Answer to Question #127893 in Mechanics | Relativity for Atharv Shukla

Question

A horizontal jet of water coming out of a pipe of area of cross section 20cm hits a vertical wallwith a velocity of 10ms and rebounds with the same speed . The force exerted by the water on the wall is

Answer 1

Explanations & Calculations

• To do this sum consider about some volume of water having a mass “m(kg)” hitting the wall at the given velocity “v” and rebounds after a period “t (s)”.
• Due to the change of momentum within this period, a force (F) is exerted on that volume of water (same is exerted on the wall at the point of hit).
• Applying the equation related to momentum change,

$\qquad\qquad \begin{aligned} \small F &= \small \frac{mv-mu}{t}\\ &= \small \frac{mu-m(-u)}{t}\\ &= \small \frac{2mu}{t}\\ & = \small 2u\big(\frac{m}{t}\big) \end{aligned}$

• This $\large \frac{m}{t}$ is called the rate of mass flow which equals to,

$\qquad\qquad \begin{aligned} &= \small \frac{Ax\rho}{t} &= A\rho\big(\frac{x}{t}\big) &=A\rho u \end{aligned}$   : $\rho =$ density of water

• Therefore,

$\qquad\qquad \begin{aligned} \small F &= \small 2u\times Au \rho\\ &= \small 2Au^2 \rho\\ &= \small2\times20\times10^{-4}m^2\times(10ms^{-1})^2 \times10^3kgm^{-3}\\ &= \small \bold{40N} \end{aligned}$