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

AI Thread Summary
A 1.4 kg ball dropped from 2.68 m loses 1/12 of its energy upon impact, leading to a calculation of its bounce height. The initial potential energy is calculated as 36.77 N, with the energy lost being approximately 3.06 N, leaving 33.71 N for the rebound. The rebound height is determined using the formula E = mgh, resulting in a height of about 2.45 m after accounting for the energy loss. The discussion emphasizes the importance of correctly applying energy conservation principles and simplifying calculations. Overall, the final bounce height is confirmed to be approximately 2.45 m.
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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.
 
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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.
 
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.
 
Raza said:
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.
 
Thank you. I did not subtract the dissipated energy from the original. Thanks to all that helped.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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