How High Will the Ball Bounce After Losing Energy on Impact?

[greyradio]

[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.

 

[Raza]

PART 1
Potential Energy
E=mgh
E=(1.4)(9.8)(2.68)
E=36.77N
Energy Lost= 36.77/12=3.06N
Total Energy Left= 36.77-3.06=33.71N
The amount of 33.71N is being carried on to the next step.

PART 2
E=mgh
33.71=(1.4)(9.8)(h)
33.71=13.72h
h=2.45m

I am not sure that my work is correct.

 

[Dick]

If it loses 1/12 of it's energy KE in the collision, then it's energy after the collision is (11/12)*KE, isn't it? And while I think you are doing this essentially correctly, you could do it a lot more simply. (2.68m)*m*g=(1/2)*m*v^2=KE. After the bounce the KE is (11/12)*KE. So (x)*m*g=(11/12)*KE, where x is the rebound height. So x*mg=(11/12)*(2.68m)*mg. You don't need to know m or g.

 

[Dick]

PART 1
Potential Energy
E=mgh
E=(1.4)(9.8)(2.68)
E=36.77N
Energy Lost= 36.77/12=3.06N
Total Energy Left= 36.77-3.06=33.71N
The amount of 33.71N is being carried on to the next step.

PART 2
E=mgh
33.71=(1.4)(9.8)(h)
33.71=13.72h
h=2.45m

I am not sure that my work is correct.

Dead right. (11/12)*2.68~2.45. Now simplify the calculation.

 

[greyradio]

Thank you. I did not subtract the dissipated energy from the original. Thanks to all that helped.