__Question 5.28__:
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}, 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__:Speed of the water stream,

*v*= 15 m/s

*A*= 10

^{–2}m

^{2}

Volume of water coming out from the pipe per second,

*V*=

*Av*= 15 × 10

^{–2}m

^{3}/s

Density of water,

*ρ*= 10

^{3}kg/m

^{3}

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 = ∆P / ∆t

= mv / t

= 150 X 15 = 2250 N

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