# Homework Help: Elastic Collision Problem

1. Dec 4, 2007

### latitude

1. The problem statement, all variables and given/known data

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.

2. Relevant equations
Emech = K + U
Emechi = Emechf
mvi = mvf
Ki = Kf

3. 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.

2. Dec 4, 2007

### 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.