linear momentum question?

Question

A superball with a mass of 47.3g is dropped from a height of 3.6m. it rebounds to a height of 3.2m.?
The acceleration of gravity is 9.8m/s/s. What is the change in its linear momentum during the collision with the floor? Answer in units of kg m/s.

Answer 1

Use kinematics to find the speed of the ball just before the collision:

2 a ( y - yo ) = v1^2 - vo^2

a = local acceleration ( 9.8 m/s^2 down )
y - yo = vertical distance traveled ( 3.6 m )
v1 = final velocity, what you're looking for
vo = initial velocity ( zero )

Solve for v1.

Now we need to know the rebound velocity, so do more kinematics using the same equation. This time we'll say

2 a ( y - yo ) = v3^2 - v2^2

a = local acceleration ( 9.8 m/s^2 down )
y - yo = vertical distance traveled ( 3.2 m )
v3 = final velocity ( zero )
v2 = initial velocity ( what you're looking for )

The initial momentum is m v1 and the final momentum is m v2. The change in momentum is final momentum - initial momentum. Note that v2 has the opposite sign as v1; one will be positive and the other negative.