A tennis ball of mass 57.0 g is held just above a basketball of mass 577 g. With

Question

A tennis ball of mass 57.0 g is held just above a basketball of mass 577 g. With their centers vertically aligned, both balls are released from rest at the same time, to fall through a distance of 1.00 m, as shown in the figure below. (a) Find the magnitude of the downward velocity with which the basketball reaches the ground. (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?

Explanation / Answer

A) we know that,

v = sqrt(2as)

= sqrt(2 * 9.807 m/s^2 * 1 m)

= 4.428769581 m/s

= 4.43 m/s ans

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B) , let
u1 = before collision velocity of the tennis ball = -4.43 m/s
u2 = before collision velocity of the basketball = 4.43 m/s
v1 = after collision velocity of the tennis ball
v2 = after collision velocity of the basketball
m1 = mass of the tennis ball = 57.0 g
m2 = mass of the basketball = 577. 0 g

Then conservation of energy says
(m1u12)/2 + (m2u22)/2 = (m1v12)/2 + (m2v22)/2

and conservation of momentum says
m1u1 + m2u2 = m1v1 + m2v2

So the solution for v1 is
v1 = (u1(m1 - m2) + 2m2u2) / (m1 + m2)
v1 = ((-4.429 m/s)(57.0 g - 577 g) + 2(577 g)(4.429 m/s)) / (57.0 g + 577 g)

V1= 11.69 m/s

now we have an upwards velocity for the tennis ball. The kinetic energy for this will be converted entirely into potential energy when the tennis ball reaches maximum height, so
KE = (1/2)m1. v12= PE = F x d = m1x g x d
Where
KE = Kinetic Energy of the tennis ball immediately after the collision
PE = Potential Energy of the tennis ball at the peak of it's bounce up
g = acceleration due to gravity
d = height the ball is at when it reaches the peak
So
(1/2)m1. v12= m1x g x d
(1/2) x v12= g x d
d = v12/(2 x g)
d = (11.69 m/s)2/(2 x 9.807 m/s2)
d = 6.97 m ans

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