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 m^2
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?
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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 = ΔP/Δt
= mv/t
= 150 x 15 = 2250 N
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