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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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Since there is no friction and the track is level, any change in the speed of the car must be due to the rain.  By including the rain in our system, we have ensured that all x-forces are internal.  Thus, the x-momentum must be a constant.  We can therefore write:
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Consider the element of length {*}Δx{*} and area *A* and density {*}ρ{*}. Its mass must therefore be
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{latex}\begin{large}\[ p^{\rm system}_{x,i}m = p^{\rm rain}_{x,i} + p^{\rm car}_{x,i} = p^{\rm system}_{x,f} =  p^{\rm rain}_{x,f} + p^{\rm car}_{x,f}rho A \Delta x \] \end{large}{latex}

We define the initial time to be the instant before it begins to rain, because we have information about the car's velocity at that point.  The final time is taken to be 2.0 seconds after the rain begins.  Before the rain has accumulated in the car, it has zero x-velocity and hence zero x-momentum.  After it has accumulated in the car, it is moving with the same speed as the train car.  Thus, we can write:\\


{latex}\begin{large} \[ m^{\rm car}v^{\rm car}_{x,i}  = \left(m^{\rm car} + m^{\rm rain}\right)v_{x,f} \] \end{large}{latex}

{note}It might be confusing that we can treat the rain, which falls individually as drops, as if it were one solid object.  The reason we can get away with this is the assumption that all the rain drops fall with exactly the same x-momentum, and that they end up at rest with respect to the car.  Because all the drops experience the same momentum change we can simply group them together and consider the momentum of the "rain block".{note}

All that remains is to determine the mass of the rain.  We can do this by noting that the density of water is 1000 kg/m{color:black}^3^{color} and that the water has filled the car to 2.0 cm deep, indicating a collected mass of:

{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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