# Conservation of energy of water flowing out of a small hole at the bottom of a bottle

There is a bottle filled with water with a small hole at the bottom. Water flows out from the hole. The height of the water surface from the bottom is $$h$$ and gravitational acceleration is $$g$$. The book says that according to the conservation of energy, $$mgh=mv^2/2$$. Pay attention to the $$m$$ here. The book does not specify whose $$m$$ it is.

How to comprehend the equation $$mgh=mv^2/2$$? Does the $$m$$ here refer to the unit mass or the overall mass of water?

This is an example of Torricelli's Law. The $$m$$ here refers to any mass of water that flows out from the hole. Since both kinetic and potential energy are proportional to mass, the mass cancels out.

This result can also be obtained directly from Bernoulli's equation.

$$m$$ is the mass of considered volume of the liquid. It varies. So you can see, in your equation $$mgh=\frac 12 mv^2$$, $$m$$ cancels out from the both sides. It's your choice to decide whether it is unit mass or the overall mass of water (not recommended) or the mass of a selected volume. Though it doesn't matter in your final result, which I guess is $$v$$. (But it is better to take $$m$$ as the mass of a tiny amount of water flows through the hole)

To be perfectly correct, I recommend you to consider $$m$$ as the mass of a tiny amount of water (as I have mentioned before). The reason is this: since $$h$$ is the height to the surface from the bottom, the gravitational potential energy of overall mass of water is $$mg\frac h2$$, because the centre of mass situated at the height of $$\frac h2$$ from the bottom. To make your equation valid, you are considering a tiny amount of water at the surface (think it as a droplet for better understanding) which has potential energy of $$mgh$$. That is the tiny amount of water (or droplet) which gains $$v$$ exit velocity at the hole. So it is not suitable to consider $$m$$ as the mass of overall water, because $$v$$ varies with the decrease of $$h$$.

• So how to comprehend that equation? If the 'm' refers to the tiny amount of water at the surface, the equation doesn't seem to hold because the h here is not the height it falls.
– Mars
Jul 30, 2021 at 13:15
• So you have to know Bernoulli's equation. Then you will better understand.
– ACB
Jul 30, 2021 at 13:18
• @Mars, It seems you have misunderstood the question. That tiny amount of water does not fall freely to get the v velocity. It goes along a streamline from the surface to the hole.
– ACB
Jul 30, 2021 at 13:26
• That makes sense, thank you.
– Mars
Jul 30, 2021 at 13:33

I would say it refers to the mass of the whole water overall (volume of the water), as the water is what is flowing into the whole, that $$m$$ value in the whole is the same as the $$m$$ value in the bottle (assuming all the water flows out).

Hence if the water in the bottle is filled to a certain height in the bottle it has potential energy = $$mgh$$ which is transferred into kinetic energy = $$\frac 12 mv²$$ (where both $$m$$ values are the same)

In Bernoulli's equation, the density $$\rho$$ (mass per unit volume) appears in place of the m in your book's equation. Then there is no more ambiguity.