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{excerpt:hidden=true}Analyzing a continuous momentum flux (water from a hose).{excerpt}

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|!Hose Spraying Water.jpg!|
|Photo from Wikimedia Commons
Original by Doclector|


When you spray water from a garden hose you feel a backward force as you hold the nozzle. Similarly, anything you spray the water at feels a force in the direction away from the spray. What is the origin of this force, and what determines its magnitude?

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h4. Solution

{toggle-cloak:id=sysa} *System:*  {cloak:id=sysa}Element of water stream as a [point particle] acting on an article, also treated as a [point particle]. {cloak}

{toggle-cloak:id=inta} *Interactions:* {cloak:id=inta}We will *ignore the vertical direction*, so that the only interaction is the force due to the changes in momentum of the water and the item hit..

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{toggle-cloak:id=moda} *Model:*  {cloak:id=moda}[Momentum and External Force].{cloak}

{toggle-cloak:id=appa} *Approach:*  

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{toggle-cloak:id=diaga} {color:red} *Diagrammatic Representation* {color}

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|!Momentum Transport 01.PNG!|
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Imagine a stream of water as a cylinder of uniform cross-sectional area *A* and density {*}ρ{*}. We consider an elemental unit of this that is {*}Δx{*} long. It travels, as does the rest of the stream, horizontally at velocity *v* (We will ignore the downward force of gravity here).


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{toggle-cloak:id=matha} {color:red} *Mathematical Representation* {color}

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Consider the element of length {*}Δx{*} and area *A* and density {*}ρ{*}. Its mass *m* must therefore be
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{latex}\begin{large}\[ m = \rho A \Delta x \] \end{large}{latex}
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Since it travels with velocity *v*, its momentum is thus 
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{latex}

\begin{large} \[ \vec{p} = \rho A \Delta x \vec{v} \] \end{large}{latex}
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{latex}\begin{large} \[ \Delta \vec{p} = \rho A \Delta x \vec{v} \] \end{large}{latex}
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If this happens in a time {*}&

Delta;t{*}, then the change in momentum with time (which is just the *Average Force*) is
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{latex}\begin{large} \[ \vec{F_{avg}} = \frac{\Delta \vec{p}}{\Delta t} = \rho A \vec{v} \frac{\Delta x}{\Delta t} \] \end{large}{latex}
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But {*}Δx/Δt{*} = *v* , so the [magnitude] of the Average

 Force is thus
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{latex}\begin{large} \[ F_{avg} = \rho A v^{2} \]\end{large}{latex}
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{latex}\begin{large}\[ m^{\rm rain} = \rho^{\rm water}V^{\rm rain} = (\mbox{1000 kg/m}^{3})(\mbox{10 m}\times\mbox{3 m}\times\mbox{0.02 m}) = \mbox{600 kg} \]\end{large}{latex}

where ρ^water^ is the density of water and _V_^rain^ is the volume of the accumulated rain.

We can now solve to find:

{latex}\begin{large} \[ v_{x,f} = \frac{m^{\rm car}v^{\rm car}_{x,i}}{m^{\rm car} + m^{\rm rain}} = \mbox{1.6 m/s}\]\end{large}{latex}

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