1. Yo guys... long time no... read :?

Anyways, I'm stuck on this physics problem and I'm thinking it will be pretty easy for some of you guys.

It is related to momentum and impulse.

This is not a Homework problem. It is one of the odd problems (with the answer in the book) and it is similar to one of the homework problems. I need somewhere to start, I can't seem to figure this out.

A ball of mass 0.150kg is dropped from rest from a height of 1.25m. It rebounds from the floor to reach a height of 0.960m. What impulse was given to the ball by the floor?

My understanding is that impulse is in kg*m/s.

Somehow I came up with 3.57 kg*m/s which is apparently wrong.

No class for three days...

P.S. Since when has this site had spell check?? That's kinda handy.  2.

3. You need to find the velocity as it bounces off the floor. You can use conservation of energy to do that by solving for the potential energy at the height of its bounce (mgh) and equating that to the kinetic energy it has just as it comes off the floor (1/2 mv^2). My answer is 0.644.  4. Yeah, I thought it might have something to do with that. We just spent a week on work and energy and I was real busy and didn't do the homework. I guess it's coming back to haunt me.

Thanks, I'll let you know if I got anymore questions.  5. Impulse is the same as the change in momentum as the object rebounds from the floor:

Impulse = mv<sub>1</sub> - mv<sub>2</sub>

where v<sub>1</sub> is the velocity just before impact and v<sub>2</sub> is the velocity just after impact.

As has been mentioned, these velocities can be found by equating potential and kinetic energies but, to do this, you need a value for the acceleration due to gravity.

mgh =(1/2)mv<sub>1</sub><sup>2</sup> with h = 1.25m, and a similar expression for v<sub>2</sub> with h = 0.96m.

For g = 9.81 m s<sup>-2</sup>, I get the impulse to be 1.394 kg m s<sup>-1</sup>.

(Note that v<sub>1</sub> and v<sub>2</sub> are in opposite directions so that the values of mv<sub>1</sub> and mv<sub>2</sub> should be added, not subtracted)

However, my use of calculators is awful, so please check it out.  6. Argh.. I forgot the definition of impulse. Oh well, it's only been about 40 years.   7. Harold,

if it's any consolation, I also had to refresh my memory about "impulse". It's about fifty years since I was introduced to it and, during those fifty years, I don't think I have ever had refer to it.   8. Originally Posted by Old Fool
Impulse is the same as the change in momentum as the object rebounds from the floor:

Impulse = mv<sub>1</sub> - mv<sub>2</sub>
You mean mv<sup>2</sup><sub>1</sub> - mv<sup>2</sup><sub>2</sub>... right? Originally Posted by Old Fool
where v<sub>1</sub> is the velocity just before impact and v<sub>2</sub> is the velocity just after impact.

As has been mentioned, these velocities can be found by equating potential and kinetic energies but, to do this, you need a value for the acceleration due to gravity.

mgh =(1/2)mv<sub>1</sub><sup>2</sup> with h = 1.25m, and a similar expression for v<sub>2</sub> with h = 0.96m.

For g = 9.81 m s<sup>-2</sup>, I get the impulse to be 1.394 kg m s<sup>-1</sup>.

(Note that v<sub>1</sub> and v<sub>2</sub> are in opposite directions so that the values of mv<sub>1</sub> and mv<sub>2</sub> should be added, not subtracted)

However, my use of calculators is awful, so please check it out.
Yep, that's what the book says... now I just have to do it on my own.  9. Originally Posted by DaBOB
You mean mv<sup>2</sup><sub>1</sub> - mv<sup>2</sup><sub>2</sub>... right?
No. Momentum is mv.  10. So, I guess what I don't understand is what the kinetic energy on the way back up is.

Potential energy on the way down would be 18.4-J. Which would make the velocity equal to 4.95-m/s and the impulse would be 7.43-kg m/s.

Now what?  11. I think you misplaced a decimal point. You should have gotten 1.84 instead of 18.4. Solve for v1 using that number, multiply by the mass and that is the momentum on the way down.

For the KE and momentum on the way up, use the height it rises to in place of the height it fell from, and do the same thing.  12. O.k....

KE<sub>1</sub> = mgh<sub>1</sub> = (0.15kg)(9.8ms<sup>-2</sup>)(1.25m) = 1.84J

1.84J = 1/2mv<sub>1</sub><sup>2</sup> = .5(.15)(v<sub>1</sub>)<sup>2</sup>

v<sub>1</sub> = [ 2(1.84)(.15)<sup>-1</sup>]<sup>-2</sup> = 4.95m/s

p<sub>1</sub> = mv<sub>1</sub> = (0.15kg)(4.95m/s) = 0.743kg m/s

------------------

KE<sub>2</sub> = 1.41J

v<sub>2</sub> = 4.34m/s

p<sub>2</sub> = 0.651kg m/s

-----------------

p<sub>1</sub> - (-p<sub>2</sub>) = I

I = 0.743kg m/s + 0.651kg m/s = 1.39kg m/s YAY!!!

----------------

So does the normal force of the ground equal the gravitational force? And is that why you use the same "mgh" (or nh) for the kinetic energy on the way back up? Or is it more like (m)(g)(-h) on the way back up? Meaning, the ball is going a negative distance.

Maybe I should just do more problems and the the answer will come to me....

Thanks a ton. Now I can go to the next problem.   13. Originally Posted by DaBOB
YAY!!!
Congrats. Originally Posted by DaBOB
So does the normal force of the ground equal the gravitational force? And is that why you use the same "mgh" (or nh) for the kinetic energy on the way back up? Or is it more like (m)(g)(-h) on the way back up? Meaning, the ball is going a negative distance.
If you mean the normal force due to the collision of the ball with the ground, no, that force is a lot higher than the gravitational force but it acts over a very short time. We used KE and PE as a way to solve for the velocity before it hit the ground and after it bounces back up. Did you learn about work energy yet? The work required to lift the ball to height h is equal to the weight multiplied by the height through which it is lifted. The weight is mg. The height is h, so the work is mgh and this becomes the potential energy of the ball at that height. Now if you drop the ball that potential energy is converted to kinetic energy. On the way back up it starts out with kinetic energy and this gets converted to potential energy as it rises.

This gives us an easy way to find the velocity, but we could also have used newton's formulas v=at and s=1/2at^2, where acceleration a=g, the acceleration due to gravity, and s is the distance fallen. Come to think of it that would have been just as easy.  14. Originally Posted by Harold14370
The work required to lift the ball to height h is equal to the weight multiplied by the height through which it is lifted.
aah, for some reason I wasn't thinking of mg being weight when it obviously is. I guess I got too caught up in the problem. That makes a lot more sense now.

domo arigatou...  Bookmarks
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