The impact of a rope versus the stagnation pressure of a fluid

AI Thread Summary
The discussion focuses on comparing the forces exerted by a moving rope and a column of water when they collide with a scale. The force from the rope, derived from its linear density and velocity, is expressed as λv², while the stagnation pressure from the water is given by ½ρv². Participants explore the reason for the factor of ½ in the water's equation, noting that it relates to the direction of the water's velocity after impact. The conversation also touches on the nature of the collision, questioning whether the rope rebounds elastically and how that affects the measured kinetic energy. Ultimately, the analysis reveals that the force calculated for the rope pertains solely to stopping its momentum, without assuming any rebound.
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Let \lambda be a linear density of a rope which is moving into a scale at velocity v. The additional force on the scale due to the collision is given as
\frac{d p}{d t} = v\frac{d m}{d t} = \lambda v^2

Where as the stagnation pressure from stopping a column of water in excess of static pressure is

\frac{1}{2}\rho v^2

We can easily compare the forms by, for example multiplying by the width of the column to obtain a linear density of the fluid, or consider hitting the scale with a continuum of infinitesimal ropes. It seems the \frac{1}{2} factor would remain different.

So what is the explanation for this relative factor of \frac{1}{2}? I have tossed around a few ideas but I'm curious what you may think.
 
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For water, the velocity of the water after hitting the tube is sideways, so thus only the kinetic energy of the forward flow is measured.

For the rope, are you assuming an elastic collision with the rope rebounding with the velocity -v.
 
256bits said:
For water, the velocity of the water after hitting the tube is sideways, so thus only the kinetic energy of the forward flow is measured.
I assume by sideways you mean radially outward, orthogonal to the original flow. Why do you think that is the case? What would happen when the "sideways" water hit the sides of the pipe? Why would that mean kinetic energy is measured?

256bits said:
For the rope, are you assuming an elastic collision with the rope rebounding with the velocity -v.

I don't agree.

It appears specifically we have found the force required to stop the rope from moving and nothing more. We did not assume the rope was to rebound with velocity -v. We assumed our force had to take a certain amount of momentum per second and bring it to a stop, nothing more.
 
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