|
Photo from Wikimedia Commons |
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?
Solution
System:
Interactions:
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..
Model:
Approach:
Diagrammatic Representation
|
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).
Mathematical Representation
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:
\begin
[ p^
_
= p^
_
+ p^
_
= p^
_
= p^
_
+ p^
_
] \end
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:
\begin
[ m^
v^
_
= \left(m^
+ m^
\right)v_
] \end
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:
\begin
[ m^
= \rho^
V^
= (\mbox
^
)(\mbox
\times\mbox
\times\mbox
) = \mbox
]\end
where ρwater is the density of water and Vrain is the volume of the accumulated rain.
We can now solve to find:
\begin
[ v_
= \frac{m^
v^
_{x,i}}{m^
+ m^{\rm rain}} = \mbox
]\end