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' = (V)/(t)*\Delta t_(total)`

Where,

`\Delta t_(total)` It is the period of time we want to calculate total rainfall, that is

`V' = (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`