Question

A water pipe has an internal diameter of $10\text{cm}$. Water flows through it at the rate of $20\text{m/sec}$. The water jet strikes normally on a wall and falls dead. Find the force on the wall (in $\text{newtons}$). (round off to the nearest integer)
[${\rho }_w=1000{\text{kg/m}}^3$]

Answer 1

Mass of water flowing through the tube per second $m=Av\rho$
where $A=$ area of cross section and $v=$ velocity of water of density $\rho$.

Momentum change/second for water striking the wall $=m(0-v)=-Av\rho v=-A{v}^2\rho$

We know that, force acting on the jet is equal to the rate of change of momentum [Newton's 2nd law].

& Force exerted by the jet on the wall is equal and opposite to this [Newton's 3rd law].

Thus, the force on the wall $=A{v}^2\rho$
$=\pi \times {\left(\frac{5}{100}\right)}^2\times (20{)}^2\times 1000=3141.59\approx 3142\text{N}$