# Magnitude of the Force of a kick to a basketball

1. Dec 11, 2006

### smedearis

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

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

3. The attempt at a solution
I know what Pinitial is, but what is Pfinal? Where does the theta come in, force or final velocity?

2. Dec 11, 2006

The impulse equals change in momentum, not kinetic energy, you may want to consider this, first.

3. Dec 11, 2006

### smedearis

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. Dec 11, 2006

It is practical to work with vectors. Assume that the initial direction of the velocity of the ball is $$\vec{i}$$, and the direction of the force (i.e. impulse) $$\vec{j}$$. Further on, use $$\vec{v}_{2} = v_{2}\cos(30)\vec{i}+v_{2}\sin(30)\vec{j}$$. Form a vector equation based on $$\vec{I} = m\vec{v}_{2}-m\vec{v}_{1}$$, where $$\vec{v}_{2}$$ is the final, and $$\vec{v}_{1}$$ the initial velocity. After 'solving for' $$\vec{i}$$ and for $$\vec{j}$$, you can calculate both the final velocity and the force.

5. Dec 11, 2006

### smedearis

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

6. Dec 11, 2006

I'm sorry, I meant 24 degrees, my mistake. Do you follow now?

7. Dec 11, 2006

### smedearis

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. Dec 11, 2006

Ok, I'll assume you're familiar with basic vector algebra. First of all, do you agree with the equation I wrote: $$\vec{I} = m\vec{v}_{2}-m\vec{v}_{1}$$? (*)
Further on, you know that the impulse equals $$\vec{I}=\vec{F}\cdot t$$. Let the magnitude of the force (which is unknown, but that doesn't matter) be F. So now you have $$\vec{I}=F \vec{j} \cdot t$$. The direction perpendicular to $$\vec{j}$$ is $$\vec{i}$$, so the initial momentum equals $$m \vec{v}_{1} = m 3.9 \vec{i}$$. 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 $$\vec{v}_{2} = v_{2}\cos(24)\vec{i}+v_{2}\sin(24)\vec{j}$$.