# Conservation of energy in bouncing ball

• My name
In summary: If you use observer in definite stationary, you will get 1/2mv'^2 + 1/2Mv''^2 as kinetic energy after collision, which equals to 1/2mv^2. And this value is the same as kinetic energy before collision.
My name
Throwing a ball which is (m)kg on the ground
If the velocity of the ball just before it reaches the ground is (v)m/s
Assume it is a elastic collision
The velocity of the ball when it leaves the ground should be (-v)m/s
According to conservation of momentum,the ground has a change of momentum = 2mv(pointing downward)
Then the total KE in the system =
The ball's KE+the ground's KE
The total KE after the collision > The total KE before the collision

Is there something I missed?
And sorry for the grammar...

Nope you are *almost correct* but the kinetic energy of the ground will be negligible (2mv)^2/2M where M is the mass of ground. In real world if we take into account the KE of the ground then the ball will not have -v velocity after the collision but just abit smaller so the difference in the kinetic energy of the ball give us the kinetic energy of the ground.

My name
Delta² said:
Nope you are *almost correct* but the kinetic energy of the ground will be negligible (2mv)^2/2M where M is the mass of ground. In real world if we take into account the KE of the ground then the ball will not have -v velocity after the collision but just abit smaller so the difference in the kinetic energy of the ball give us the kinetic energy of the ground.

Oh...thanks a lot

Background:
I want to define something, and use your example to answer your question. First, I assume the ground has 0 momentum just before the ball crashes on it, you are the observer standing on the ground. Neglect gravity effect because the whole process happens in very short time. Assume upward as positive.

Analysis:
When the ball crashes on the ground with momentum -mv. Then when it bounces upward, it will have mv momentum. However, remember one thing, you are an observer standing on the ground. What it means is that, the mv is actually mv'(v' is a vector pointing upward) if you are observer in definite stationary state. However, because the ground also moves downward, with momentum Mv''(v'' is a vector pointing downward), so, you also move downward with v'' if you are observer standing on the ground. The mv is actually related to relative velocity,
v=v'-v''.

Execution:
Formula we have: v = v' - v'' -> v = v' + v'' ; Mv'' - mv' = mv + 0
Evaluate (1/2)Mv''2 + (1/2)mv'2.
Tricks in evaluation: (1/2)Mv''2 = (1/2)(Mv'')(v''). By Mv'' = mv' + mv and v'' = v - v',
(1/2)(Mv'')(v'')
= (1/2)(mv'+mv)(v-v')
= (1/2)(mv'v-mv'2+mv2-mvv') = (1/2)(mv2) - (1/2)(mv'2)

Back to original formula:

(1/2)Mv''2 + (1/2)mv'2 = (1/2)mv2 -(1/2)mv'2 + (1/2)mv'2 = (1/2)mv2 (original kinetic energy of the falling ball)

Conclusion: Energy is conserved!

Last edited:
Delta² said:
Nope you are *almost correct* but the kinetic energy of the ground will be negligible (2mv)^2/2M where M is the mass of ground. In real world if we take into account the KE of the ground then the ball will not have -v velocity after the collision but just abit smaller so the difference in the kinetic energy of the ball give us the kinetic energy of the ground.
it is the problem of observer. the v will not decrease in reality even if the ground is low in mass.

sunmaggot said:
it is the problem of observer. the v will not decrease in reality even if the ground is low in mass.

But then if we choose an observer in a stationary frame (not in the ground but in a frame where the ground is stationary before the collision) then for that observer the kinetic energy will not be conserved.

Delta² said:
But then if we choose an observer in a stationary frame (not in the ground but in a frame where the ground is stationary before the collision) then for that observer the kinetic energy will not be conserved.
I have already calculated that. for both observer, energy is conserved. please look at my post above

To tell you the truth i ve a problem following your post, but for the observer i am saying, it will be

Kinetic Energy before collision =1/2mv^2

Kinetic Energy after collision=1/2mv^2+(2mv)^2/2M.

As i see KE is not conserved for that observer . Where am i wrong?

Delta² said:
To tell you the truth i ve a problem following your post, but for the observer i am saying, it will be

Kinetic Energy before collision =1/2mv^2

Kinetic Energy after collision=1/2mv^2+(2mv)^2/2M.

