Magnitude of the Force of a kick to a basketball

AI Thread Summary
The discussion focuses on calculating the final velocity and force exerted on a basketball after a kick. The initial momentum is established using the formula Pinitial = 0.5 * 0.5 * (3.9^2). Participants emphasize the importance of using vector components to analyze the kick's effect, particularly with the angle of 24 degrees. The impulse-momentum theorem is highlighted, stating that impulse equals the change in momentum, leading to the equation I = m(v2 - v1). Clarifications are sought regarding the application of trigonometric functions and the direction of forces involved in the calculations.
smedearis
Messages
11
Reaction score
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
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 \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.
 
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: \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}.

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