Five Balls Bounce

All the balls start hitting each other after having gained the same velocity $v= \sqrt{2gh}$ , where $h=1m$ is the distance of the bottom ball from the ground. The best configuration of masses is one where $m_1 >> m_2 >> \dots>>m_5$ since it allows the highest fraction of kinetic energy to be associated with the fifth ball (a massive object hitting another with smaller mass will make it bounce off with the highest possible velocity). One hit can be thought as a superposition of the following two (bouncing off a wall and the bouncing off a massive object of a body at rest), as showed in the following picture: The velocity of the second ball will be $2\cdot v+v=3v$ , that of the third one $2\cdot(3v)+1=7v$ ,...,the fifth one $31v$ . Conservation of energy gives $\frac{1}{2} \cdot 31^2 \cdot (\sqrt{2gh})^2=gH$ , hence $d=H-h=961m^{-1}h^2-h=960m$ .

Note:considering $v=1$ in general $v_{n+1}=2v_n+1$ , equivalent to $v_{n+1}-2v_n=1$ , satisfied by $v_n=2^n-1$ . This consideration can be used to produce a formal proof in conjunction with mathematical induction.

Suppose that the balls have masses $m_1 < m_2 < m_3 < m_4 < m_5$ from the top down. We can solve this problem by considering a sequence of collisions, imagining that $m_5$ collides with the floor first, then $m_5$ collides with $m_4$ , then $m_4$ collides with $m_3$ , then $m_3$ collides with $m_2$ , and finally $m_2$ collides with $m_1$ , where all five balls are initially travelling downwards with speed $v = \sqrt{2g}$ . Let $v_5 = v$ be the upward speed of $m_5$ after its collision with the floor, let $v_4$ be the upward speed of $m_4$ after its collision with $m_5$ , let $v_3$ be the upward speed of $m_3$ after its collision with $m_4$ , let $v_2$ be the upward speed of $m_2$ after its collision with $m_3$ , and let $v_1$ be the upward speed of $m_1$ after its collision with $m_2$ . Of course, the final speeds of $m_2,m_3,m_4,m_5$ after all the collisions will not be the same as $v_2,v_3,v_4,v_5$ , but $v_1$ will be the final speed of $m_1$ , and that is what we want to know.

If a particle of mass $m$ is moving downwards with speed $v$ , and collides with a particle of mass $M > m$ that is moving upwards with speed $V$ then using conservation of momentum and Newton's Law of Restitution (with $e=1$ to conserve energy) we deduce that the particle $m$ has upward speed $W$ after the collision, where $W \; = \; \frac{2MV + (M-m)v}{M+m}$ and since $(2V+v) - W \; = \; \frac{2mv}{M+m}$ we deduce that $W < 2V+v$ .

Thus we deduce that $v_4 < 2v_5 + v = 3v$ , $v_3 < 2v_4 + v = 7v$ , $v_2 < 2v_3 + v = 15v$ , while $v_1 < 2v_2 + v = 31v$ . Thus, if the particle $m_1$ rises to a height of $h$ metres above its starting position, then $h < 31^2 - 1 = \boxed{960}$ . However, it is important to note that all these inequalities are strict , and so it is not possible for $h$ to be equal to $960$ .

We can achieve any height less than $960$ m, however. Considering the $m,M$ ball collision above, we obtain $W = 2V+v$ in the limit as $\tfrac{m}{M} \to 0$ . The "outcome" $v_5=v$ , $v_4=3v$ , $v_3=7v$ , $v_2 = 15v$ , $v_1 = 31v$ is an approximation of the case $m_1 \ll m_2 \ll m_3 \ll m_4 \ll m_5$ where we consider that each ball is very much more massive than the one above it. As an approximation , then, we say that each lower ball's speed is unaffected by its collision with the ball above it. However, in the real world, it is not possible to choose $m_5$ infinitely greater than $m_4$ , $m_4$ infinitely greater than $m_3$ , $m_3$ infinitely greater than $m_2$ , and $m_2$ infinitely greater than $m_1$ , which is what would be required to achieve $960$ metres.