Question

If a rainstorm drops 5 cm of rain over an area of 13 km2 in the period of 3 hours, what is the momentum (in kg · m/s) of the rain that falls in five seconds? Assume the terminal velocity of a raindrop is 10 m/s. (Enter the magnitude. The density of water is 1,000 kg/m3.)

Answer 1

To solve the problem it is necessary to apply the concepts related to Conservation of linear Moment.

The expression that defines the linear momentum is expressed as

P=mv

Where,

m=mass

v= velocity

According to our data we have to

v=10m/s

d=0.05m

$A=13*10^6m^2$

Volume $(V) = A*d = (15*10^6)(0.03) = 3.9*10^5m^3$

t = 3hours=10800s

$\rho = 1000kg/m^3$

From the given data we can calculate the volume of rain for 5 seconds

$V' = \frac{V}{t}*\Delta t_{total}$

Where,

$\Delta t_{total}$ It is the period of time we want to calculate total rainfall, that is

$V' = \frac{3.9*10^5}{10800}*5$

$V' = 1.805*10^2m^3$

Through water density we can now calculate the mass that fell during the 5 seconds:

$m' = V'*\rho$

$m' = 1.805*10^2*1000$

$m' = 1.805*10^5m^2$

Now applying the prevailing equation given we have to

$P=m'v$

$P = (1.805*10^5)(10)$

$P = 1.805*10^6 Kg.m/s$

Therefore the momentum of the rain that falls in five seconds is $1.805*10^6 Kg.m/s$