A ball of mass .14 kg is dropped from rest from a height of 1.25 m.It rebounds f

Question

A ball of mass .14 kg is dropped from rest from a height of 1.25 m.It rebounds from the floor to reach a height of .960 m. Whatimpulse was given to the ball by the floor?

This was one of the questions on my last physics test, I a disputein how he worked it out, but he won't see to our questions untilMonday. If you could please point out my lapse in logic, I'dgreatly appreciate it; I'm itching to know where I went wrong.

He worked it out like this :

He split the motion into two parts, ball at rest to ball hittingthe floor, and ball from floor to final height. He calculated theimpulse using:
I=delta P
P= mv
Kf+Uf=Ki+Ui

So considering part 1:
Ki=0 (rest) and Ui= mgh1
Kf= .5mv^2 and Uf=0
he solved for v1

Part 2:
Ki= .5mv^2 and Ui=0
Kf= 0 and Uf= mgh2
solved for v2

then plugged into I= mv2-mv1

My question is why is it that when the ball hits the floor it stillhas a kinetic energy? Given my understanding, it isn't moving. Nowa classmate told me that it could be the KE right before it hits,but then wouldn't it have a potential energy as well?

I solved it considering both actions as one, doing:

I= mvf-mvi
Where mvi is the ball dropped from rest, and mvf is the ball afterit bounces and reaches its final height. Now I said ok, mvi=0 givenit was dropped from rest. So I did Kf+ Uf= Ki+Ui where Ki=0

So: mgh1 = .5mvf^2+ mgh2
Solved for v, then multiplied by mass for impulse.

Any help would be greatly appreciated.

Alex

Explanation / Answer

Thanks for the reply. I just don't understand how it can have an initial velocity when it was released from rest. =/

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.