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Excerpt
hiddentrue

Analyzing a continuous momentum flux (falling water).

A railroad car that is open on the top is rolling freely (without friction) along a straight, level section of track when it begins to rain heavily at a steady rate. The rain collects in the car at a rate of 1.0 cm per minute. The dimensions of the car are 10.0 m long by 3.0 m wide and the mass of the (empty) car is 1200 kg.

Composition Setup
 Deck of Cards
idprobdeck
 Wiki Markup
{excerpt:hidden=true}Analyzing a continuous momentum flux (falling water).{excerpt}

A railroad car that is open on the top is rolling freely (without friction) along a straight, level section of track when it begins to rain heavily at a steady rate.  The rain collects in the car at a rate of 1.0 cm per minute.  The dimensions of the car are 10.0 m long by 3.0 m wide and the mass of the (empty) car is 1200 kg.


{composition-setup}{composition-setup}

{deck:id=probdeck}

{card:label=Part A}

h4. Part A

If the car was initially moving at a rate of 2.5 m/s and the rain is falling straight down, what is the speed of the car 2.0 minutes after the rain has begun?  

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:*  

{cloak:id=appa}

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

{cloak:id=diaga}
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.

!TrainRain.png!
{cloak:diaga}

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

{cloak:id=matha}
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}
 Card
labelPart A
Part A
If the car was initially moving at a rate of 2.5 m/s and the rain is falling straight down, what is the speed of the car 2.0 minutes after the rain has begun? 
Solution
 Toggle Cloak
idsysa
 System: 
 Cloak
idsysa
Train car as a  plus the rain collected in 2.0 minutes as a . 
 Toggle Cloak
idinta
 Interactions: 
 Cloak
idinta
We will ignore the vertical direction, so there are no relevant external forces (gravity and the normal force each provide no x-component).
 
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.
 
 Toggle Cloak
idmoda
 Model: 
 Cloak
idmoda
.
 Toggle Cloak
idappa
 Approach: 
 Cloak
idappa
 Toggle Cloak
iddiaga
  Diagrammatic Representation 
 Cloak
iddiaga
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.
Image Added
 Cloak
diaga
diaga
 Toggle Cloak
idmatha
  Mathematical Representation 
 Cloak
idmatha
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} \[ m^{\rm car}v^{\rm car}_{x,i}  = \left(m^{\rm car} + m^{\rm rain}\right)v_{x,f} \] \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}
 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".
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/m3 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}
where ρwater is the density of water and Vrain 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}

{cloak:matha}
{cloak:appa}

{card}{card:label=Part B}

h4. Part B

Consider again the situation of Part A.  Suppose that the rain is falling with a speed of 10.0 m/s, and that it comes to rest with respect to the train car immediately upon impact.  At the instant the rain begins to hit the car, what is the force exerted on the train car by the rain?

h4. Solution

{toggle-cloak:id=sysb} *System:*  {cloak:id=sysb}The rain entering the car in a time _t_, treated as a [point particle].{cloak}

{toggle-cloak:id=intb} *Interactions:* {cloak:id=intb}External forces from gravity and from the railroad car.  We assume the collision force is much larger than the force of gravity during the collision.{cloak}

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

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

{cloak:id=appb}
{note} Note that we have chosen to model the water, even though we are asked for the force from the water on the car.  During the collision, the collision forces on the water dominate and so we can solve for them.  The collision forces on the car, however, are not dominant.  The y-component of the collision force is completely balanced by the normal force.  For this reason it is hard to solve for the force using the car's dynamics.  Instead, we will use [Newton's 3rd Law|Newton's Third Law] to relate the collision force on the rain to the force from the rain on the car.{note}

We will make the approximation that enough raindrops are hitting the car that the rain can essentially be treated as a continuous flow of water.  The force that acts on the water is equal to the time derivative of its momentum.  Because of the discontinuous nature of the impact (the rain is assumed to stop instantaneously) it is impossible to express the rain's momentum as a continuous function of time.  For this reason, we must re-express the law of interaction:

{latex}\begin{large}\[ F_{x} = \frac{dp_{x}}{dt} = \lim_{t \rightarrow 0} \frac{p_{x}(t) - p_{x}(0)}{t} = \lim_{t \rightarrow 0} \frac{m^{\rm rain}(t)v^{\rm car} - m^{\rm rain}(t)(\mbox{0 m/s})}{t}\]
\[F_{y} = \frac{dp_{y}}{dt} = \lim_{t \rightarrow 0} \frac{p_{y}(t) - p_{x}(0)}{t} = \lim_{t \rightarrow 0} \frac{m^{\rm rain}(t)(\mbox{0 m/s}) + m^{\rm rain}(t)v^{\rm rain}}{t}\]\end{large}{latex}

{note}Note that the rain is falling downward, which we have chosen to be the negative _y_ direction.{note}

The mass of rain that enters the car in a time _t_ can be expressed in terms of the rate at which the rain accumulates in the car, known to be:

{latex}\begin{large}\[ r = \frac{\mbox{1.0 cm}}{\mbox{1.0 min}} = \frac{\mbox{0.01 m}}{\mbox{60 s}} = \mbox{0.00017 m/s} \]\end{large}{latex}  

{warning}Because rain falls in drops, not as a continuous sheet, the rate of height accumulation is *not* the same as the rain's speed through the air!{warning}

In terms of this rate _r_, we can write:

{latex}\begin{large}\[ m^{\rm rain}(t) = \rho^{\rm water}rlwt \]\end{large}{latex}

where _l_ and _w_ are the length and width of the train car.

We now have:

{latex}\begin{large}\[ F_{x} = \lim_{t \rightarrow 0}  \frac{\rho^{\rm water}rlwtv^{\rm car}}{t} = \rho^{\rm water}rlwv^{\rm car} = 12.5 N \]
\[ F_{y} = \lim_{t \rightarrow 0} \frac{\rho^{\rm water}rlwtv^{\rm rain}}{t} = \rho^{\rm water}rlwv^{\rm rain} = 50 N \]\end{large}{latex}

{tip}Both components should be positive, since the rain is _acquiring a positive_ x-velocity and _losing a negative_ y-velocity.{tip}

The total force on the rain is therefore:

{latex}\begin{large} \[ F^{\rm rain} = \mbox{12.5 N}\hat{x} + \mbox{50 N}\hat{y} = \mbox{52 N at 76}^{\circ}\mbox{ counterclockwise from the +x axis}\]\end{large}{latex}

Thus, the force on the _car_ is 52 N at 256° counterclockwise from the +x axis.
{cloak}
{card}
{deck}




 Cloak
matha
matha
 Cloak
appa
appa
 Card
labelPart B
Part B
Consider again the situation of Part A. Suppose that the rain is falling with a speed of 10.0 m/s, and that it comes to rest with respect to the train car immediately upon impact. At the instant the rain begins to hit the car, what is the force exerted on the train car by the rain?
Solution
 Toggle Cloak
idsysb
 System: 
 Cloak
idsysb
The rain entering the car in a time t, treated as a .
 Toggle Cloak
idintb
 Interactions: 
 Cloak
idintb
External forces from gravity and from the railroad car. We assume the collision force is much larger than the force of gravity during the collision.
 Toggle Cloak
idmodb
 Model: 
 Cloak
idmodb
.
 Toggle Cloak
idappb
 Approach: 
 Cloak
idappb