= Two-ball elastic bounce
In the separated-impact approximation, an upper ball of <mass> $m_1$ falling at <speed> $u$ meets a lower ball of <mass> $m_2$ after the latter rebounds elastically from a fixed floor. Their outgoing upward <velocities> are $(3m_2-m_1)u/(m_1+m_2)$ and $(m_2-3m_1)u/(m_1+m_2)$. These follow from <conservation of momentum> and reversal of relative <velocity>. The upper ball receives a fraction $r(3-r)^2/(1+r)^3$ of the initial <kinetic energy>, where $r=m_1/m_2$; subtracting this fraction from unity gives $(1-3r)^2/(1+r)^3$. Hence complete transfer occurs at $r=1/3$, whereas a very light upper ball rebounds to nine times the common drop height.
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