What Is The Meaning Of This Problem

Discussion in 'Physics & Math' started by neelakash, Nov 1, 2007.

  1. neelakash Registered Senior Member

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    A 1 gm rifle bullet takes 0.5 seconds to reach the target.Regarding the bullet as a point mass, and neglecting the air resistance and the mottion of the earth,find the order of magnitude of the spread of successive shots at the target with optimum condition for aiming and firing.

    I cannot understand what I am to find out in this problem.This problem appears in QM book of Schiff and possibly,it is a problem of uncertainty principle.

    The book has no answer list at the back for which I am confused...whether to find the time interval or something else...
    What is meant by: "the spread of successive shots"?

    Can anyone please tell?
     
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  3. neelakash Registered Senior Member

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    There was a suggestion that we are asked to find the spread in the position of the bullet in the problem.

    Then,


    ∆E~[ћ/(2∆t)]~ћ

    This uncertainty in KE of the bullet implies the bullet has some deviation from the target due to uncertainty in energy.

    For ∆E~E and ∆p~p, we must have

    (∆p)²~2mE=2mћ

    => (∆p)~√ћ
    => (∆x)~√ћ

    I have not taken care with that 1 gm mass.Essentially the problem is like this... as I believe.
     
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  5. Enmos Valued Senior Member

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    Please Register or Log in to view the hidden image!

    This is exactly what I meant in the other thread

    Please Register or Log in to view the hidden image!



     
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  7. neelakash Registered Senior Member

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    I have told in the other thread...
     
  8. James R Just this guy, you know? Staff Member

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    neelakash:

    Sounds like what is intended is application of the uncertainty principle, as you say.
     
  9. neelakash Registered Senior Member

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    Can anyone tell if my analysis is correct?
     
  10. Reiku Banned Banned

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    I can not be sure. At first i thought it might have been referring to the quantum wave function... now i'm not sure... maybe it is all about uncertainty.
     
  11. neelakash Registered Senior Member

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    491
    I think I need to clarify my method a bit more:

    I am just giving the name ∆t to the time in which the bullet reaches its target...It is not that we have made error of ∆t in calculating the elapsed time.

    Thus,∆t=0.5 second

    Then,the limited time available to us restricts us from knowing the precise value of KE of the bullet.We can know it with error ∆E.

    ∆E>ћ (this ∆energy varies with off-target v²)

    This ∆E is the off-target energy of the bullet.This corresponds to a non-zero component of off-target momentum ∆p (this momentum varies with off-target v).

    ∆p>√(2mћ) or, ∆p~√(2mћ) (for the sake of the problem)

    =>∆y~ ћ/[2√(2mћ)]~ √(ћ/8m)

    Which is roughly √(ћ/m)

    Actually many books use (h/4π)=(ћ/2);some use only ћ...But essentially, the spread looks to be of the order of √(ћ/m)

    What the only problem appears to me in this development is what is the case when ∆E->0,we have ∆p->0 =>∆y-> infinity!!!

    But,there is an intrinsic problem in assuming ∆E->0 => ∆t->infinity contrary to the problem.
     

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