordinate (O) at (0,0);
        \coordinate (Y) at (0,4);
        \coordinate (X) at (3,0);

        \draw[-latex, red, thick] (Y) -- ($(O)!0.025!(Y)$);
        \draw[-latex, red, thick] (O) -- ($(O)!0.975!(X)$);
        \draw[-latex, red, thick] (Y) -- ($(Y)!0.975!(X)$);

        \node[below] at ($(O)!0.5!(X)$) {$v_{wedge}+\frac12v_{wedge}$};

        \node[left] at ($(O)!0.5!(Y)$) {$v_{particle, \downarrow}$};

        \pic [draw, angle radius=1cm, angle eccentricity=1.25, "$\theta$"] {angle = Y--X--O};
        
    \end{tikzpicture}
\end{center}

We know that the particle must remain on the slope, and so $v_{particle,\downarrow} = \frac{4}{3} \frac{3}{2} v_{wedge} = 2v_{wedge}$.

In conclusion, we have:

\begin{align*}
&& 8amg &= \frac12 m v_{wedge}^2 + \frac12 (2m)\left ((-\tfrac12 v_{wedge})^2 + (2v_{wedge})^2 \right ) \\
&&&= \frac{19}{4}mv_{wedge}^2 \\
\Rightarrow && v_{wedge}^2 &= \frac{32}{19}ag
\end{align*}.

To calculate the time the ball bounces for, note that:

$s = ut + \frac12 at^2 \Rightarrow 0 = 2v_{wedge} - \frac12 gt \Rightarrow t = \frac{4v_{wedge}}{g}$. During this time, the wedge (and ball) who horizontally are moving apart with speed $\frac32 v_{wedge}$ we have they move apart by:

\begin{align*}
&& DE &= \underbrace{\frac32 v_{wedge}}_{\text{speed they move apart}} \cdot  \underbrace{\frac{4v_{wedge}}{g}}_{\text{time they are moving apart for}} \\
&&&= \frac{6}{g} v_{wedge}^2 \\
&&&= \frac{6}{g}\frac{32}{19}ag \\
&&&= \frac{192}{19}a
\end{align*}