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Homework Help: What to calculate in this problem

  1. Nov 1, 2007 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

    3. The attempt at a solution

    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?
  2. jcsd
  3. Nov 1, 2007 #2


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    If you are shooting at the center point of a target, what is meant is the distribution of distances from the center point. It is the uncertainty principle. When the bullet leaves the rifle it's confined to the opening at the end of the barrel. This gives it an unpredictable component of momentum not directed at the target.
  4. Nov 1, 2007 #3

    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)~√ћ
    => (∆x)~√ћ

    I have not taken care with that 1 gm mass.Essentially the problem is like this... as I believe.
  5. Nov 2, 2007 #4
    Dick, can you please tell me if my method is correct?
  6. Nov 2, 2007 #5


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    Uh, I was actually thinking of a somewhat different problem, and I'm finding this one a lot more confusing. I don't think what you are doing is correct. But I don't know the correct way of thinking about it either. The only thing I can think of is that if you can regard the 0.5sec as a delta(t) (can you?) then you can get a delta(E). The only way I can think of to relate that to a spread is equating it to mg*delta(h), where h is the height of the bullet at the target. If anyone knows the correct line of thinking on this problem we would LOVE to hear from you.
  7. Nov 2, 2007 #6
    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.
  8. Nov 2, 2007 #7


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    It looks like a lot of hand waving to me. But whatever you do, I guess you'll get something with an hbar in it that has the right dimensions. Speaking of which, you are writing down a lot of stuff which doesn't appear to have the correct dimension. Like E=hbar. You mean E=hbar/sec, right? I wouldn't drop the units coming from the 0.5sec.
  9. Nov 2, 2007 #8


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    I don't see how to give the answer in terms of only what is provided. There is nothing about the size of the barrel??

    It seems to me like an application of the position-momentum uncertainty principle, not the energy-time one.

    what I would be inclined to do would be to find the spread of momentum along x (where x is the coordinate aligned with the width of the barrel, i.e. the gun is pointing along y) using

    [tex] \Delta x \Delta p_x \approx \hbar/2 [/tex]

    using [itex] \Delta x = w [/itex] where w is the width of the barrel, you get an estimate for the value of [itex] v_x [/itex] (using [itex] \Delta v_x \approx v_x [/itex]).

    Now, the spread of the bullets at the target will simply be [itex] v_x \times 0.5 second[/itex]

    That's what comes to my mind but would require the width of the barrel
  10. Nov 2, 2007 #9


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    That's how I thought the problem should go, as well. Apparently Schiff has something else in mind.
  11. Nov 2, 2007 #10


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    If [itex]\Delta x[/itex] and [itex]\Delta p [/itex] are the initial uncertainties in position (height) and (vertical) momentum, then at later times the wave packet spreads out, and the time-dependent position uncertainty is

    [tex]\Delta x(t) = \sqrt{(\Delta x)^2 + (\Delta p)^2 t^2/m^2}.[/tex]

    We also have [itex]\Delta p \le \hbar/(2\Delta x)[/itex]. So, plug that in to the above, and find the value of [itex]\Delta x[/itex] that minimizes [itex]\Delta x(t)[/itex]. The result is [itex]\Delta x(t) = (\hbar t/m)^{1/2}[/itex].
  12. Nov 2, 2007 #11
    I have not done any mistake in dimension...you might have.

    Firstly, for ∆E~ћ, the dimension is (energy x time)=(energy x time).

    Then, ∆E joule~ћ joule

    =>(∆p)²=2mћ which has the units of kg²m²/s²

    Anything more?

    If you have anything to say for ∆y~√(ћ/m),please check the dimension of Uncertainty principle...
  13. Nov 2, 2007 #12
    I am looking yo what other people say just a bit later.For now, let me show:

    ∆p ∆y>ћ

    But,p=m(∆y/∆t) and ∆p~p


    =>m(∆y)²>ћ(∆t) [from uncertainty principle]

    This also gives the value of ∆y>√(ћ∆t/m) where putting the value of ∆t you can see it.

    This calculation looks better for we can avoid a cumbersome calculation which involves E.
  14. Nov 2, 2007 #13
    nrqed, there is a clasic problem of this type concrerning the Stern Gerlach experiment. But since we are not given anything of this sort,we are helpless...

    Avodyne's formula is working well.But,I did not know it...where can I get it?
  15. Nov 2, 2007 #14


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    The formula I gave makes sense classically: it's the root-mean-square combination of the initial posititon uncertainty and the later position uncertainty induced by the momentum uncertainty due to the classical motion.

    I don't remember which book(s) do the quantum calculation, but it's a nice problem. We want to solve the free-particle Schrodinger equation with an initital wave function [itex]\psi(x,0)=A\exp(-x^2/4(\Delta x)^2)[/itex]. Guess that the time dependent wave function is of the form [itex]\psi(x,t)=A(t)\exp(-B(t)x^2/4)[/itex], with [itex]B(0)=1/(\Delta x)^2[/itex]. Plug into the time-dependent Schrodinger equation, factor out [itex]\psi(x,t)[/itex]; the result has terms with [itex]x^2[/itex] and terms without [itex]x[/itex]. Matching the coefficients of [itex]x^2[/itex] yields a differential equation for B(t) that you can solve. Then [itex]|\psi(x,t)|^2\propto \exp[-\Re B(t)x^2/2][/itex]. You should find [itex]\Re B(t) = 1/[\Delta x(t)]^2[/itex].
  16. Nov 2, 2007 #15
    From statistics...we might have the sqrt of the sum of squares of standard deviations which are uncorelated with each other...right??? I have a little knowlede about it and think so...

    I am viewing the formula from this aspect...
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