How Does Reflecting a Particle Stream Affect the Force on a Surface?

[Mr Davis 97]

Homework Statement

A one-dimensional stream of particles of mass $m$ with density $\lambda$ particles per unit length, moving with speed $v$, reflects back from a surface, leaving with a different speed $v'$, as shown. Find the force on the surface

Homework Equations

The Attempt at a Solution

The incoming momentum of the particle stream is $\lambda m v^2$ and the outgoing momentum is $\lambda m v'^2$, thus the change in momentum of the particle stream by the mirror is $\lambda m (v^2 + v'^2)$. Thus the force on the mirror is also $\lambda m (v^2 + v'^2)$.

Is this correct?

 

[mfb]

Nearly.

and the outgoing momentum is $\lambda m v'^2$

How did you get this?

Is the particle density still the same for the outgoing stream?

 

[Mr Davis 97]

Well, $\lambda$ is particles per unit length. If we multiply by m we get the mass of the stream per unit length, which is $\lambda m$ If we multiply this by v we get the momentum per unit length, and if we multiply by v again we get the momentum per second, which is exactly the incoming change in momentum. So $\lambda m v^2$ is the incoming change in momentum. Now it seems that the outgoing change in momentum would be $\lambda m v'^2$, but you are saying that the particle density may not be the same for the outgoing stream, but I don't see how the velocity of the particles changes the number of particles per unit length... Wouldn't they all be the same length from each other, except going slower?

 

[PeroK]

Well, $\lambda$ is particles per unit length. If we multiply by m we get the mass of the stream per unit length, which is $\lambda m$ If we multiply this by v we get the momentum per unit length, and if we multiply by v again we get the momentum per second, which is exactly the incoming change in momentum. So $\lambda m v^2$ is the incoming change in momentum. Now it seems that the outgoing change in momentum would be $\lambda m v'^2$, but you are saying that the particle density may not be the same for the outgoing stream, but I don't see how the velocity of the particles changes the number of particles per unit length... Wouldn't they all be the same length from each other, except going slower?

Have you ever watched a motor race? Like the formula 1. What happens when the cars get to the fastest part of the track? And the slowest?

 

[mfb]

What is the conserved quantity in the stream? It is not the density.

 

[Mr Davis 97]

Is the momentum conserved?

 

[PeroK]

Is the momentum conserved?

Given you are calculating the force caused by the change in momentum, that can hardly be conserved.

You didn't answer my question about the racing cars. It's the time interval not the spatial interval that remains constant. Think about it.

 

[Voicu]

The initial momentum is P1=λmv^2t .
(you that that by P1=M*v , M=λd , d is the distance between the last particle that hits the mirror during t and mirror , d=v*t ->P1=λmv^2t ) .
Each particle must be reflect (same M) -> P2=(-M*v')=(-λmvv't) .
F=|P2-P1|/t=λm(v^2+vv') .
I hope I helped.
Sorry for my poor english.

 

[haruspex]

The initial momentum is P1=λmv^2t .
(you that that by P1=M*v , M=λd , d is the distance between the last particle that hits the mirror during t and mirror , d=v*t ->P1=λmv^2t ) .
Each particle must be reflect (same M) -> P2=(-M*v')=(-λmvv't) .
F=|P2-P1|/t=λm(v^2+vv') .
I hope I helped.
Sorry for my poor english.

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