Quantum Mechanics Uncertainty problem

[CatWoman]

Homework Statement

An object of mass m is dropped from a tower of height d. Show that according to
quantum mechanics the closest the object can fall on average to the base of the
tower is:
(d/g)^(1/4) (h/m)^(1/2)

Homework Equations

I don't know - that is the problem to know what to use.

The Attempt at a Solution

Using v^2=u^2+ 2as => velocity when mass reaches ground is v=√2gd

so momentum p=mv=m√2gd
using Uncertainty Principle ∆x∆p=h => ∆x=h/∆p=h/(m√2gd)

This is not the right answer so what have I done wrong?

 

[HallsofIvy]

You are confusing p with [itex]\Delta p$.$[/itex]

 

[CatWoman]

I do not know how to find delta p. I have often got confused with the Uncertainty Principle and the difference between a value and it's uncertainty. I thought that maybe the value is uncertain until you measure or calculate it, then the measured or calculated value is the degree of uncertainty. But this is obviously wrong. I havn't been able to find a decent explanation. Then even if I find delta p, I am confused between what would then be meant by x or delta x.

 

[Doc Al]

Start by assuming some initial uncertainty in horizontal position Δx_i. What minimum spread in horizontal velocity does that imply? Use that to deduce the horizontal spread when object hits the ground.

 

[CatWoman]

So before the mass falls delta x = h/m delta v ?
Then how is that used to calculate delta x after the fall?
Many thanks for your suggestions so far.

 

[Doc Al]

Do this: Assume some initial spread in x (delta x). Now calculate the spread in velocity. Use that calculated delta v to figure out the new spread in x when it reaches ground level.

 

[CatWoman]

I do not know how to start other than delta v = h / (m . delta x)
I then put delta v into v^2=u^2+ 2as substituting delta v in place of v, then converted back to delta x using delta x = h / (m . delta v)

I think I am barking completely up the wrong tree with this method, especially using the classical equations of motion, but can't see anything better to do!

 

[Doc Al]

I do not know how to start other than delta v = h / (m . delta x)

Right: Δv = h/(mΔx_i)

I then put delta v into v^2=u^2+ 2as substituting delta v in place of v, then converted back to delta x using delta x = h / (m . delta v)

Careful: Δv is the spread in horizontal velocity. Use this to figure out the associated spread in horizontal position at the bottom of the drop. (Hint: How long does it take the mass to fall? How far does it move sideways in that time?)

 

[CatWoman]

I used s = ut + 1/2 at^2 to get t=(2d/g)^1/2
then delta x = t . delta v = (2d/g)^1/2 (h/m delta x)
then rearranging to get delta x = (h/m)^1/2 (2d/g)^1/4.

So the answer is almost right except I have a factor of 2 in the answer that I cannot get rid of as the question says I should get: delta x = (h/m)^1/2 (d/g)^1/4.

Many thanks for your help with this.

 

[Doc Al]

I used s = ut + 1/2 at^2 to get t=(2d/g)^1/2
then delta x = t . delta v = (2d/g)^1/2 (h/m delta x)
then rearranging to get delta x = (h/m)^1/2 (2d/g)^1/4.

Looks good.

So the answer is almost right except I have a factor of 2 in the answer that I cannot get rid of as the question says I should get: delta x = (h/m)^1/2 (d/g)^1/4.

I don't see how they got rid of the factor of 2. What textbook is this from? (Assuming it's a textbook problem.)

 

[CatWoman]

Hi, no it's not from a textbook, it's on a worksheet so it's probably a mis-print then. I really appreciate your guidance with this problem.