# B Why does a basketball bounce?

1. Mar 18, 2017

### gex

Hi all,

My understanding for why a basketball bounces is that the earth acts against the ball, causing it to deform due to the ball having a downwards inertia and wanting to continue travelling down. This deformation reduces the volume inside the ball, in turn increasing the air pressure which pushes against the inside wall of the ball. What I do not understand is why this ball leaves the ground after the deformation. Can someone explain this to me?

2. Mar 18, 2017

### PeroK

There are lots of freely available web pages on this:

https://van.physics.illinois.edu/qa/listing.php?id=103

3. Mar 18, 2017

### gex

So the extra air pressure inside the ball pushes against the earth, resulting the earth to push back against the ball and causing it to lift off of the ground? Why does the force of the earth pushing against the ball not result in extra deformation of the ball but instead for lift off?

4. Mar 18, 2017

### PeroK

One concept that is important is equilibrium. Suppose you put a basketball on the ground. Gravity pushes it down, it deforms slightly, the air pressure is increased to the point where the force of the air pressure equals gravity. The ball is in equilibrium. It doesn't bounce.

But, if you drop the ball, it compresses past the point of equilibrium when it hits the ground. It must do this as extra upward force is needed not just to hold the ball (as in the case above) but also to slow it down. Now, when it has stopped, this extra upward force is still acting and is greater than gravity, so the ball begins to accelerate upwards and leaves the ground.

All this happens so fast that it's hard to see. It's the same principle, however, as landing on a trampoline - where you can see what happens. The higher you jump, the more you deform the trampoline when you land, the greater the maximum upward force of the trampoline and the higher it propels you back up.

5. Mar 18, 2017

### ZapperZ

Staff Emeritus
Why don't you start with something simpler? Do you have a problem understanding why, if I drop a rock onto a spring, the rock can bounce right back up?

Zz.

6. Mar 18, 2017

### gex

Oh, so the upward force of the ground would be the kinetic energy of the ball upon impact / the compression distance? And this force would continue acting on the ball even after its velocity reaches 0, therefore accelerating it until it leaves the ground?

7. Mar 18, 2017

### gex

Dropping a rock on the spring would create elastic energy due to Hooke's law yes?

8. Mar 18, 2017

### ZapperZ

Staff Emeritus
Upon compression, sure. But you didn't answer my question.

Zz.

9. Mar 18, 2017

### PeroK

I didn't expect you to know about Hooke's law. It's roughly the same for a basketball, with a very large spring constant. It would take quite a large force to compress a basketball by a few millimetres.

10. Mar 18, 2017

### gex

The thing is I roughly understand this process, the conversion of KE to EP back to KE energy, this is for an assignment in which I am aiming for an A on, which means I need to go into more detail. So the spring constant of the basketball would be dependent on the air pressure inside the ball yes? meaning that a lower air pressure inside the ball would require more compression and more energy lost to heat reducing the balls coefficient of restitution? What I still don't understand is the physics behind why the ball leaves the ground after compression. Sorry if I'm going in circles here but I just can't wrap my head around this.

11. Mar 18, 2017

### gex

To answer your question, the rock would compress the spring until the force of the spring acting the rock exceeds its downwards force, causing an acceleration in the opposite direction yes?

12. Mar 18, 2017

### ZapperZ

Staff Emeritus
And what I don't understand here is why you are not understanding it. It is why I asked you for the simpler form of a spring compression. You never answered if you had any problem understanding the rock-spring scenario that I asked earlier. BTW, it would make a lot easier to know to whom you are responding to if you quote the exact post.

So, for the second and last time before I decide to bow out of this one, do you also have a problem understanding why the rock would bounce back up after it drops onto the spring?

Zz.

13. Mar 18, 2017

### ZapperZ

Staff Emeritus
Yes, and do you understand why it would bounce back up and leave the spring?

See? This is why it would have been easier if you had started quoting posts from the beginning.

Zz.

14. Mar 18, 2017

### gex

My apologies, I am very new to posting on forums.

Thank you for your persistence ZapperZ, so the rock would leave the spring because the spring would change direction due to having an upwards force greater than the weight of the ball. This is where it gets hazy for me, so the rock would leave the spring because it would get an acceleration equal to the force of the spring divided by its mass? Or would the rock leave the spring due to the upwards inertia it gains from the spring changing direction?

I'm not too sure about this but the above scenarios are the two possible explanations that came to me.

15. Mar 18, 2017

### PeroK

Why doesn't the basketball stick to the ground? The same reason a tennis ball doesn't stick to your racket or a baseball to a bat.

The underlying reason why some things bounce and some don't is essentially whether they can permanently change their shape (under the forces involved). A piece of putty can change its shape. A flat basketball can too. But a pumped up basketball cannot change shape expect under permanent force. So, once compressed, like the spring, it will force itself back into shape given the chance.

16. Mar 18, 2017

### gex

What would this force equal? Is the only way to calculate this force to determine the ball's spring constant and compression distance? And is this tendency to return to its original shape what gives the basketball an upwards force? So that it can leave the ground and return to its spherical shape once again?

17. Mar 18, 2017

### ZapperZ

Staff Emeritus
But this where you need to question yourself on "what causes all those projectile problems that I've solve before to have an initial velocity?"

The compressed spring has a force that pushes the rock upwards. It initially causes the rock to slow down, then momentarily stopped at maximum compression, and then it pushes it back up. If energy isn't lost (or the loss is negligible), then at maximum extension, the rock will be pushed with the same speed as when it first comes in contact with the spring on the way down. That's why it will leave the spring.

As for the ball, you can think of the ball as "a spring with mass". The flexible, elastic skin, when filled with air, mimics a less-efficient-spring. The deformation is similar to a spring compression.

Zz.

18. Mar 18, 2017

### PeroK

This is where I bow out. If you don't get it by now, there's nothing more I can do to help you.

19. Mar 18, 2017

### gex

Thank you for for helping PeroK, if anyone is willing the answer my above questions I will appreciate it very much.

20. Mar 18, 2017

### A.T.

It's rather non-linear so you cannot model it with a constant.

Yes.

21. Mar 19, 2017

### sophiecentaur

I will get on my hobby horse again and suggest that this should be described just in terms of Energy Transfer and avoid the nasties of the Maths of the actual motion involved.
Firstly, (to recap stuff already posted) assume that the collision loses no energy. It's easier that way and they design most balls to lose as little energy as possible (ignore squash balls here). Also, can we have the collision against a wall, rather than the floor and have a massless spring on a mass M? (As simple as we can make it). The Kinetic Energy of the ball (Mv2/2 will all be converted into strain energy in the spring. At this stage, the mass is stationary and ready to move off in the other direction. The energy stored in the spring will be kx2/2 (notice the identical forms of the mathematical expressions for the two energies) where x is the amount the spring has been compressed and k is the spring constant. The force F against the wall, at maximum compression will be kx (that's how k is defined). So:
Mv2/2 = kx2/2
and F = kx
Those two equations tell you all you need to know for the ideal case.
People always want to know the "Force of the Collision" but there's no answer to that question, as it stands.
Notice that the value of F will depend on the spring constant so a high k will involve a large F and a low k will involve a low F for any given approach speed. So you can have any force in this experiment, if you choose the k of your spring / ball / floor etc. but the rebound speed will still be the same as the approach speed - as long as the collision is lossless.
Edit: Regarding the OP. As well as the air being compressed, the skin of the ball will also stretch, storing energy. If you use a ball full of ('incompressible') water, that will also bounce and the only distortion that can occur will be in the rubber skin. This will involve a high k and, hence, a much higher Force.

22. Mar 19, 2017