Momentum and impulse of a volleyball

[*intheclouds*]

OK I have 1 more question for today...

Homework Statement

A 0.45kg volleyball travels witha horizontal velocity of 3.2m/s over the net. You jump up an hit the ball back witha horizontal velocity 7 m/s. If the contact time is 0.047s, what is the average force on the ball?

Homework Equations

Here are all the equations that were in our notes for this section.
p (momentum) There was triangle P and triange t in my notes, so I wrote "change in"

p=mv
f=ma
f*"change in"t="change in"p
f*"change in"t=mv_f-mv_i
"change in" P=mv_f-mv_i

The Attempt at a Solution

Again, I wasnt really sure what equation to use...

[(.45kg)(-3.8m/s)-(.45kg)(3.2m/s)]/(.047s)

I got -67.0, but it wasn't right...

Please help...=]

 

[asleight]

OK I have 1 more question for today...

Homework Statement

A 0.45kg volleyball travels witha horizontal velocity of 3.2m/s over the net. You jump up an hit the ball back witha horizontal velocity 7 m/s. If the contact time is 0.047s, what is the average force on the ball?

Homework Equations

Here are all the equations that were in our notes for this section.
p (momentum) There was triangle P and triange t in my notes, so I wrote "change in"

p=mv
f=ma
f*"change in"t="change in"p
f*"change in"t=mv_f-mv_i
"change in" P=mv_f-mv_i

The Attempt at a Solution

Again, I wasnt really sure what equation to use...

[(.45kg)(-3.8m/s)-(.45kg)(3.2m/s)]/(.047s)

I got -67.0, but it wasn't right...

Please help...=]

\textbf{f}=\Delta \textbf{p}/\Delta t.

 

[*intheclouds*]

Ok the change in p would be...
(.45kg)(3.2m/s)-(.45kg)(7m/s)?
Im not really sure if i could just use the numbers from the problem or if i had to change the velocities or if they were in the correct order...

 

[asleight]

Ok the change in p would be...
(.45kg)(3.2m/s)-(.45kg)(7m/s)?
Im not really sure if i could just use the numbers from the problem or if i had to change the velocities or if they were in the correct order...

\Delta \textbf{p} = m(\textbf{v}_f-\textbf{v}_i)/t, where the final velocity is 7 m/s in the negative direction and the initial velocity is 3.2 m/s in the positive direction.

 

[*intheclouds*]

so it would be:
.45kg(-7+3.2)/.047??

 

[asleight]

so it would be:
.45kg(-7+3.2)/.047??

Close.

The final velocity is -7, the initial velocity is 3.2. But, the change in momentum is given by the change in velocity. Id est, final minus initial.

 

[*intheclouds*]

Oh, so:
.45kg(3.2-7)/.047??

 

[asleight]

Oh, so:
.45kg(3.2-7)/.047??

Final, -7, minus initial, +3.2, =>

0.45kg (-7m/s - +3.2m/s)/0.047s =

0.45kg (-10.2m/s) / 0.047s =...

 

[*intheclouds*]

Wow. Thank you again. You are amazing...=]

 

[asleight]

Wow. Thank you again. You are amazing...=]

No big deal.