Question

A jet of water with a cross-sectional area $a$ is striking against a wall at an angle $\theta$ to the horizontal and rebounds elastically. If the velocity of water jet is $v$ and the density is $\rho$, the normal force acting on the wall is:

• A. $2a{v}^2\rho \cos \theta$
• B. $a{v}^2\rho \cos \theta$
• C. $2av\rho \cos \theta$
• D. $av\cos \theta$

Answer 1

The correct option is A $2a{v}^2\rho \cos \theta$

Consider a differential element of water of length $dl$ in the water jet before collision.

Corresponding differential volume will be $dV=adl$

and differential mass will be $dm=\rho dV=\rho adl$

Momentum of the differential element along the direction of motion will be:

$dp=vdm=\rho avdl$

Only, change in horizontal component of momentum contributes to the normal force acting on the wall.

Initial horizontal component of momentum of the differential element is given by:

$p_{x,i}=\rho avdl\cos \theta$

Final horizontal component of momentum of the differential element is given by:

$p_{x,f}=-\rho avdl\cos \theta$

Change in momentum along the $x-$ direction is:

$d{p}_x=d{p}_{x,f}-d{p}_{x,i}=2\rho avdl\cos \theta$

Force acting by wall is rate of change of momentum of water.

$F=\frac{d{p}_x}{dt}=\frac{d\left(2\rho avdl\cos \theta \right)}{dt}$

$F=2\rho a{v}^2\cos \theta$ ($v=dl/dt$)