Momentum and impulse of a ball dropped

[indietro]

Homework Statement

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?

Homework Equations

Jx = [tex]\Delta[/tex]p
Fmax*[tex]\Delta[/tex]t = Jx

The Attempt at a Solution

so i though that
Jx = [tex]\Delta[/tex]p
= mv_f - mv_i
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!

 

[indietro]

anyone??

 

[cepheid]

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.

 

[indietro]

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 [tex]\Delta[/tex] t). It has 0 F until t₁ and that force is applied until t₂ (it is a triangle with the point at Ft_max). and then the force is 0 again

 

[indietro]

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