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We will consider three separate collisions: the bottom ball with the second ball, the second with the third, and the third with the top. We will assume that all the balls are slightly separated in flight, so that each collision becomes a separate interaction. The largest ball (Ball 1) strikes the ground first, and undergoes a perfectly elastic collision with the ground, so that the magnitude of its velocity is unchanged, but it completely reverses direction, as shown in Schematic for the First Collision  under "Diagrammatic Representation" below.

For each collision, we will consider the system to be made up of the two balls that are colliding. We will assume that each collision is instantaneous, so that external forces will provide negligible . We are told to assume that the collisions are elastic, so we will assume there is no non-conservative work done on any of the systems.

plus .

\begin{large}\[ m_{A}v_{A,y,i} + m_{B}v_{B,y,i} = m_{A}v_{A,y,f} + m_{B}v_{B,y,f}\]
\[ \frac{1}{2}m_{A}v_{A,y,i}^{2} + \frac{1}{2}m_{B}v_{B,y,i}^{2} = \frac{1}{2}m_{A}v_{A,y,f}^{2} + \frac{1}{2}m_{B}v_{B,y,f}^{2} \]\end{large}

\begin{large}\[ v_{A,y,f} = \frac{2m_{B}}{m_{A}+m_{B}}v_{B,y,i} + \frac{m_{A}-m_{B}}{m_{A}+m_{B}}v_{A,y,i}\]\end{large}

\begin{large}\[ v_{B,y,f} = \frac{2m_{A}}{m_{A}+m_{B}}v_{A,y,i} + \frac{m_{B}-m_{A}}{m_{A}+m_{B}} v_{B,y,i}\]\end{large}

If this is not obvious, simply apply our collision formula to the collision between ball 1 and the earth, treating the earth as a ball of infinite mass.

\begin{large}\[ v_{2,1} = \frac{2m_{1}}{m_{2}+m_{1}}v + \frac{m_{2}-m_{1}}{m_{1}+m_{2}}(-v) = \frac{3m_{1}-m_{2}}{m_{1}+m_{2}}v\]\end{large}

\begin{large}\[ v_{3,2} = \frac{2m_{2}}{m_{3}+m_{2}}\left(\frac{3m_{1}-m_{2}}{m_{1}+m_{2}} v\right) + \frac{m_{3}-m_{2}}{m_{3}+m_{2}}(-v) = \frac{7m_{2}m_{1}-m_{2}^{2}-m_{1}m_{3}-m_{2}m_{3}}{(m_{1}+m_{2})(m_{2}+m_{3})} \:v\]\end{large}

\begin{large}\[ v_{4,3} = \frac{2m_{3}}{m_{4}+m_{3}}\frac{7m_{2}m_{1}-m_{2}^{2}-m_{1}m_{3}-m_{2}m_{3}}{(m_{1}+m_{2})(m_{2}+m_{3})} \:v + \frac{m_{4}-m_{3}}{m_{4}+m_{3}}(-v) \]
\[= \frac{15m_{1}m_{2}m_{3} -m_{2}^{2}m_{3}-m_{1}m_{3}^{2}-m_{2}m_{3}^{2} - m_{1}m_{3}m_{4}-m_{1}m_{2}m_{4}-m_{2}^{2}m_{4}-m_{2}m_{3}m_{4}}{(m_{1}+m_{2})(m_{2}+m_{3})(m_{3}+m_{4})}\:\:v\]\end{large}

\begin{large}\[ v_{4,3} = 5.0 v\]\end{large}

The manufacturer claims that the ball will rise to over 5 times its initial height. What does this claim imply about the accuracy of our assumptions?

\begin{large}\[ K_{i} = \frac{1}{2}(m_{1}+m_{2}+m_{3}+m_{4})v^{2} \]\end{large}

\begin{large}\[ \frac{K_{f,4}}{K_{i}} = \frac{\frac{1}{2}m_{4}(5v)^{2}}{\frac{1}{2}(m_{1}+m_{2}+m_{3}+m_{4})v^{2}} = 0.91\]\end{large}

The red ball carries away about 90% of the initial kinetic energy of the system!

\begin{large}\[ v_{3,4} = \frac{2m_{4}}{m_{4}+m_{3}}(-v) + \frac{m_{3}-m_{4}}{m_{4}+m_{3}}\left(\frac{7m_{2}m_{1}-m_{2}^{2}-m_{1}m_{3}-m_{2}m_{3}}{(m_{1}+m_{2})(m_{2}+m_{3})}\right) \:v
= 0.79v\]\end{large}

\begin{large}\[ \frac{K_{f,3}}{K_{i}} = \frac{\frac{1}{2}m_{3}(0.79v)^{2}}{\frac{1}{2}(m_{1}+m_{2}+m_{3}+m_{4})v^{2}} = 0.057 \]\end{large}

Compare these percentages to the percent of the kinetic energy carried by each ball before contacting the ground.