Calculating Bounce Height Using Conservation of Energy

[NasuSama]

Homework Statement

Assume there is no air friction. A ball of mass m = 3.39 kg is dropped from rest at height h = 1.51 m. It hits the ground, losing energy ΔE = 1/11 of its total energy in the collision. How high will the ball bounce upward before it comes momentarily to rest?

Homework Equations

U = mgh
KE = ½mv²

The Attempt at a Solution

mgh = ½mv_f² [At the start before the first bounce]
v_f = √(2gh) ≈ 5.44 m/s

½mv_f² = 10/11 * mgh_n where h_n is the new height
h_n = 11/20 * v_f²/g ≈ 1.66 m, which is wrong.

 

[SHISHKABOB]

well so the moment it hits the ground it has some velocity which you determined

right after it hits the ground it loses 1/11 of that energy

what you have here: ½mv_f² = 10/11 * mgh_n where h_n is the new height

is setting the energy right before impact equal to 10/11ths of some other amount of energy

so if you move the 10/11 to the other side you get

$\frac{11}{10}\frac{1}{2}mv_{f}^{2} = mgh_{n}$

what does that look like to you

 

[NasuSama]

well so the moment it hits the ground it has some velocity which you determined

right after it hits the ground it loses 1/11 of that energy

what you have here: ½mv_f² = 10/11 * mgh_n where h_n is the new height

is setting the energy right before impact equal to 10/11ths of some other amount of energy

so if you move the 10/11 to the other side you get

$\frac{11}{10}\frac{1}{2}mv_{f}^{2} = mgh_{n}$

what does that look like to you

It's what I already did. Still wrong answer.

 

[SHISHKABOB]

I know, but think about what that equation means.

$\frac{11}{10}\frac{1}{2}mv_{f}^{2} = mgh_{n}$ has kinetic energy on the left and potential energy on the right

the new potential energy is what we want and it's on the right

but on the left we have the kinetic energy right before impact *multiplied by 11/10*

if you multiply something by 11/10 does it get bigger or smaller?

 

[nasu]

Homework Statement

Assume there is no air friction. A ball of mass m = 3.39 kg is dropped from rest at height h = 1.51 m. It hits the ground, losing energy ΔE = 1/11 of its total energy in the collision. How high will the ball bounce upward before it comes momentarily to rest?

Homework Equations

U = mgh
KE = ½mv²

The Attempt at a Solution

mgh = ½mv_f² [At the start before the first bounce]
v_f = √(2gh) ≈ 5.44 m/s

½mv_f² = 10/11 * mgh_n where h_n is the new height
h_n = 11/20 * v_f²/g ≈ 1.66 m, which is wrong.

First of all, you don't need all the intermediate steps. Why do you need to calculate the speed at impact?

Second, the energy after collision is less than that before.
You wrote your equation as it is larger!
You have KE_after=(1-1/11) KE_before =10/11 KE_before.
And then, the final potential energy (mgh) is equal to which one, the KE_before or KE_after?

 

[NasuSama]

I know, but think about what that equation means.

$\frac{11}{10}\frac{1}{2}mv_{f}^{2} = mgh_{n}$ has kinetic energy on the left and potential energy on the right

the new potential energy is what we want and it's on the right

but on the left we have the kinetic energy right before impact *multiplied by 11/10*

if you multiply something by 11/10 does it get bigger or smaller?

Bigger.

 

[NasuSama]

I know, but think about what that equation means.

$\frac{11}{10}\frac{1}{2}mv_{f}^{2} = mgh_{n}$ has kinetic energy on the left and potential energy on the right

the new potential energy is what we want and it's on the right

but on the left we have the kinetic energy right before impact *multiplied by 11/10*

if you multiply something by 11/10 does it get bigger or smaller?

SO this means you need...

mgh = 11/10 * mgh_1! Oops!

 

[apamirRogue]

You do not need to even worry about the velocity that the ball has when it hits the ground. The only information that you need to consider is that E_initial=(11/10)E_final. So think of what the energy is due to when the ball is "momentarily at rest" and you should have your answer.

 

[SHISHKABOB]

SO this means you need...

mgh = 11/10 * mgh_1! Oops!

so basically you want to take 10/11ths of the energy right before impact and use conservation of energy to determine how high that kinetic energy will take the ball

 

[NasuSama]

so basically you want to take 10/11ths of the energy right before impact and use conservation of energy to determine how high that kinetic energy will take the ball

Thanks! You are right this time.