Ball colliding with inclined stationary plane

Question

A particle with mass m1 is dropped from a height onto an incline plane creating the angle $\alpha$ with the ground (the slope is frictionless). The coefficient of restitution is e. Find the angle $\beta$ that the ball's velocity after the bounce creates with the line perpendicular to the slope.

Here is how I want to solve it:

Call $v_1$ the velocity of the ball before the collision and $v_1'$ the velocity after. Let also $v_2$ be the velocity of the slope before, and $v_2'$ the velocity after. Each of these velocities have one component parallel to the slope (denoted by p) and one normal to the slope (denoted by n). For example, $v_{1,p}$ is the component of $v_1$ that is parallel to the plane.

If I only include the ball and the slope in my system (frame), then the slope can be considered stationary, and $v_{2,p} = v_{2,n} = v_{2,p}' = v_{2,n}' = 0$.

The line of impact is in the normal direction, so the formula for e is $$e = \frac{v_{2,n}' - v_{1,n}'}{v_{1,n} - v_{2,n}} = \frac{-v_{1,n}'}{-v_{1,n}} = \frac{v_1'\cos\beta}{v_1\cos\alpha}$$

Now I will use the conservation of momentum in the direction parallel to the plane: $$m_1 v_{1,p} + m_2 v_{2,p} = m_1 v_{1,p}' + m_2 v_{2,p}' \implies v_{1,p} = v_{1,p}' \\ \implies v_1 \sin \alpha = v_1' \sin \beta \\ \implies \frac{v_1'}{v_1} = \frac{\sin \alpha}{\sin\beta}.$$

When I put this expression into the formula for $e$, I get $$e = \frac{\tan \alpha}{\tan \beta} \implies \tan \beta = \frac{\tan \alpha}{e}$$ which is the correct answer. However, here comes my question. Why can't I use the law of conservation of momentum on the direction normal to the slope? The calculations are: $$- m_1 v_{1,n} + m_2 v_{2,n} = m_1 v_{1,n}' + m_2 v_{2,n}' \implies -v_{1,n} = v_{1,n}' \implies \frac{v_{1,n}'}{v_{1,n}} = -1$$

And if I were to put this result into the formula for $e$, I would get that $e = -1$ which doesn't make any sense at all. The only reason (as far as I can see) I get two different answers depending on which direction I use when applying conservation of momentum, is that in the normal direction the momentum is not conserved, possible because of the presence of an external force (gravity from earth). But if the force from gravity points vertically downwards, it has one component normal to the slope and one component parallel to the slop. Thus, there is an external force in both the normal and parallel directions, and neither should conserve momentum! So, why is momentum conserved in the parallel direction but not in the normal?

Answer 1

The inclined plane exerts a force in the normal direction which gives a pulse in the normal direction, so the momentum in normal direction is reversed.

Answer 2

$\def \b {\mathbf}$

with these equations

$$m\,(\b V_1-\b U_1)=-\lambda\, \b n\\ M\,\b V_2=\lambda \,\b n\\ \left(\b V_2-\b V_1-\epsilon\,\b U_1\right)\cdot \b n=0$$

where $~\b U_1~$ is the start velocity and $~\b V_i~$ are the velocities after the collision.

$$\begin{align*} & \b U_1=-v_1\begin{bmatrix} \sin(\alpha)\\ \cos(\alpha) \\ \end{bmatrix}\quad, \b n=\begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}\\ \end{align*} $$

you obtain the velocities after the collision $~\b V_1~,\b V_2~$ in x,y system . taking $~M\mapsto \infty~$ thus:

$$\b V_2=\b 0\quad, \begin{align*} \b V_1=v_1\,\begin{bmatrix} -\sin(\alpha) \\ \epsilon\,\cos(\alpha) \\ \end{bmatrix}\\ \end{align*} $$ $$\tan(\beta)=-\frac{\epsilon\,\cos(\alpha)}{\sin(\alpha)}$$

the conservation of the linear impulse is:

$$\left[m\,(\b V_1-\b U_1)+ M\,V_2\right]\bigg|_{M\mapsto \infty}=\b 0$$