Question

A water jet,whose cross sectional area is 'a' strikes a wall making an angle theta with the wall and rebounds with same elastically.The velocity of water of density 'd' is v.
Force exerted on wall is
​

Answer 1

$\huge\mathcal{\underbrace{SOLUTION:-}}$

✍️ See the attachment picture for direction of components of velocity .

Let,

• v = velocity

• d = density

• V = volume

• m = mass

$\mathcal{\purple{Force\:=\:Rate\:of\:change\:in\:momentum\:}}$

$\rm{\implies\:Force\:=\:\dfrac{dp}{dt}\:}$

Where,

• dp = change in momentum .

✍️ The two components of velocity, ‘vcosθ’ are cancelled to each other . Because, in which direction water came, at that direction they left .

✍️ And the two components of velocity, ‘v sinθ’ are added to each other . Because, in which direction water came, that's opposite direction they left .

So,

$\rm{\implies\:change\:in\:momentum\:(dp)\:=\:m\:v\sin\theta\:-\:(-\:m\:v\sin\theta)\:}$

$\rm{\implies\:dp\:=\:m\:v\sin\theta\:+\:m\:v\sin\theta\:}$

$\rm{\red{\implies\:dp\:=\:2\:m\:v\sin\theta\:}}$

✍️ Now, calculating the force exerted on the wall .

$\rm{\implies\:F\:=\:\dfrac{dp}{dt}\:}$

$\rm{\implies\:F\:=\:\dfrac{d(2\:m\:v\sin\theta)}{dt}\:}$

$\rm{\implies\:F\:=\:2\:v\sin\theta\:\times\:\dfrac{dm}{dt}\:}$

• mass = density(d) × volume(V)

$\rm{\implies\:F\:=\:2\:v\sin\theta\:\times\:\dfrac{d({\green{d}}V)}{dt}\:}$

$\rm{\implies\:F\:=\:2\:(v\sin\theta)\:{\green{d}}\:\times\:\dfrac{dV}{dt}\:}$

• Volume(V) = Area(a) × Length(l)

$\rm{\implies\:F\:=\:2\:(v\sin\theta)\:{\green{d}}\:\times\:\dfrac{d(a\:l)}{dt}\:}$

$\rm{\implies\:F\:=\:2\:(v\sin\theta)\:{\green{d}}\:a\:\times\:\dfrac{d\:l}{dt}\:}$

• velocity(v) = $\mathcal{\dfrac{displacement}{time}\:}$

✍️ So, Here $\rm{\dfrac{d\:l}{dt}\:=\:velocity(v)\:}$

$\rm{\implies\:F\:=\:2\:(v\sin\theta)\:{\green{d}}\:a\:v\:}$

$\bigstar\:\rm{\boxed{\blue{\implies\:Force\:=\:2\:a\:d\:v^2\sin\theta\:}}}$