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?
Solution 1
Speed of the water stream, v = 15 m/s
Cross-sectional area of the tube, A = 10–2 m2
Volume of water coming out from the pipe per second,
V = Av = 15 × 10–2 m3/s
Density of water, ρ = 103 kg/m3
Mass of water flowing out through the pipe per second = ρ × V = 150 kg/s
The water strikes the wall and does not rebound. Therefore, the force exerted by the water on the wall is given by Newton’s second law of motion as:
F = Rate of change of momentum = `(triangleP)/(trianglet)`
`= "mv"/t`
= 150 x 15 = 2250 N
Solution 2
In one second, the distance travelled is equal to the velocity v.
Volume of water hitting the wall per second, V = av where a is the cross-sectional area of the tube and v is the speed of water coming out of the tube.
V = 10-2 m2 x 15 ms-1 = 15 x 10-2 m3 s-1
Mass of water hitting the wall per second = 15 x 10-2 x 103 kg s-1 = 150 kg s-1 [v density of water = 1000 kg m-3] Initial momentum of water hitting the wall per second
= 150 kg s-1 x 15 ms-1 = 2250 kg ms-2 or 2250 N Final momentum per second = 0
Force exerted by the wall = 0 – 2250 N = – 2250 N Force exerted on the wall = – (- 2250) N = 2250 N.
