What Are the Outcomes of a Tennis Ball and Basketball Elastic Collision?

[latitude]

Homework Statement

A tennis ball of mass 57 g is held just above a basketball of mass 590 g. With their centres verticaly aligned, both are released from rest at the same moment, to fall through a distance of 1.2 m.
a) Find the magnitude of the downward velocity w/ which the basketball reaches the ground
b) Assume 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. What is the velocity of the tennis ball immediately after the collision?
c) Find the height to which the tennis ball rebounds.

Homework Equations

Emech = K + U
Emechi = Emechf
mvi = mvf
Ki = Kf

The Attempt at a Solution

Well, I think I did a) right; I used
Emechi = Ki + Ui
= 1/2mv^2 + mgh
= 1/2(0.59 g)(0)^2 + (0.59)(9.81)(1.2)
= 6.94 J
Emechi = Emechf
6.94 J = 1/2(0.59)v^2 + (0.59)(9.81)(0)
v = 4.85 m/s

But yeah. I can't seem to get b. I've tried using the formula v1i - v2i = -(v1f - v2f)
but I don't know the tennis ball's initial velocity. Is there a way to find this using the whole conservation of momentum and kinetic energy spiel? Thanks.

 

[HallsofIvy]

In an "elastic collision", kinetic energy is conserved. The basketball's upward speed is the same as the downward speed you got in (a). Since the tennis ball has exactly the same acceleration as the basket ball and falls for the same length of time (since the tennis ball is held "just above" the basketball, the time beween the basketball hitting the ground and the tennisball hitting the basketball is neglegible) the speed of the tennis ball downward is the same as the speed of the basketball when they collide- only different signs. Since the answer you give for (a) is positive, I assume that you are taking "downward" to be positive. Just before they collide, the basketballs velocity is -4.85 m/s and the tennis ball's velocity is +4.85 m/s. Now use conservation of momentum and conservation of energy.

 

[ogb p]

I would like to first commend your attempt at solving the problem and using the appropriate equations. Your solution for part a) seems correct.

For part b), you are correct in using the conservation of momentum and kinetic energy equations. However, you are missing one crucial piece of information - the initial velocity of the tennis ball. In order to solve for this, we can use the fact that the two balls have the same initial height and are released at the same time. This means that they both have the same initial potential energy, and therefore the same initial kinetic energy.

Using this information, we can set up the equations:

m1vi1 = m2vi2 (conservation of momentum)
1/2m1vi1^2 = 1/2m2vi2^2 (conservation of kinetic energy)

Substituting in the masses of the two balls, we get:

0.057vi1 = 0.59vi2
0.057vi1^2 = 0.59vi2^2

Solving for vi1 and vi2, we get:

vi1 = 10.35vi2
vi1^2 = 10.35vi2^2

Plugging these values into the original equations, we get:

m1(10.35vi2) = m2vi2
1/2m1(10.35vi2)^2 = 1/2m2vi2^2

Simplifying and solving for vi2, we get:

vi2 = 4.85 m/s

Therefore, the initial velocity of the tennis ball is also 4.85 m/s.

For part c), we can use the same equations and substitute in the initial velocities we just found. However, this time, the final velocity of the tennis ball will be in the opposite direction, as it is rebounding upward.

m1vi1 = m2vi2
1/2m1vi1^2 = 1/2m2vi2^2

Substituting in the masses and initial velocities, we get:

0.057(4.85) = 0.59(-vi2)
0.057(4.85)^2 = 0.59(-vi2)^2

Simplifying and solving for -vi2, we get:

-vi2 = 0.47 m/s

Therefore, the velocity of the