Change in momentum, impulse force calculations

[Yousufshad]

Homework Statement

A 178.0g ball is dropped from a height of 2.99m, bounces on a hard floor, and rebounds to a height of 1.36m. The impulse received from the floor is shown below.

(Is a picture of a graph, Y-axis is force, X-axis is time. No numbers are labelled and is simply a line starting at 0 and goes to a peak and back down to zero like a triangle)

What maximum force does the floor exert on the ball if it is exerted for 2.00ms

Homework Equations

Impulse = Change in Momentum

Momentum = mv
Ui = mgy
KE = 1/2 mv^2

The Attempt at a Solution

[/B]
Ui = mgy
=5.2157
1/2 mv^2 = 5.2157
v=7.655m/s

(to find velocity just as ball hits the ground)

Pi = (0.178)(7.655)
=1.3626

(Initial Momentum)

Finding velocity just as it bounces back from floor
Uf = KEi
v=5.1629m/s

Pf = 0.919

(final momentum)

Change in momentum = 0.433

DeltaP(change in momentum) = 1/2 Fmax (0.002)

Fmax = 443N (not correct answer) where did I go wrong?

 

[gneill]

Momentum is a vector quantity. You haven't taken into account the change in direction at the bounce.

 

[TSny]

Did you take into account that momentum is a vector quantity, so it has direction? How does the final direction of momentum compare to the initial direction of momentum?

[EDIT: oops, my post is redundant to gneill's.]

 

[Yousufshad]

Momentum is a vector quantity. You haven't taken into account the change in direction at the bounce.

Ouch, thanks a lot that negative sign fixed things up!

DeltaP =2.2816 got me the right answer!

 

[Yousufshad]

Did you take into account that momentum is a vector quantity, so it has direction? How does the final direction of momentum compare to the initial direction of momentum?

Yeah that fixed things up :)