An initially empty beaker, in the shape of a cylinder with cross sectional area $A$ , is left out in the rain. The raindrops hit the beaker vertically downward with speed $v$ . The rain continues at a constant rate, so the height of the water in the beaker $h\left(t\right)$ increases with time $t$ at a rate $\frac{dh}{dt}=u$ , where $u$ is small compared to $v$ . The raindrops quickly come to rest inside the beaker, so we can neglect any kinetic energy of the water that has collected in the beaker. Let $\rho$ denote the density of water (i.e., the mass per unit volume).

If the beaker is placed on a weighing scale, while the beaker is still in the rain, the impact of the raindrops on the beaker will cause the reading on the scale to be larger than the weight of the beaker and the water it contains. By how much is the reading on the scale increased by the impact of the raindrops? (Neglect the effect of raindrops that hit the scale directly.)

The mass in the beaker is given by the product of the density $\mathrm{\rho }$ and the volume $\mathrm{V} = \mathrm{Ah}$ that is filled, i.e.

$\mathrm{m}=\mathrm{\rho Ah}$

When the height changes at a rate $\frac{\mathrm{dh}}{\mathrm{dt}}=\mathrm{u}$ , the mass changes at the rate

$\frac{\mathrm{dm}}{\mathrm{dt}}=\mathrm{\rho A}\frac{\mathrm{dh}}{\mathrm{dt}}=\mathrm{\rho Au}$

The height of the center of mass of the water in the beaker is at $y_{cm}= \frac{h}{2}$ (at least if we assume incompressible fluid, i.e. a constant density, which is a very good approximation for water). The rate at which the height of the center of mass increases is

$\frac{{\mathrm{dy}}_{\mathrm{cm}}}{\mathrm{dt}}=\frac{1}{2}\frac{\mathrm{dh}}{\mathrm{dt}}=\frac{\mathrm{u}}{2}$

The center of mass is moving upward because more water is added to the beaker; i.e. as time evolves we are talking about a whole sequence of physical systems, not about the center of mass of a fixed given system. At any given time, the water in the beaker is at rest and hence its momentum vanishes. It is therefore incorrect to conclude that the water in the beaker has a total vertical momentum.

The raindrops enter the beaker with a speed $v$ and then quickly come to rest inside the beaker. Consider a short time interval $∆t$ , and consider the system that consists of the beaker, all the water that has landed in the beaker by the beginning of the time interval, plus the water that will come to rest inside the beaker during the time interval. During the time interval $∆t$ the height of the water in the beaker will increase by $∆\mathrm{h}=\mathrm{u}∆\mathrm{t}$ , so the volume of water that will come to rest during the time interval is $∆\mathrm{V}=\mathrm{A}∆\mathrm{h}=\mathrm{Au}∆\mathrm{t}$ , shown shaded in the diagram on the right. The mass of this water is $∆\mathrm{M}=\mathrm{\rho }∆\mathrm{V}=\mathrm{\rho Au}∆\mathrm{t}$ . Taking the vertical direction as the $y$ -direction, the momentum of this water at the beginning of the interval is $\overrightarrow{{\mathrm{P}}_{\mathrm{i}}}=\left[0,-\mathrm{v}∆\mathrm{M},0\right]\left[0,-\mathrm{\rho Auv}∆\mathrm{t},0\right]$ . The final momentum is zero, after the raindrops come to rest, so $\overrightarrow{P_f}-\overrightarrow{P_i}=-\overrightarrow{P_i}$ . The total external force applied to the system is equal to the rate of change of its momentum, so

$\overrightarrow{{\mathrm{F}}_{{\mathrm{ext}}_{\mathrm{tot}}}}=\frac{∆\overrightarrow{\mathrm{P}}}{∆\mathrm{t}}=\frac{∆\overrightarrow{{\mathrm{P}}_{\mathrm{i}}}}{∆\mathrm{t}}\left[0,\mathrm{\rho Auv},0\right]$

From the diagram one can see that $\overrightarrow{F_{{ext}_{tot, y}}}=N-M\left(t\right)g$ where $M\left(t\right)$ is the total mass of the beaker and the water in it at time $t$ . So,

$\mathrm{N}=\mathrm{M}\left(\mathrm{t}\right)\mathrm{g}+\mathrm{\rho Auv}$

By Newton's third law the beaker exerts a force of equal magnitude on the scale, which determines its reading. The first term is just the weight of the system, while the question asks by how much the reading is increased by the impact of the raindrops. Thus the second term in the above expression is the answer to the question: $∆\mathrm{N}=\mathrm{\rho Auv}$

Since, scales often read in units of mass rather than force, it would be equally correct to say that the mass reading would be increased by $∆\mathrm{M}=\frac{\mathrm{\rho Auv}}{\mathrm{g}}$