### Q.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?

**Answer:-**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 = ∆P / ∆t

= *mv* / *t*

= 150 × 15 = 2250 N