Motion of an object inside an open box

Question:
As shown in the figure, an open box is put on frictionless floor; then, an object is placed in the middle of the box. There is no friction between the object and the surface of the box. The masses of the object and the box are equal. The coefficient of restitution between the object and inside box is $e \ (0<e<1)$.
(1) Let the object move at $v_i$ in the positive $x$ direction. Find the velocity of the object related to the floor after colliding the inside box.
(2) What is the position of the middle of the box after the second collision in the other side?

Answer:
(1) There is no external force; therefore, we can consider conservation of momentum. Let $v$ and $u$ be the velocities of the object and the box, respectively. The subscripts, $i$ and $f$, represent initial and final. Thus, we have
$$\begin{equation} mv_i = mv_f + mu_f \end{equation}$$
The coefficient of restitution is defined as the relative velocity after collision divided by the relative velocity before collision. Namely,
$$\begin{equation} e = \frac{v_f - u_f}{0 - v_i} \end{equation}$$
From (1) and (2), we have
$$\begin{eqnarray} v_i &=& v_f + u_f \\ -v_i e &=& v_f - u_f \end{eqnarray}$$
Let's find $v_f$ in terms of $v_i$ by eliminating $u_f$.
$$\begin{equation} v_f = \frac{1-e}{2}v_i \end{equation}$$

(2) First, we find the final velocity of the box by eliminating $v_f$ from (3) and (4).
$$\begin{equation} u_f = \frac{1+e}{2}v_i \end{equation}$$
The time from the first to the second collision is the distance, $L$, divided by the relative velocities.
$$\begin{equation} t = \frac{L}{u_f - v_f} = \frac{L}{ev_i} \end{equation}$$
Therefore, the distance that the box moved is
$$d = t \times u_f = \frac{L}{ev_i} \times \frac{1+e}{2}v_i = \frac{1+e}{2e}L$$

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