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Let's take a small coke bottle (a plastic one). Now fill the bottle entirely with water not even letting a small air bubble, however we cannot stop forming a tiny bubble. Now this water bottle when dropped will bounce of the floor as shown in the image.enter image description here

I think that it became elastic due to the shock waves produced in water due to sudden impulse on the bottle.

However, if the air bubble inside the bottle is somewhat large then it would lose it bouncy nature. I don't know how these waves make this. Could you help me to understand the exact reason behind this?

Even the bouncy nature is dependent of configuration of fall. In the given figure if the bottle is allowed to leave at the below given configuration the bottle is not a bouncy one. How could you help me in understanding this phenomenon?enter image description here

Also, I found an interesting fact about this. Not only this orientation, but also it depends on the stiffness of the material. I expected that when stiffness decreases then it will bounce more. But after observations on it, I got result that when stiffness increases then bouncy nature increases to a peak value then after increasing the stiffness the bouncy nature decreases. But why?

Why this phenomenon is seen in bottles with liquids but not gases? What if it is a solid!

I even found a new thing that bounciness also depend upon the length of the bottle. I had guessed that “The more spherical it is, the more bouncier it is”. That is length is inversely proportional to bounciness. Is this assumption correct?

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    $\begingroup$ I got result that when stiffness increases then bouncy nature increases to a peak value then after increasing the stiffness the bouncy nature decreases - This is interesting, but how did you experimentally change the stiffness, take a bottle of thicker plastic? $\endgroup$ – Aleksey Druggist Oct 7 '19 at 10:49
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    $\begingroup$ After observing that it has bouncy nature. I and my friends played with it for sometime. Then the bottle became softer, and bounced height also got reduced. But if I take a thick bottle intially, I didn't see any thing like before. $\endgroup$ – Sandesh Goli Oct 7 '19 at 10:58
  • $\begingroup$ But also, great to get a reason for it. What might be the reason!!! $\endgroup$ – Sandesh Goli Oct 7 '19 at 10:59
  • $\begingroup$ I'm not sure, but perhaps the whole thing is the viscosity of the water. A plastic bottle can be considered as a spring with a bending stiffness $k$, mass $m$ in the presence of damping force. If the stiffness is too high, the damping time when hitting the surface is short. The velocity components arising in the fluid upon hitting are higher and, accordingly, the internal friction (damping) is higher and the impact is inelastic with no bouncy. That is, a stiffer collision with the surface is less elastic $\endgroup$ – Aleksey Druggist Oct 7 '19 at 12:27
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It's because if there's an air bubble alot of energy is used to momentarily compress the air on the collision, and subseqent disturbances on the water-air boundary cause a lot of energy disperssion. If you just have a bottle full of water, the dispersion of energy is much smaller and the bounce is more elastic.

As MArtin pointed out in his comment, when a bubble of air is present, a bottle itself may be deformed more easily (since the air is easier to compress than water, it takes less energy for it to change its volume), and the deformation of bottle helps to disperse the energy.

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    $\begingroup$ That’s a good answer but I wonder: have you missed something more obvious? When the absolutely full bottle is dropped, the incompressibility of water prevents it deforming; when there is an air bubble, its compressibility allows the bottle to deform momentarily, and the crumpling and restoration of the walls is likely to use up more energy than anything that happens in the fluid phases. Compare how much bouncier a ball is when fully inflated. $\endgroup$ – Martin Kochanski Jul 9 '19 at 5:05
  • $\begingroup$ You're right, the deformation of the bottle will also help to disperse the energy. $\endgroup$ – Adam Latosiński Jul 9 '19 at 9:40

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