A 22 oz basketball with circumference 29.5 inches is dropped from rest from a he

Question

A 22 oz basketball with circumference 29.5 inches is dropped from rest from a height of 2 meters (2 meters from the floor to its center of mass). A small ball with mass 35 g is initially positioned on top of the basketball, and the two are dropped together simultaneously. The basketball collides elastically with the floor, and the small ball collides elastically with the basketball.

(a) What is the speed of the small ball immediately after colliding with the basketball?

(b) What is the maximum height reached by the small ball after the collision?

Explanation / Answer

velocity of basketball before collision to floor = sqrt(2gH)= sqrt ( 2*9.8*2) = 6.26 m/s

velocity of basketball after collision to floor = 6.26 m/s (elastic collision)

velocity of small before collision to basketball = sqrt(2gH) = sqrt ( 2*9.8*(2-0.75)) = 4.94 m/s ( 29.5 inch = 0.75 m)

momentum conservation

m1u1 + m2u2 = m1v1 + m2v2 ( it's linear ...no square )

0.035*4.94 + 0.623.69*6.26 = 0.035v1+ 0.623.69v2 ( mass in kg , velocity in m/s ....SI unit)

=> 4.072 = 0.035v1+ 0.623.69v2

=> 116.36 = v1 + 17.81v2

v1-v2 = (4.94+6.26) = 11.2 ( velocity of separation = e( velocity of appraoch) , e= 1 in this case )

substract eq 2 from 1

116.36-11.2 = 18.81v2

=> v2 = 5.59 m/s

=> v1 = 5.59+11.2 = 16.79 m/s

b) max. height = v1^2/2g = (16.79)^2/(2*9.8) = 14.38 m

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