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\begin{large}\[ m = \rho A \Delta x \] \end{large}

\begin{large} \[ \vec{p} = \rho A \Delta x \vec{v} \] \end{large}

{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}Train car as a [point particle] plus the rain collected in 2.0 minutes as a [point particle]. {cloak}

{toggle-cloak:id=inta} *Interactions:* {cloak:id=inta}We will *ignore the vertical direction*, so there are no *relevant* external forces (gravity and the normal force each provide no x-component).

{note}The _y_ direction can be neglected here because the normal force of the ground guarantees that the train car will not accelerate in the _y_ direction.  One advantage of the vector nature of momentum is that even though the y-momentum of our system is clearly not conserved, the x-momentum _is_ conserved.{note} {cloak}

{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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As remarked above, because the rain has only a y-velocity initially, we can treat the process of collecting the rain as a typical collision in the x-direction, with momentum conserved.  Thus, we can effectively diagram the situation as shown below.


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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:

{latex}\begin{large}\[ p^{\rm system}_{x,i} = p^{\rm rain}_{x,i} + p^{\rm car}_{x,i} = p^{\rm system}_{x,f} =  p^{\rm rain}_{x,f} + p^{\rm car}_{x,f} \] \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} \[ m^{\rmDelta car}v^{\rm car}_{x,i} \vec{p} = \left(m^{\rm car} + m^{\rm rain}\right)v_{x,frho A \Delta x \vec{v} \] \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} \[ \vec{F_{avg}} = \frac{\Delta \vec{p}}{\Delta t} = \rho A \vec{v} \frac{\Delta x}{\Delta t} \] \end{large}

\begin{large} \[ vF_{x,favg} = \frac{m^{\rm car}v^{\rm car}_{x,i}}{m^{\rm car} + m^{\rm rain}} = \mbox{1.6 m/s}\rho A v^{2} \]\end{large}{latex}

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