#### 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} 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 1

Speed of the water stream, *v* = 15 m/s

Cross-sectional area of the tube, *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 = `(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} m^{2} x 15 ms^{-1} = 15 x 10^{-2} m^{3} s^{-1}

Mass of water hitting the wall per second = 15 x 10^{-2} x 10^{3} 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.