Cube bouncing off a wall
An elastic cube sliding without friction along a horizontal floor hits a vertical wall with one of its faces parallel to the wall. The coefficient of friction between the wall and the cube is $\mu$. The angle between the direction of the velocity $v$ of the cube and the wall is $\alpha$. What will this angle be after the collision (see Figure P.1.3 for a bird's-eye view of the collision)?
This is Problem 1.3 in A Guide to Physics Problems, Part 1: Mechanics, Relativity, and Electrodynamics. There are things I don't understand in the solution provided in the book.
There are two forces acting on the cube. One is the normal reaction $N(t)$, perpendicular to the wall, and the other is the force of friction $F_{fr}(t)$ parallel to the wall
We expect that, as a result of the collision, the cube’s velocity $\textbf v$ will change to $\textbf v'$. In the direction perpendicular to the wall, the collision is elastic, i.e., the velocity in the $\textbf{x}$ direction merely changes sign: $v'_x=-v_x=-v\sin(\alpha)$. 512
Therefore, the momentum changes by $-2mv\sin(\alpha)$ in the $x$ direction. This change is due to the normal reaction $N(t)$
So, according to <PERSON>’s second law: $$\Delta p_x = -\int_0^\tau N(t) dt = -2mv\sin(\alpha)$$
where $\tau$ is the collision time. If there were no friction, the parallel velocity component would not change and the angle would remain the same. However, in the actual case, the $y$ component changes and $$\Delta p_y = -\int_0^{\min(\tau',\tau)}F_{fr}(t)=-\int_0^{\min(\tau',\tau)}\mu N(t) dt$$
Here $\tau'$ is the time at which the velocity $v_y$ goes to $0$. So $$mv_y'=mv_y -\int_0^{\min(\tau',\tau)}\mu N(t) dt$$
The first thing I can't grasp is why the collision is elastic in the $x$ direction. Admitting this, I'm ok with the equation $\displaystyle \Delta p_x = -\int_0^\tau N(t) dt = -2mv\sin(\alpha)$.
Besides, I don't get why $\displaystyle \Delta p_y = -\int_0^{\min(\tau',\tau)}F_{fr}(t)$. It looks a lot like the previous equation the author got on the $x$-axis, but I don't understand why he considers $\tau'$ as the time at which the velocity $v_y$ goes to $0$, and why the upper bound of the integral has to be $\min(\tau',\tau)$. 512