How Do You Calculate the Impulse of a Ball During a Collision with the Floor?

[crono_]

Homework Statement

A 0.480-kg ball is dropped from rest at a point 1.10 m above the floor. The ball rebounds straight upward to a height of 0.880 m. Taking the negative direction to be downward, what is the impulse of the net force applied to the ball during the collision with the floor?

m = 0.480 kg

h₀ = 1.10 m

h_f = 0.880 m

v_o = ?

v_f = ?

a = 9.80 m/s²

Homework Equations

Impulse-Momentum Theorem

J = mv_f - mv_o

v_f² = v_o² + 2ay

The Attempt at a Solution

As we need the velocity of the ball before and after the collision, we can use v_f² = v_o² + 2ay. But I'm wondering, would we solve it once for the v_f using h_o = 1.10 m? As that would determine the final velocity of the ball just before it hits the floor. And then solve it again, but this time for v_o using h_f = 0.880 m? The would determine the initial velocity of the ball as soon as it rebounds up? Right...?

Doing it this way, I came up with:

v_f = 4.64 m/s

v_o = 4.15 m/s

Putting these into the Impulse-Momentum Theorem, v_f from above would now be the v_o as it is the velocity right before impact. This velocity would also be negative as it is pointing downwards. Then v_o from above would be the new v_f as it is the velocity right after impact. Did that make sense?


J = mv_f - mv_o

J = 0.480 kg (4.15 m/s) - [ - 0.480 kg (4.64 m/s)]

J = 1.992 kg*m/s - (- 2.227 kg*m/s)

J = 4.22 kg*m/s

Ummm...Does that look right? I've tried this question a few different ways but haven't been successful yet.

Thanks!

 

[Doc Al]

Looks good to me!