Question

An ambitious physics major decides to check out the Uncertainty Principle for macroscopic systems. She goes to the top of the UD tower and drops a marble of mass m to the ground, trying to hit one of the cracks between bricks on the mall. To aim her marble, she teeters precariously directly over the desired crack and uses a very sophisticated apparatus of the highest possible precision, which she has borrowed from the General Physics Lab. Alas, try as she might, she cannot hit the crack.
Required:
Prove that the marble will inevitably miss the crack.

Answer 1

Answer:

The order = $\mathbf{\sqrt{\dfrac{h}{2 \pi m}} \sqrt[4]{\dfrac{H}{g}} \sqrt[4]{\dfrac{1}{2}} }}$

Explanation:

To miss the crack at a given distance is apparently not the same as the uncertainty that occurred in the distance while falling from the tower. However, it is believed that the uncertainties in both cases appear to be the same.

So, let's work it out together

According to Heisenberg's uncertainty principle:

$\Delta s. \Delta p =\dfrac{h}{2} =\dfrac{h}{4 \pi}$

Also; if we recall from the equation of motion that:

$v = u + at ---(1) \\ \\ v^2 - u^2 = 2as --- (2) \\ \\ s = ut + \dfrac{1}{2}at^2 --- (3)$

So, if u = 0 and a = g

Then;

$v = gt --- (1) \\ \\ v^2 = 2gs - - - ( 2) \\ \\ s = \dfrac{1}{2}gt^2 --- (3)$

From (2)

Making (s) the subject, we have:

$s = \dfrac{v^2}{2g}$

$s = \dfrac{p^2}{2gm^2}$

By differentiation;

$ds = d (\dfrac{p^2}{2gm^2})$

$ds = \dfrac{2pdp}{2gm^2}$

$\Delta \ s = \dfrac{p \Delta p}{gm^2 }$

where;

$\Delta p = \dfrac{h}{4 \pi \Delta \ s}$ from uncertainty principle

This implies that:

$\Delta s = \dfrac{p(\dfrac{h}{4 \pi \Delta s }) }{gm^2}$

$\Delta s = p(\dfrac{h}{4 \pi gm^2 }) \times \dfrac{1}{ \Delta s}}$

$(\Delta s)^2 = \dfrac{hmv} {4 \pi gm^2 }$

here;

v = 2gH

So;

$(\Delta s)^2 = \dfrac {h \sqrt{2gH} }{4 \pi gm }$

$\mathbf{(\Delta s)^2 = \sqrt{\dfrac{h}{2 \pi m}} \sqrt[4]{\dfrac{H}{g}} \sqrt[4]{\dfrac{1}{2}} }$

Thus, the order = $\mathbf{\sqrt{\dfrac{h}{2 \pi m}} \sqrt[4]{\dfrac{H}{g}} \sqrt[4]{\dfrac{1}{2}} }}$