A tennis ball of mass 57.0 g is held just above a basketball of mass 617 g. With

Question

A tennis ball of mass 57.0 g is held just above a basketball of mass 617 g. With their centers vertically aligned, both balls are released from rest at the same time, to fall through a distance of 1.15 m, as shown in the figure below.

(a) Find the magnitude of the downward velocity with which the basketball reaches the ground.
m/s

(b) Assume that an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down. Next, the two balls meet in an elastic collision. To what height does the tennis ball rebound?
m

Explanation / Answer

the speed of v of both ball just before the basketbll reaches the ground

vy^2 = uy^2 + 2gh

uy = 0

v = sqrt(2gh) =sqrt(2*9.8*1.15) = 4.75 m/s

part b )

velocity of both ball before collision

tennis ball velocity = vt = -v

basketball velocity = vb = v

after the elastic collsion as follows

momentum conservation

mt*vti + mb*vbi = mt*vtf + mb*vbf

mt*vtf + mb*vbf = (mb- mt)v .... (1)

condition for perfectly elastic collision

vti - vbi = -(vtf - vbf)

vbf = vtf + vti - vbi = vtf - 2v .. (2)

subsitute the eq(2) in eq(1)

mt*vtf + mb (vtf - 2v) = (mb - mt)v

the upward speed of the tennis ball immediately after the collision is

vtf = (3mb - mt)*v/(mt+mb) = (3mb - mt)*sqrt(2gh)/(mt+mb)

the vertical displacement

dy = vyf^2 - vyi^2 / 2ay

dy = (0 - vtf^2) / 2(-g)

dy = [(3mb - mt)/(mt+mb)]^2* h

dy = 8.15 m

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