Magnitude of the Force of a kick to a basketball

In summary, when kicking a basketball, the force is perpendicular to the initial direction of the ball's motion. The impulse is the change in momentum, which is equal to the net force times the time it was applied.
  • #1
smedearis
11
0

Homework Statement


A 0.5 kg basketball is rolling by you at 3.9 m/s. As it goes by, you give it a quick kick perpendicular to its path. Your foot is in contact with the ball for 0.002 s. The ball eventually rolls at an angle of theta = 24 degrees from its original direction.


Homework Equations


DeltaP=Fnet*DeltaT
Pinitial=.5*0.5*3.9^2



The Attempt at a Solution


I know what Pinitial is, but what is Pfinal? Where does the theta come in, force or final velocity?
 
Physics news on Phys.org
  • #2
smedearis said:
DeltaP=Fnet*DeltaT
Pinitial=.5*0.5*3.9^2

The impulse equals change in momentum, not kinetic energy, you may want to consider this, first.
 
  • #3
ok, impulse =Fnet*DeltaT=DeltaP(change in momentum), right?

am i still assuming that p=mv, so m*Vfinal-m*Vinitial=DeltaP
where would i find the Vfinal?
 
  • #4
smedearis said:
ok, impulse =Fnet*DeltaT=DeltaP(change in momentum), right?

am i still assuming that p=mv, so m*Vfinal-m*Vinitial=DeltaP
where would i find the Vfinal?

It is practical to work with vectors. Assume that the initial direction of the velocity of the ball is [tex]\vec{i}[/tex], and the direction of the force (i.e. impulse) [tex]\vec{j}[/tex]. Further on, use [tex]\vec{v}_{2} = v_{2}\cos(30)\vec{i}+v_{2}\sin(30)\vec{j}[/tex]. Form a vector equation based on [tex]\vec{I} = m\vec{v}_{2}-m\vec{v}_{1}[/tex], where [tex]\vec{v}_{2}[/tex] is the final, and [tex]\vec{v}_{1}[/tex] the initial velocity. After 'solving for' [tex]\vec{i}[/tex] and for [tex]\vec{j}[/tex], you can calculate both the final velocity and the force.
 
  • #5
I don't think I'm following you on this. how and why are you using cos(30) and sin(30)?
 
  • #6
smedearis said:
I don't think I'm following you on this. how and why are you using cos(30) and sin(30)?

I'm sorry, I meant 24 degrees, my mistake. Do you follow now?
 
  • #7
I'm still not understanding.
To find Vf, I use Vi*sin(24)+Vi*sin(24), b/c previously, you said that i need to multiply the trigs by the Vf, but i don't know what that is.

how do i know the direction of the force and velocity?
 
  • #8
smedearis said:
I'm still not understanding.
To find Vf, I use Vi*sin(24)+Vi*sin(24), b/c previously, you said that i need to multiply the trigs by the Vf, but i don't know what that is.

how do i know the direction of the force and velocity?

Ok, I'll assume you're familiar with basic vector algebra. First of all, do you agree with the equation I wrote: [tex]\vec{I} = m\vec{v}_{2}-m\vec{v}_{1}[/tex]? (*)

Further on, you know that the impulse equals [tex]\vec{I}=\vec{F}\cdot t[/tex]. Let the magnitude of the force (which is unknown, but that doesn't matter) be F. So now you have [tex]\vec{I}=F \vec{j} \cdot t[/tex]. The direction perpendicular to [tex]\vec{j}[/tex] is [tex]\vec{i}[/tex], so the initial momentum equals [tex]m \vec{v}_{1} = m 3.9 \vec{i}[/tex]. Now we have the condition satisfied that the force acted (i.e. the impulse) in a direction perpendicular to the initial direction of the ball's motion, right? Now, all we have to do is plug these values back into the equation (*), along with [tex]\vec{v}_{2} = v_{2}\cos(24)\vec{i}+v_{2}\sin(24)\vec{j}[/tex].

You may want to draw a simple sketch, It should help, too.
 

1. What is the magnitude of the force of a kick to a basketball?

The magnitude of the force of a kick to a basketball depends on various factors such as the velocity of the kick, the mass and size of the basketball, and the contact surface area between the foot and the ball. It can range from a few pounds of force to over a hundred pounds.

2. Does the angle of the kick affect the magnitude of force?

Yes, the angle of the kick can affect the magnitude of force. A kick that is perpendicular to the surface of the ball will result in a greater force compared to a kick that is at an angle.

3. How does the weight of the basketball impact the magnitude of force?

The weight of the basketball does not directly impact the magnitude of force, but it can affect the acceleration of the ball. A heavier ball will require more force to accelerate compared to a lighter ball.

4. Can the material of the basketball affect the magnitude of force?

Yes, the material of the basketball can affect the magnitude of force. A softer and more elastic material will absorb some of the force of the kick, resulting in a lower magnitude of force compared to a harder and less elastic material.

5. Is there a maximum or minimum amount of force needed to kick a basketball?

There is no specific maximum or minimum amount of force needed to kick a basketball. It depends on the desired speed and distance the ball needs to travel. However, there is a certain amount of force needed to overcome the weight and friction of the ball in order to move it.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
23
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
6K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
4K
Back
Top