Magnitude of the Force of a kick to a basketball

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Homework Help Overview

The problem involves a 0.5 kg basketball rolling at a speed of 3.9 m/s, which is kicked perpendicular to its path, resulting in a change in direction at an angle of 24 degrees. Participants are discussing the implications of impulse and momentum in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to determine the final momentum and velocity of the basketball after the kick, questioning how the angle affects these calculations. There is discussion about using vector components and the relationship between impulse and momentum.

Discussion Status

Some participants are exploring the vector approach to resolve the problem, while others express confusion regarding the application of trigonometric functions and the direction of forces and velocities. There is an ongoing exchange of ideas without a clear consensus on the next steps.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the information they can reference or the methods they can use. There is a noted misunderstanding regarding the angle used in calculations, which has led to some confusion in the discussion.

smedearis
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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?
 
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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.
 
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?
 
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.
 
I don't think I'm following you on this. how and why are you using cos(30) and sin(30)?
 
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?
 
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?
 
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.
 

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