Force exerted on the palm of your hand by a beam of light

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Homework Help Overview

The original poster is investigating the force exerted on the palm of the hand by a beam of light from a 1.0W flashlight, considering two scenarios: one where the light is absorbed and another where it is reflected. The problem also involves calculating the mass of a particle that would exert the same force at Earth's surface.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the change in momentum when light reflects off the hand, questioning the reasoning behind the momentum calculations and whether certain terms should cancel out. There is exploration of the implications of defining positive and negative directions in momentum calculations.

Discussion Status

The discussion is focused on clarifying the concept of momentum change in the context of light reflecting off a surface. Some participants have provided guidance on how to approach the problem, particularly regarding the sign conventions used in momentum calculations. There appears to be a productive exchange of ideas, with participants confirming each other's reasoning.

Contextual Notes

Participants are navigating the complexities of vector quantities and the conventions used in physics, which may lead to different interpretations of the problem. There is an emphasis on consistency in applying these conventions throughout the discussion.

Eric Diaz
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Compute the force exerted on the palm of you hand by the beam from a 1.0W flashlight. (a) if your hand absorbs the light, and (b) if the light reflects from your hand.

What would the mass of the particle that exerts that same force in each case would be if you hold it at Earth's surface?

On the problem I used E = pc to solve part (a). My lack of understanding lies in part (b). Where I am told that the momentum change is twice the amount. I do not understand.

[P][/total] = [P][/1] + [P][/2]

Shouldn't [P][/1] cancel with [P][/2] since the momentum of [P][/2] is moving in the opposite direction?
 
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Eric Diaz said:
Shouldn't [P][/1] cancel with [P][/2] since the momentum of [P][/2] is moving in the opposite direction?

The momentum was P and changed to be -P. Thus, the change in momentum was the change from +P to -P and that's 2P... The difference between 3 and -3 is 6.
 
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Nugatory said:
The momentum was P and changed to be -P. Thus, the change in momentum was the change from +P to -P and that's 2P... The difference between 3 and -3 is 6.
So i should look at this as +P - (-P) = 2P ?
 
Eric Diaz said:
So i should look at this as +P - (-P) = 2P ?
Yes.

Footnote:
Or you could look at it as (-P) - P = -2P, depending on whether you're defining the towards-your-hand direction to be the positive direction or the negative direction. Either way, the change in momentum is going to be the final momentum minus the initial momentum. The important thing with vector quantities like forces, momenta, velocities is that whatever convention you use, you use it consistently. Note that in this problem the force exerted on (and acceleration and resulting velocity of) your hand will have a positive sign under one convention and a negative sign under the other, but either way it will point from the palm to the back of your hand.
 
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Nugatory said:
or (-P) - P = -2P, depending on whether you're defining the towards-your-hand direction to be the positive direction or the negative direction, but either way, the change in momentum is going to be the final momentum minus the initial momentum. The important thing with vector quantities like forces, momenta, velocities is that whatever convention you use, you use it consistently.
THANK YOU! I get it now.
 

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