Momentum and impulse of a volleyball

In summary: Just remember to keep track of your units and signs. Good luck with your other questions!In summary, the conversation discusses a problem involving a 0.45kg volleyball with a horizontal velocity of 3.2m/s traveling over a net and being hit back with a horizontal velocity of 7m/s. The question asks to find the average force on the ball given a contact time of 0.047s. The conversation goes on to discuss various equations and solutions, with the final correct equation being f = Δp/Δt. The final calculation is shown to be 0.45kg (-10.2m/s) / 0.047s, resulting in a final answer of approximately -96.4 N
  • #1
*intheclouds*
19
0
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=mvf-mvi
"change in" P=mvf-mvi

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...=]
 
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  • #2
*intheclouds* said:
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=mvf-mvi
"change in" P=mvf-mvi

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...=]

[tex]\textbf{f}=\Delta \textbf{p}/\Delta t[/tex].
 
  • #3
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
*intheclouds* said:
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...

[tex]\Delta \textbf{p} = m(\textbf{v}_f-\textbf{v}_i)/t[/tex], 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
so it would be:
.45kg(-7+3.2)/.047??
 
  • #6
*intheclouds* said:
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.
 
  • #7
Oh, so:
.45kg(3.2-7)/.047??
 
  • #8
*intheclouds* said:
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 =...
 
  • #9
Wow. Thank you again. You are amazing...=]
 
  • #10
*intheclouds* said:
Wow. Thank you again. You are amazing...=]

No big deal.
 

1. What is momentum?

Momentum is a measure of an object's mass and velocity. In simpler terms, it is the amount of motion an object has.

2. How is momentum calculated?

Momentum is calculated by multiplying an object's mass by its velocity. The equation for momentum is p = m * v, where p is momentum, m is mass, and v is velocity.

3. What is the difference between momentum and impulse?

Momentum and impulse are closely related concepts, but they are not the same. Momentum is a measure of an object's motion, while impulse is a measure of the change in an object's momentum. In other words, impulse is the force applied to an object over a period of time, resulting in a change in its momentum.

4. How does momentum affect the movement of a volleyball?

The momentum of a volleyball is directly related to its movement. The more momentum a volleyball has, the harder it will be to stop or change its direction. This is why when a player spikes the ball, it has a higher velocity and more momentum, making it harder for the other team to defend against.

5. Can the momentum of a volleyball be changed?

Yes, the momentum of a volleyball can be changed. This can happen through interactions with other objects, such as when it is hit by a player or when it collides with the floor or a wall. The change in momentum can also be controlled by the players, as they can apply forces to the ball to change its direction and speed.

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