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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 - Physics

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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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.

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NCERT Class 11 Physics Textbook
Chapter 5 Laws of Motion
Q 28 | Page 113
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