A stream of water flowing horizontally with a speed . . .

Question : A stream of water flowing horizontally with a speed of 15 m/s gushes out of a tube of cross-sectional area 10-2 m2, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?


Doubt by Krish

Solution : 
velocity (v) = 15 m/s
Area of cross section (A) = 10-2 m2
We know, 
Distance = speed × time
Distance covered by the water in 1 sec
Distance = (15 m/s)×(1 s)
Distance = 15 m

We also know, 
Volume = Distance × Area
So, volume of water coming out of the pipe per second
= 15× 10-2
= 0.15 m3/s
Now 
Mass of the water coming out of the pipe per second
Mass = Density × Volume
Mass (m) = (103 kg/m3)×(0.15 m3/s)
[∵ Density of water = 103 kg/m3]
Mass = 150 kg/s

Using the Second Law of Motion
F= dp/dt
F=m|(v-u)|/t
F=(150)|(0-15)|/(1)
F=150×15
F=2250 N

Hence, force exerted on the wall by the impact of water is 2250 N.