Question

A water pipe has an internal diameter of $10\text{cm}$. Water flows it at the rate of $20\text{m/sec}$. The water jet strikes normally on a wall and falls dead. Find the force( in N upto two decimal places) on the wall.(Assume $\pi =3.14$)

Answer 1

Mass of water flowing through the tube per second = $Av\rho$
where A = area of cross section and v = velcoity of water of density $\rho$.
Momentum change = $Av\rho v=A{v}^2\rho$
Thus, the force on the wall $=A{v}^2\rho =\pi \times {\left(\frac{5}{100}\right)}^2\times (20{)}^2\times 1000=3140\text{N}$