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

|!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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Let's assume that this element of the stream strikes an object (a wall, say), and breaks up, dissipating in all directions. the stream element loses all of its momentum in the process. The change in momentum of the stream element is thus
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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 {*}Δt{*}, 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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