# Impulse and Momentum problem HELP!

1. Jan 25, 2005

### BlasterV

Impulse and Momentum problem!! HELP!

Joe Varsity kicks a football of mass 1 kg. As his foot makes contact with the ball, it exerts a force which gradually increases to a maximum value over milliseconds, then falls immediately to zero, as shown in the graph above. The force is given by the equation

Force = (250 N) * (t / tsub0)^2

where tsub0 equals 1 millisecond.

What is the impulse given by Joe's foot to the ball? Don't forget the units!

I did this: Integral of 0 to .005 of ( 250 N * (t/tsub0)^2 ) dt
Evaluate from 0 to .005 of (( 250 N * (t/tsub0)^3) / 3 )
Impulse = (250N * ((.005/.001)^3) /3 ) = 10416.67 J
This is wrong :(

The ball was originally sitting on the tee, motionless. What is its speed immediately after Joe kicks it?

Without a correct impulse I can't even hope to do this.

But in terms of variables I don't know the equation to find this either :(

2. Jan 25, 2005

### vincentchan

where is your .005 came from? the problem didn't say the duratioin of force. but I believe it is .001 sec
and one more thing, the unit of impulse is not J

3. Jan 25, 2005

### BlasterV

Thanks, J was the problem ,dont respond anymore guys I got it :)

4. Jan 25, 2005

### Andrew Mason

It would help immeasureably if you would provide the graph or give an adequate description. The area under the graph is the impulse. Divide impulse by the mass and you get speed of the ball.

It is not clear how long the force continues. It appears from your attempted solution that it lasts for .005 seconds or 5 milliseconds. If that is the case:

$$m\Delta v = \int_0^{.005} \frac{250}{.000001}t^2dt$$

Since the anti-derivative is $\frac{1}{3}t^3$:

$$m\Delta v = \frac{1}{3}2.5e8*.005^3 = 10.42$$

$$v = 10.42/1 = 10.42 m/sec$$

But that is, as I say, assuming the force applies for 5 milliseconds, which is not clear from your question.

AM