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

[huybinhs]

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?

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?

 

[collinsmark]

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.

 

[huybinhs]

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 ?

 

[collinsmark]

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.

 

[huybinhs]

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?

 

[collinsmark]

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)

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.

 

[huybinhs]

Great! I got part a CORRECTLY.

Please continue help me out part b ;)

 

[collinsmark]

Great! I got part a CORRECTLY.

Great!

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.

 

[huybinhs]

Is it 32 / 000.8 = 4000 N ?

Note: 32 N*m = impulse : is the correct answer for part a

 

[huybinhs]

Thank u! I got it all CORRECTLY ;)

 

[collinsmark]

Is it 32 / 000.8 = 4000 N ?

There you go.

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.