Basic Heisenberg uncertainty principle stuff

[rtsswmdktbmhw]

Homework Statement

A bullet is shot from a rifle.

1) if the position of the centre of mass of the bullet perpendicular to its motion is known to have an accuracy of 0.01 cm, what is the corresponding uncertainty in its momentum?

2) If the accuracy of the bullet were determined only by the uncertainty principle, by how much might the centre of the bullet miss a pinpoint target 300 m away?

Homework Equations

$\Delta x \Delta p_{x}\geq\frac{\bar h}{2}$ (or $\Delta x \Delta p_{x}\approx\frac{\bar h}{2}$?)

The Attempt at a Solution

1) If the uncertainty is perpendicular to the motion, then can't we have 0 uncertainty? But the question asks for 'corresponding uncertainty' so it sounds like I'm supposed to get some answer. I'm not sure whether to plug into the equation and get a number. But that doesn't sound right because then I would be getting $\Delta p_{y}$ if using 0.01 cm, right?

2) Since the position of the bullet has an uncertainty of 0.01 cm, would this simply be how much the centre of the bullet might miss the pinpoint target?

Sorry if there is anything obvious or simple. I have just done what seems like a crash course on introductory QM in the past week or two. Things might not have clicked properly yet...

 

[jtbell]

$\Delta x \Delta p_{x}\geq\frac{\hbar}{2}$ (or $\Delta x \Delta p_{x}\approx\frac{\hbar}{2}$?)

(tip: in LaTeX, use "\hbar" for $\hbar$.)

That's just along the x direction. There are exactly similar equations for the y and z directions: $\Delta y \Delta p_y \geq\frac{\hbar}{2}$ and $\Delta z \Delta p_z \geq\frac{\hbar}{2}$ Remember momentum is a vector quantity, with x, y and z components.

If you assume the gun is aimed along the x direction, then you can use either y or z for the perpendicular direction.

1) if the position of the centre of mass of the bullet perpendicular to its motion is known to have an accuracy of 0.01 cm, what is the corresponding uncertainty in its momentum?

Here the "corresponding uncertainty in its momentum" refers to the component pependicular to the direction that the gun is aimed.

 

[rtsswmdktbmhw]

Oh, okay, so I should be finding the y component uncertainty. So it's simple plug and chug with the equation? Is it correct to replace the inequality with the approximately equal sign, or should I be leaving it as an inequality?

For question 2, then, is it a matter of finding y component velocity using p=mv, then simply finding the maximum inaccuracy (deviation from pinpoint target) at 300 m using dist=vel*time?

 

[jtbell]

Right on both counts. Strictly speaking it should be kept as an uncertainty, although most of the time we're just interested in the order of magnitude (power of ten), so lots of times people use the "approximately equals." The actual uncertainty is likely to be somewhat bigger than what $\hbar/2$ gives, but not as much as ten times bigger.

 

[rtsswmdktbmhw]

Okay, thanks for your explanations! And also for the tip for $\hbar$.