Momentum and impulse of a ball dropped

1. Oct 18, 2009

indietro

1. The problem statement, all variables and given/known data
A 231.0 g ball is dropped from a height of 2.60 m, bounces on a hard floor, and rebounds to a height of 1.37 m. The impulse received from the floor is shown below. What maximum force does the floor exert on the ball if it is exerted for 2.00 ms?

2. Relevant equations
Jx = $$\Delta$$p
Fmax*$$\Delta$$t = Jx

3. The attempt at a solution
so i though that
Jx = $$\Delta$$p
= mvf - mvi
but for this question you have to add the momentums.. is that because at first the momentum is pointing down and then after the collision the momentum is pointing up so they have opposite signs??

thanks for the help!

2. Oct 19, 2009

indietro

Re: Impulse/Momentum

anyone??

3. Oct 19, 2009

cepheid

Staff Emeritus
Re: Impulse/Momentum

It would help if we could actually see the impulse "shown below".

To answer your question: you are still subtracting the momenta. It's just that momentum is a vector quantity (direction is important). Therefore, since one of the momenta has the opposite sign, subtracting a negative = adding.

Think about it intuitively. If you bounce a ball off a wall and it comes off having the same speed as it impacted with, then its change in momentum is not zero. In fact, its change in momentum is quite large (equal to TWICE the magnitude of the momentum it originally had). This is because it has reversed directions.

4. Oct 19, 2009

indietro

Re: Impulse/Momentum

sorry i didnt include the picture just that last time they took so long to approve it. but it is an impulse graph (F vs $$\Delta$$ t). It has 0 F until t1 and that force is applied until t2 (it is a triangle with the point at Ftmax). and then the force is 0 again

5. Oct 19, 2009

indietro

Re: Impulse/Momentum

but thank you for the explanation, now i understand.. so because momentum is a vector quantity the sign matters :)!