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 2nd 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.