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A stream of water flowing horizontally with a speed of 15 m/s2 gushes out of a tube of cross - sectional area 102 m2, and hits at a vertical wall nearby. What is the force (in newtons) exerted on the wall by the impact of water, assuming it does not rebound?


Solution

Figure shows a column of water of length 15 m and cross - section area 102 m2 at rest. Let us calculate the mass of water flowing out per second and hitting the wall. When the stream of water has a speed of 15 m/s, the particle of water at section B would have travelled a distance of 15 m in one second and arrived at the section A. The water contained in this column would have gone out in one second and hit the wall. 

Thus, the volume of water coming out of A in one second = volume of water contained in a cylinder of length 15 m and area of cross - section 0.01 m2
= Length×Area of cross - section
=15×102=0.15 m3
Mass of water gushing out per second
= Volume of water×Density of water
=0.15×1000=150 kg
[density of water =1000 kg/m3]
|Change in momentum| =22500=2250 N
Since change in momentum per second is the force, hence
force exerted on the wall will be 2250 N.

Physics

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