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  • If we increase or decrease the internal pressure in a basketball for example, does that affect the weight or the mass of the ball?

  • I am calculating the energy dissipated as the ball bounces having different internal pressures. I am using the formula: Energy dissipated= mg(h1-h2)
    should I take into account the mass of the air or just the mass of the ball as a constant throughout my calculations?

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As the air inside the ball is going to stay inside the ball it is the mass of the ball and the air within it which is constant and that should be used for your mass in working out the loss of gravitational potential energy.
For a full-sized basketball with a mass of about $620 \,\rm g$ if the air was at $15^\circ \rm C$ and at the recommended air pressure of about $1\frac 12$ atmospheres then the mass of air inside a basketball is about $13\, \rm g$.
All you need to do is measure the mass of the basketball with the air inside it.

There is another factor which you should consider and that is the work done due to viscous drag on the basketball.
The viscous drag $= \frac 1 2 C_{\rm D} A \rho v^2$ depends on the frontal area of the basketball $A$, the density of the air $\rho$, the speed of the basketball $v$ and the drag coefficient $C_{\rm D}$ which can be taken as being equal to approximately $0.5$ in this case.
What you need to decide is whether the force on the basketball is significant with the weight of the basketball.
If it is then you will have to take that into consideration in your calculations which is not so easy as the speed (and hence the viscous drag) changes as the ball falls and then rebounds.

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    $\begingroup$ Regarding drag: good rule of thumb is that it is significant if the total mass of air displaced is comparable to the mass of the object. Your calculated mass inside is about 2% of the total; when you move five ball diameters the mass of displaced air is about 10% of the total (very rough in-my-head calculation). So if you drop sufficient height it would seem drag cannot be ignored. The equation of motion for quadratic drag has a closed former so you can actually calculate the energy dissipated. Google "quadratic drag" and find the hyperphysics link. $\endgroup$ – Floris Feb 18 '17 at 17:36
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If one includes the air as part of the mass of the ball, then the more air, the more ball mass. Increasing the air pressure certainly increases the amount of air contained in the ball, and therefore would increase the ball mass. Air at one-atmosphere pressure has a mass of about $1.2 \space kg\space m^{-3}$. It goes up fairly linearly as pressure goes up. That is, if you double the pressure it doubles the mass (until the pressure is so high there are other changes in the air). Also, don't forget that when you compress air, it heats it up. So you'd have to decide whether you want to measure it that way or wait for it to cool down.

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  • $\begingroup$ It depends whether the pressure increases because you are pumping up the ball or because the ball is getting hotter. $\endgroup$ – Floris Feb 18 '17 at 17:37

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