As i see KE is not conserved for that observer . Where am i wrong?
if you use the observer on the ground to think, you will conclude that after the ball crashes on the ground and bounces upward, there is no energy transferred into the ground because you see no change in velocity. If you use observer in definite stationary, you will get 1/2mv'^2 + 1/2Mv''^2 as kinetic energy after collision, which equals to 1/2mv^2. And this value is the same as kinetic energy before collision. The problem in your analysis is, you use observer on the ground to describe the energy of the ball, but you use observer in stationary state to describe the energy of ground, which makes problems. you can only use one observer for the whole event.

Delta² said:
As i see KE is not conserved for that observer . Where am i wrong?
I don't think you are wrong, If the observer is non-inertial during the collision (because he is fixed to the accelerated Earth), kinetic energy isn't conserved in his rest frame.

A.T. said:
I don't think you are wrong, If the observer is non-inertial during the collision (because he is fixed to the accelerated Earth), kinetic energy isn't conserved in his rest frame.
I don't think you are correct. the Earth always has 0 kinetic energy if you are moving with the Earth all the time. you only know the KE of the Earth if you are in definite stationary state

sunmaggot said:
I don't think you are correct. the Earth always has 0 kinetic energy if you are moving with the Earth all the time.
I never claimed otherwise.
sunmaggot said:
definite stationary state
There is no such thing.

A.T. said:
I never claimed otherwise.

There is no such thing.
you don't know what is imaginary experiment?

sunmaggot said:
he problem in your analysis is, you use observer on the ground to describe the energy of the ball, but you use observer in stationary state to describe the energy of ground,which makes problems. you can only use one observer for the whole event.
I think in post #2 he used the inertial frame where the Earth is initially at rest for the whole event, which is perfectly valid.

A.T. said:
He used the inertial frame where the Earth is initially at rest for the whole event, which is perfectly valid.
yup, then why would he say there is KE for the earth? if you use observer on the ground, there is no KE for the earth.

sunmaggot said:
yup, then why would he say there is KE for the earth?
Because the Earth is only initially at rest in that inertial frame. After collision it moves. Since the frame is inertial total KE is conserved (assuming perfectly elastic collision).

The frame where the Earth is always at rest throughout the collision, is not inertial, so kinetic energy isn't conserved in that frame.

A.T. said:
Because the Earth is only initially at rest in that inertial frame. After collision it moves. Since the frame is inertial total KE is conserved (assuming perfectly elastic collision).

The frame where the Earth is always at rest throughout the collision, is not inertial, so kinetic energy isn't conserved in that frame.
you are making the same mistake. If you are observer on the earth, you won't see the Earth moving, there is no KE for the earth. The Earth is always at rest because observer will not see any speed for the earth. then why would KE be not conserved? you should think, the Earth will not gain KE because you are observer on the earth! don't use different observers in same event to create troubles!

sunmaggot said:
you are making the same mistake. If you are observer on the earth, you won't see the Earth moving, there is no KE for the earth.
I never said otherwise, so I don't know what mistake you are referring to.

A.T. said:
I never said otherwise, so I don't know what mistake you are referring to.
Mistake about using stationary frame to describe the earth, using ground frame to describe the ball.

A.T. said:
I never said otherwise, so I don't know what mistake you are referring to.
actually I said that on the above posts :D but maybe you didn't see those posts.

sunmaggot said:
Mistake about using stationary frame to describe the earth, using ground frame to describe the ball.
I never did that.

A.T. said:
Because the Earth is only initially at rest in that inertial frame. After collision it moves. Since the frame is inertial total KE is conserved (assuming perfectly elastic collision).

The frame where the Earth is always at rest throughout the collision, is not inertial, so kinetic energy isn't conserved in that frame.
Did you say this? "After collision it moves", that means the frame is stationary, by my calculation in early post, energy is still conserved.

sunmaggot said:
"After collision it moves", that means the frame is stationary
If the frame remains inertial, the Earth starts moving after the collision.

A.T. said:
If the frame remains inertial, the Earth starts moving after the collision.
and the ball will not move with v, it moves with a speed v' where v' < v. if you say the ball moves in v, you are using frame on the ground. see your mistake now?

sunmaggot said:
and the ball will not move with v, it moves with a speed v' where v' < v
That's correct. In the inertial frame the ball will transfer KE to the Earth, while the total KE remains constant.

sunmaggot said:
if you say the ball moves in v, you are using frame on the ground.
We can use this frame too, but it's not inertial, so we cannot expect total KE to be conserved in this frame.

A.T. said:
That's correct. In the inertial frame the ball will transfer KE to the Earth, while the total KE remains constant.We can use this frame too, but it's not inertial, so we cannot expect total KE to be conserved in this frame.
which frame are you using...? come on, think carefully please. If you say the ground moves, then you are using stationary frame,. If you say the ball moves with speed v, the ground doesn't move, you are observer on the ground...

