A small ball of mass m is aligned above a larger ball of mass M = 0.72 kg (with

Question

A small ball of mass m is aligned above a larger ball of mass M = 0.72 kg (with a slight separation, as with the baseball and basketball of Figure 9-67a), and the two are dropped simultaneously from height h = 1.8 m. (Assume the radius of each ball is negligible relative to h.)

(a) If the larger ball rebounds elastically from the floor and then the small ball rebounds elastically from the larger ball, what value of m results in the larger ball stopping when it collides with the small ball?
kg
(b) What height does the small ball then reach (Figure 9-67b)?
m

Explanation / Answer

use conservation of mechanical energy

initially KE = 0 ; PE = Mgh

final KE = 1/2*Mv^2

v = sqrt(2gh)

vMf = (M-m)*vMi/(M+m) + 2m*vmi/(M+m)

= (M-m)*sqrt(2gh)/(M+m) - 2m*sqrt(2gh)/(M+m)

vMf = (M-3m)*sqrt(2gh)/(M+m)

for this to be zero

m = M/3

m = 0.72/3 = 0.24 kg

part b )

substitute M = 3m

vmf = 2*sqrt(2gh)

here mechanical energy conserve to h'

1/2*m*vmf^2 = mgh'

h' = vmf^2/2g = 4h

h = 1.8m

h' = 7.2 m

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