- #1

Np14

- 27

- 2

## Homework Statement

A 0.500 kg ball is dropped from rest at a point 1.20 m above the floor. The ball rebounds straight upward to a height of 0.700 m. What are the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor?

## Homework Equations

F

_{NET}t = mv - mv

_{o}

k = 1/2mv

^{2}

U

_{g}= ma

_{g}h

## The Attempt at a Solution

I first solved U

_{g}= K as the ball's gravitational energy will be converted to kinetic energy in an elastic collision. Point A is the highest point (1.2 m); point B is the highest point the ball bounces back up (0.7m).

ma

_{g}h = 1/2mv

^{2}

(0.5)(9.8)(1.2) = (1/2)(0.5)(V

_{B})

^{2}

That gives V

_{B}= 4.85 m/s.

F

_{NET}t = mv - mv

_{o}

F

_{NET}t = (0.5)(4.85) - (0.5)(0)

I tried to solve for Impulse using the first equation (the impulse momentum theorem) and got +2.42 N(s) upward. The answer is +4.28 N(s) upward. What did I do incorrectly?