A stream of water flowing horizontally with a speed of 15 m s^-1 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?

Given:

Speed of water stream, v = 15 ms^-1

Cross-sectional area of the tube, A = 10^-2 m²

Density of water, ρ = 1000 Kgm^-3

Volume of water leaving the tube per second = A× v

V = A× v …(1)

V = 10^-2 m²× 15 ms^-1

V = 0.15 m³s^-1

Mass of water flowing out of the pipe = density of fluid× volume

M = ρ× V …(2)

M = 1000 Kgm^-3× 0.15 m³s^-1

M = 150 Kgs^-1

Since the water doesn’t rebound after striking the Wall, hence according to Newton’s 2^nd law , we can write,

F = Rate of Change of momentum

F = change in momentum / time

F = |mv-mu|/ t …(3)

Where,

M = mass of water

v = final velocity = 0

u = initial velocity

t = time = 1sec

the equation (3) becomes,

F = mu/t

By putting values in the above equation,

F = 150 Kgs^-1× 15 ms^-1

F = 2250 N.

The water exerts a force of 2250 N on the wall.