Question

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?

Answer 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

Answer 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.