8.6: Collision against a wall
A special case is if a ball \(A\) collides against a wall. In that case there is normally no momentum conservation, because the wall is connected to the ground, which can exert external forces on the system. Let us consider a reference frame in which the wall has zero velocity \(\overrightarrow{\boldsymbol{v}}_{W}=\overrightarrow{\mathbf{0}}\). Because the mass of the wall \(m_{W}\) is very large compared to the mass of the ball \(\left(m_{W} \gg m_{A}\right)\) and \(\overrightarrow{\boldsymbol{v}}_{W}=\overrightarrow{\mathbf{0}}\) the chosen reference frame is a CoM-frame as can be shown as follows:
\[\overrightarrow{\boldsymbol{v}}_{G}=\frac{m_{A} \overrightarrow{\boldsymbol{v}}_{A}+m_{W} \overrightarrow{\boldsymbol{v}}_{W}}{m_{A}+m_{W}} \approx \frac{m_{A} \overrightarrow{\boldsymbol{v}}_{A}}{m_{W}} \approx \overrightarrow{\mathbf{0}} \tag{8.49} \label{8.49}\]
This shows that ball \(A\) is in a CoM-frame and therefore collides against the wall just like in Figs. 8.4. From Equation 8.27, we find that the final velocity of ball \(A\) with initial velocity \(\overrightarrow{\boldsymbol{v}}_{A 0}\), after colliding with a wall parallel to the \(y\)-axis is:
\[\overrightarrow{\boldsymbol{v}}_{A 2}=-e v_{A 0, x} \hat{\boldsymbol{\imath}}+v_{A 0, y} \hat{\boldsymbol{\jmath}} \tag{8.50} \label{8.50}\]