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

(3 m_{2} - m_{1}) u / (m_{1} + m_{2})

and

(m_{2} - 3 m_{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 - 3 r)^{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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