A tennis ball of mass 57.0 g is held just above a basketball of mass 580 g.?

a) You should round your answer to at most 3 significant digits because that is the best you can get with the data in this problem (possibly even only 2 significant digits, depending on how precise that value is for the mass of the basketball). Since there was no initial velocity, you can use v = sqrt(2as) = sqrt(2 * 9.807 m/s^2 * 1 m) = 4.428769581 m/s = 4.43 m/s as you basically have.

b) Because this is an elastic collision, we must use both the conservation of momentum and the conservation of energy to solve the problem. (Note the first answer is incorrect as it incorrectly assumes that the after collision velocity of the basketball is negligible.) So, 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 = 589 g

Then conservation of energy says

m1u1^2/2 + m2u2^2/2 = m1v1^2/2 + m2v2^2/2

and conservation of momentum says

m1u1 + m2u2 = m1v1 + m2v2

The only two true unknowns in these two equations are v1 and v2 and with two equations you can solve them. However, the solution is long and messy (I already did one of these types of problems earlier today), so I am going to cheat a little and give you the answer. (Note, if you need to show all your work, a good tip is to change the frame of reference so that one of the initial velocities is zero and then convert back after you have the solution – see my source.)

So the solution for v1 is

v1 = (u1(m1 – m2) + 2m2u2) / (m1 + m2)

v1 = ((-4.429 m/s)(57.0 g – 580 g) + 2(580 g)(4.429 m/s)) / (57.0 g + 580 g)

v1 = ((-4.429 m/s)(-523 g) + (1160 g)(4.429 m/s))/(637 g)

v1 = ((2316 m/s) + (5138 m/s))/637

v1 = (7454 m/s)/637 = 11.7 m/s

(FYI, v2 is about 2.84 m/s, and you can plug these figures back into the two conservation equations and you will see that they are correct.)

Anyway, now we have an upwards velocity for the tennis ball. The kinetic energy for for this will be converted entirely into potential energy when the tennis ball reaches maximum height, so

KE = (1/2)m1*v1^2 = PE = F*d = m1*g*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*v1^2 = m1*g*d

(1/2)*v1^2 = g*d

d = v1^2/(2*g)

d = (11.7 m/s)^2/(2 * 9.807 m/s^2)

d = 6.98 m

(You can also use the standard kinematic equations to get this distance, the result is the same.)

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