A.T. said:
That's correct. In the inertial frame the ball will transfer KE to the Earth, while the total KE remains constant.We can use this frame too, but it's not inertial, so we cannot expect total KE to be conserved in this frame.
why must you think that the ground will move when you are observer on the ground? The ground will not move. No KE for the ground!

A.T. said:
That's correct. In the inertial frame the ball will transfer KE to the Earth, while the total KE remains constant.We can use this frame too, but it's not inertial, so we cannot expect total KE to be conserved in this frame.
maybe I simply end this.
Why isn't energy conserved? when the ball crashes on the ground with speed v, I see the ball bounces upward with speed v when I am on the ground. Why do you think the ground has KE? From stationary frame, I see the ground moves. Energy of the ball + energy of the ground is larger than original energy!

However you used two frames in one event, this is how the issue starts.

sunmaggot said:
Why isn't energy conserved? when the ball crashes on the ground with speed v, I see the ball bounces upward with speed v when I am on the ground.
I checked again, and I think you are right on this point. In this special case even the non-inertial ground frame does conserve KE (just not momentum). I was thinking too generally about non-inertial frames, which usually don't conserve KE.

sunmaggot said:
From stationary frame, I see the ground moves. Energy of the ball + energy of the ground is larger than original energy!
If by "stationary frame" you mean "inertial frame", then I don't think the above is true. In inertial frames total KE stays constant in perfectly elastic collisions, and can be reduced in inelastic collisions. But it never increases from collisions alone.

A.T. said:
If by "stationary frame" you mean "inertial frame", then I don't think the above is true. In inertial frames total KE stays constant in perfectly elastic collisions, and can be reduced in inelastic collisions. But it never increases from collisions alone.
that was sarcasm, so you are correct, those sentences are wrong.

Very confusing for me at least.

I ll define two different (i think) frames:

Frame A: The frame where the ground is stationary before the collision.
Frame B: The ground frame where ground is stationary, before, during and after the collision.

I hope frames A and B are well defined (i sense there might be a problem with the way i define frame A, if so explain to me). Also i think Frame A is inertial before, during and after the collision, while Frame B is inertial before and after, but not during, the collision. We "have the right" to remain in frame A even after the collision right?

What i really want to ask is this:
What are the velocities of the ball and the ground , before and after the collision, in Frame A?
Same question for Frame B.

For the CM system:$$mv + MV = 0 ,\, E_{cm} = \frac{1}{2}mv^2 + \frac{1}{2}MV^2$$For V=0 we have another frame witch move by V before collision and -V after collision.
For what reason a system that in time dt change his velocity from V to -V must conserve any kinetic energy?

Delta2
Ok it is now clear to me that in Frame B it will be ##v_2=-u_2## for some reason i couldn't see that, thanks. However in Frame A it is ##|v_2|<|u_2|## correct?

## 1. What is the law of conservation of energy?

The law of conservation of energy states that energy cannot be created or destroyed, but can only be transformed from one form to another.

## 2. How does the law of conservation of energy apply to a bouncing ball?

When a ball is dropped, it gains potential energy due to its position above the ground. As it falls, this potential energy is converted into kinetic energy. When the ball hits the ground, the kinetic energy is transferred into the ground and the ball bounces back up. The total energy (kinetic + potential) remains constant throughout the bouncing process, in accordance with the law of conservation of energy.

## 3. What factors affect the conservation of energy in a bouncing ball?

The conservation of energy in a bouncing ball is affected by the elasticity of the ball and the surface it bounces on, as well as the initial height from which it is dropped. A more elastic ball and a harder surface will result in less energy being lost during the bounce, leading to a higher bounce height.

## 4. Can the law of conservation of energy be violated in a bouncing ball?

No, the law of conservation of energy is a fundamental law of physics and cannot be violated. In a perfectly elastic scenario, where no energy is lost during the bounce, the ball would theoretically bounce to the same height as the initial drop. In reality, some energy is always lost due to factors such as air resistance and friction, leading to a slightly lower bounce height.

## 5. How is the conservation of energy in a bouncing ball relevant to real-world applications?

The conservation of energy in a bouncing ball has many real-world applications, such as in sports equipment design, engineering of shock-absorbing materials, and understanding the behavior of objects in motion. It also plays a crucial role in understanding the behavior of other systems, such as pendulums and springs.

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