Energy lost by a bouncing ball

[bernard08]

1. I am investigating how the initial height a ball is dropped from affects the rebound height. We have been asked to make a prediction about what we think will happen; however, we have not been given any guidance as to what we should write.

Homework Equations

To find the % potential energy lost, I will use this equation: (h1-h2) / h1 * 100

where h1 = initial height ball is dropped from;
h2 = rebound height of ball.

The Attempt at a Solution

My prediction is, that as the initial height the ball is dropped from increases, the percentage PE (potential energy) lost by the ball will also increase. I think this is the case because at a higher height, the ball will have more PE. This will then all be converted to KE (kinetic energy) as the ball falls. Therefore, when the ball hits the ground, there will be more KE available for conversion into sound, heat and the ball's deformation so more % PE will be lost.

That is what I had written in my project, but my teacher said I need to include more DETAILED SCIENTIFIC KNOWLEDGE. But I have included all I know, and tried to research as much as possible on the internet, but I can't find anymore than what I already have. Could any of you guys please help me out with detailed scientific reasons as to why my prediction is what it is? Please, I'm really freaking out! I would be so grateful!

 

[xxChrisxx]

Coefficient of Restitution.

GO!

p.s welcome to PF

 

[bernard08]

Coefficient of Restitution.

GO!

p.s welcome to PF

Thanks! But not that complicated, lol. My teacher probably wouldn't even understand that :P
Is there anything else you know of that would make % energy lost increase? And is my prediction even correct, lol?

 

[xxChrisxx]

What year are you in? I'll try to tailor my answer.

 

[bernard08]

I'm 15, so in the UK I'm in Year 12 but I'm not sure how that system works in the US...

 

[xxChrisxx]

Ahh gcse physics project then. (good job I am english :P)

Do you know about plastic and elastic collisions?

You are thinking along the correct lines mostly, in reality a ball will deform more if it impacts with a higher speed. However the coefficient of restitution shows how elestic or plastic a collision is. It is a ratio of velocityin/velocity out and is between 0 and 1 in most practical cases.

COR is a property of a collision largely independent of speed. for example:

You dropped a football on to a hard floor. The Vin at impact was 1m/s, the Vout was 0.8m/s. The collision would have a COR of 0.8. If you dropped the ball at 2m/s the output would then be 1.6m/s. If you changed the floor to a large rubber mat, the COR may drop to 0.5 so half the speed wouls be lost in the collison.

This can also be applied to height of the bounce. COR^2 = h bounce / H dropped.

So for most practical cases, this is a linear relationship. The same percentage would be lost in the drop.When you get to very high speed collisions, or very high deformations this model starts to break down and become less accurate.

 

[bernard08]

Thanks! :) I'll add that in!
Is there anything else you know of that would cause an increased % energy loss?

 

[xxChrisxx]

Dont just add that in. I'd suggest reading about it first so you get a decent understanding. You never know I could be talking utter crap. ;)

There are no other losses really, its purely down to the geometry and materials of the two objects in the collision. Of course if you want to get really picky you can say things like, heating effects can change how the material reacts, but that's not in the majority of cases is probably above what you are trying to explain.

Heating effects tend to make the COR increase, a squash ball gets 'faster' when it gets up to temperature. This is because the COR is higher so it rebound off the wall with less speed loss.

 

[bernard08]

So really its just sound + some heat that causes a loss of energy? And the COR stuff? I'm going to kill my teacher; he's making it all so complicated!

So should I just stick with my prediction:
My prediction is, that as the initial height the ball is dropped from increases, the percentage PE (potential energy) lost by the ball will also increase. I think this is the case because at a higher height, the ball will have more PE. This will then all be converted to KE (kinetic energy) as the ball falls. Therefore, when the ball hits the ground, there will be more KE available for conversion into sound, heat and the ball's deformation so more % PE will be lost.

 

[xxChrisxx]

The percent of energy lost wouldn't increase, the percentage loss would stay the same. The overall loss would be larger with a ball dropped from a greater height, but it would be the same percentage loss.

How did your teacher describe this topic to you?

Also the COR isn't 'as well as the rest of the stuff' its all the losses grouped together.

The losses included within COR would be mostly sound and heat.

 

[bernard08]

Thats what I was thinking, but I couldn't think of how to word it about the same percentage of energy lost by sound and heat each time. Could you help me with trying to get it into words please? I basically want to say that despite the fact that there is more energy lost due to louder sound/more heat, there is also more PE to begin with so the actual % PE energy lost stays the same, e.g. with a small initial height, the energy lost would be small but the % PE energy lost would be the same as a ball dropped from a larger height.

I need to make it more scientifically detailed - argh, lol.

He didn't really explain it. He just said, 'it's an experiment. do it. evaluate your prediction.'

 

[xxChrisxx]

Ahh you did an experiment on it. Do you have your results handy? (or have you not done it yet)

Simply state that the % loss is constant across the range of results (IF it agrees with your results).

You could phrase it as, there is a linear relation between initial height and rebound height.

 

[bernard08]

Nope, no experiment yet. This is just planning. Thanks! And how could I phrase the scientific knowledge bit about sound/heat and how it makes the % energy lost constant? Thanks so much for not getting annoyed at me Chris! lol

 

[xxChrisxx]

You can state that we know the trend is likely to be linear and that we know it doesn't bonuce as high so some of the energy is transferred. The energy 'lost' is transferred to xyz.

And i'll leave you to work out the details of how you are going to write it.