Question

A water jet strikes a vertical wall elastically at angle $\theta $ with the horizontal. The cross-sectional area of the pipe is A, density of water is D and the speed of water jet is v. the normal force acting on the wall is
(A) $DAv\cos \theta $
(B) \[DA{v^2}\cos \theta \]
(C) $2DAv\cos \theta $
(D) $2DA{v^2}\cos \theta $

Answer 1

Hint: In this question, the water from the pipe will strike the wall at an angle of $\theta $ it will also bounce back at the same angle. This is a question based on conservation of linear momentum which states that the momentum before collision of two bodies and after their collision is equal.
Formula Used:
\[F = \dfrac{{dp}}{{dt}}\]

Complete step by step solution:
 We know that the linear momentum of a body is the product of its mass and velocity so,
here if the water which strikes the wall has momentum $p = mv$ and angle of $\theta $ , then resolving it into the x and y components,
then the x component of linear momentum is $mv\cos \theta $and the y component will be $mv\sin \theta $
in the x direction, the momentum on striking and after rebound will be acting in opposite direction,
hence the momentum will be $dp = mv\cos \theta - ( - mv\cos \theta ) = 2mv\cos \theta $
we know that the rate of change of momentum is the force applied
so now the force will be $F = \dfrac{{dp}}{{dt}} = \dfrac{d}{{dt}}(2mv\cos \theta ) = 2v\cos \theta \dfrac{{dm}}{{dt}}$ since the velocity and the angle does not vary, the change is observed in mass
we know that the change in mass is known as the mass rate,
hence $\dfrac{{dm}}{{dt}} = \dfrac{d}{{dt}}[D \times V]$ where $D$ is the density of the water and V is the volume
$\dfrac{{dm}}{{dt}} = \rho \dfrac{d}{{dt}}[V]$since the density of water remains same
Now when a liquid is flowing in pipe of length L and cross sectional area A then the velocity of liquid is $v = \dfrac{L}{t}$where t is the time and volume $V = A \times L$
$\dfrac{{dV}}{{dt}} = Av$ put this equation of force
$F = \dfrac{{dp}}{{dt}} = 2v\cos \theta DvA = 2A{v^2}D\cos \theta $

Hence, the correct option is D.

Note: We need to keep in mind that when a liquid which flows in a streamline motion in a non-uniform cross-sectional area, then the velocity times the area of cross-section is the same or we can say constant at every point in the tube.