Photon stopping an object travelling at 10m/s

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Homework Help Overview

The discussion revolves around the momentum of photons and their interaction with an object moving at 10 m/s. Participants explore the implications of conservation of energy and momentum in the context of elastic collisions involving photons.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants examine the relationship between the momentum of the photon and the mass of the object, questioning the assumption that the mass must be zero for the object to be stopped. They discuss the change in momentum of the photon and its implications for the conservation equations.

Discussion Status

The discussion is active, with participants questioning each other's reasoning and attempting to clarify the assumptions made regarding the mass of the object and the momentum of the photon. Some guidance is offered through hints about the change in momentum, but no consensus has been reached.

Contextual Notes

Participants are working under the constraints of conservation laws and the specific scenario of elastic reflection, which may influence their interpretations and assumptions about mass and momentum.

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Homework Statement



The momentum of a photon is ħk, where k is the wave vector. Assume an object moving at 10m/s, objects of which mass can be stopped by elastic reflection of the photon.

Homework Equations



Conservation of energy and momentum
p = ħk = h/λ

The Attempt at a Solution



Assume the photon is traveling in the opposite direction to the object. Then the mass of the object has to be 0. I feel I'm missing something?
 
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Mr.A.Gibson said:
Assume the photon is traveling in the opposite direction to the object. Then the mass of the object has to be 0.
How did you conclude this?

Hint: What's the change in momentum of the photon?
 
Doc Al said:
How did you conclude this?

conservation of momentum,
[tex] p_b+10m=p_a[/tex]
subscripts refers to photons momentum before and after collision.

Then from conservation of energy
[tex] \frac{1}{2}mv^2+cp_b=cp_a[/tex]
substitute in the first equation and velocity of the object to give
[tex] 50m+cp_b=c(p_b+10m)[/tex]
[tex] 50m=10cm[/tex]
only valid if m=0
 
Last edited:
Doc Al said:
Hint: What's the change in momentum of the photon?

2ħk = 10m

m = ħk/10

Makes sense, what did I mess up in the first place then?
 

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