# Homework Help: Momentum and impulse of a volleyball

1. Nov 9, 2008

### *intheclouds*

OK I have 1 more question for today...

1. The problem statement, all variables and given/known data

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?

2. Relevant 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=mvf-mvi
"change in" P=mvf-mvi

3. 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...

2. Nov 9, 2008

### asleight

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

3. Nov 9, 2008

### *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...

4. Nov 9, 2008

### asleight

$$\Delta \textbf{p} = m(\textbf{v}_f-\textbf{v}_i)/t$$, 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.

5. Nov 9, 2008

### *intheclouds*

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

6. Nov 9, 2008

### asleight

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.

7. Nov 9, 2008

### *intheclouds*

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

8. Nov 9, 2008

### asleight

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

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

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

9. Nov 9, 2008

### *intheclouds*

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

10. Nov 9, 2008

No big deal.