What is the impulse and force exerted on a soccer ball when kicked?

  • Thread starter Thread starter huybinhs
  • Start date Start date
  • Tags Tags
    Ball Impulse
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving the impulse and force exerted on a soccer ball when kicked. The scenario includes a soccer ball of mass 0.44 kg, initially at rest, and the duration of contact between the player's foot and the ball is 0.008 seconds. Participants are exploring the concepts of impulse and average force based on the provided force-time graph.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss calculating impulse using the area under the force-time curve and question the need for a functional form of the force. There is an exploration of deriving coefficients for a quadratic equation to represent the force as a function of time. Some participants express uncertainty about how to proceed with finding these coefficients.

Discussion Status

There has been productive engagement, with participants successfully deriving the coefficients for the force equation and confirming the impulse calculation. Guidance has been offered regarding the integration of the force function to find impulse, and there is an ongoing exploration of calculating the average force based on the impulse obtained.

Contextual Notes

Participants note that the time on the graph is in milliseconds, which is crucial for the calculations. There is also a reminder to be cautious about the units of impulse, as they are not in Newton-meters.

huybinhs
Messages
229
Reaction score
0

Homework Statement



A soccer player kicks a ball of mass 0.44 kg that is initially at rest. The player's foot is in contact with the ball for 0.008 s. The force of the kick is shown in the Figure.

a) What is the impulse imparted on the ball?

b) What is the average force exerted by the player's foot on the ball during the period of contact?

c) What is the magnitude of the ball's velocity immediately after it loses contact with the player's foot?

1-5.gif



Homework Equations



P = mv

I = E(ext) * Delta t

The Attempt at a Solution



a) I took 2 point then use I = F * Delta t. After that I substract them and it's INCORRECT! Any guide?
 
Physics news on Phys.org
huybinhs said:
a) I took 2 point then use I = F * Delta t. After that I substract them and it's INCORRECT! Any guide?
The impulse is the area under the curve. I = ∫F dt.

Is this a calculus based class?

And if so does the problem give you an equation for F as a function of t? If no equation is given, you can figure one out yourself. The curve looks like a parabola to me. Find the values of a, b, and c, such that
F = at2 + bt +c
matches the given curve. (plug in 3 different points for F and t, giving you three equations and 3 unknowns [the unknowns are a, b and c]). Once you have that you can integrate.
 
No, there is not equation given. But could u help me figure out a, b, and c? I have no ideas how to find them ?
 
huybinhs said:
No, there is not equation given. But could u help me figure out a, b, and c? I have no ideas how to find them ?
Try to plug in some F and t points into the
F = At2 + Bt + C
equation (here A, B, and C are just constants. You don't have to call them A, B, and C. You could call them anything you want for now. Eventually they will become numbers).
To make things easy, I suggest using the points [F,t] = [0,0], [6000 N, 4 sec], [0, 8 sec]

I'll do the first one for you.
0 = (A)(0)2 + (B)(0) + C
You should have a good idea what C is now.

Plug the other two points in and solve for A and B.
 
Yes, I pluged in and I got A = -3.75 * 10^8 and B = 3 * 10^ 6

Plz note time on the graph is in ms!

After this I got equation:

F = -3.75*10^8 t ^2 + 3*10^6 t

then, what's next?
 
huybinhs said:
Yes, I pluged in and I got A = -3.75 * 10^8 and B = 3 * 10^ 6

Plz note time on the graph is in ms!

After this I got equation:

F = -3.75*10^8 t ^2 + 3*10^6 t
Very nice! (and good catch about the ms vs. s) :approve:
then, what's next?
Now find the area under the curve! (Hint: you need to integrate the force from t = 0 to t = 8 ms.) That's the impulse.
 
Great! I got part a CORRECTLY.

Please continue help me out part b ;)
 
huybinhs said:
Great! I got part a CORRECTLY.
Great! :smile:
Please continue help me out part b ;)
You're going to have to show some work.

But I'll give you a hint that part b) is really simple (i.e. very simple calculation). Hypothetically speaking, if the force was constant over the same time interval (instead of being parabola shaped), what constant force over the same time interval would produce the same impulse? That's the average force.
 
Is it 32 / 000.8 = 4000 N ?

Note: 32 N*m = impulse : is the correct answer for part a
 
  • #10
Thank u! I got it all CORRECTLY ;)
 
  • #11
huybinhs said:
Is it 32 / 000.8 = 4000 N ?
There you go. :approve:
Note: 32 N*m = impulse : is the correct answer for part a
Be careful of the units. The units of impulse are not N·m.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
7K
Replies
14
Views
3K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 13 ·
Replies
13
Views
9K
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K