- #1

greyradio

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**[SOLVED] Conservation of Energy problem**

Assuming there is no air friction, A 1.4 kg ball is dropped from a height of 2.68 m. It hits the ground, losing 1/12 of its energy in the collision. How high will the ball bounce upward before it comes momentarily to rest?

E = K sys + U sys

Ef = Ei

1/2mv^2 + mgh = 1/2mv^2 + mgh

I've attempted this problem but I seem to be missing some piece of information and I have run out of ideas hopefully someone can point me in the right direction.

Since there are no non-conservative forces then W non conservative = 0 and there are no external forces so that Work is equal to zero.

change in mechanical E = Ef = Ei

Ef = Ei

1/2 m vf^2 + mghf = 1/2 m vi^2 + mghi

I assumed that the final hieght is zero and the initial height is 2.68m. I also assumed that the initial velocity is 0.

1/2 mvf^2 + mg0 = 1/2 m 0 + mghi

1/2 mvf^2 = 1/2 mghi

1/2 (1.4) vf^2 = (1.4)(9.81)(2.68)

i found vf^2 using kinematic equations = 52.5816 m^2/s^2

36.80712 = 36.80712

Since the final energy is 36.80712 and when it collides with the ground the energy disspates by 1/12.

So :

36.80712/12 = 3.06726

So I made the the inital energy of the first bounce.

Ei = Ef

3.06726 = 1/2mv^f + mgh

3.06726 = 0 + mgh

3.06726 = mgh

3.067/mg = h

.22333 = h

This is what I calculated the height to be but this is wrong.

I'm not sure what I'm missing or if I should be looking at it in a different way.