Minimum Uncertainty in Measuring Velocity of a Particle in X-direction.

omegas
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Homework Statement


A small flat circular disc of diameter 2mm and mass 5mg is lowered vertically to rest at the centre of a flat tray of width 10mm in the x direction and 50mm in the y direction. What is the minimum uncertainty in measuring the velocity of the particle in the x direction? Considering possible motion only in the x direction, determine whether it is likely that the disc would touch the side of the tray in a given time:
(a) overnight
(b) in the lifetime of the Universe (approx. 15 billion years)



Homework Equations


The only equation I can muster up is Heisenberg's
\Deltax\Deltap ~ h




The Attempt at a Solution


I have no idea how to go about this one. Not sure I understand the problem at all. Please help.
 
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Honestly, I'm not sure I understand it either. Heisenberg's uncertainty principle is the only equation I can think of, but it seems pretty useless because you're given no information about the uncertainty in position of the wavefunction.
 
What I really don't understand is that it says the disk is at rest, which I am assuming means there is no velocity, so how can I measure \Deltap?
 
Hmmm...could I just set \Deltap = h \Deltax? The change in x would be the difference in the edge of the disc and the side of the tray in the x direction, right? So \Deltax would be 4mm to the side of the tray. I know h = 6.62*10-24 and p = m v and my mass would be 5mg. So if I fenagle around I can get v = (h\Deltax) / m. Would this be right?
 
omegas said:
Hmmm...could I just set \Deltap = h \Deltax? The change in x would be the difference in the edge of the disc and the side of the tray in the x direction, right? So \Deltax would be 4mm to the side of the tray. I know h = 6.62*10-24 and p = m v and my mass would be 5mg. So if I fenagle around I can get v = (h\Deltax) / m. Would this be right?
Nope... for starters, the units don't work out for that, so it can't be a correct equation.

Also, though, what \Delta x really means is the standard deviation of a wavefunction in position space. It doesn't actually correspond to any physical distance between two objects. And in your case especially, a disk several mm in diameter is going to be an essentially classical object, which means the standard deviation of its wavefunction is extremely tiny. Definitely much smaller than the disk itself.

By the way, saying that the disk is at rest really means that any velocity it may have is below the tolerance of your measuring equipment. When you get down to the smallest scales, you can't really know that the velocity is exactly zero because there's some uncertainty in the position (and also usually there's thermal motion, of course).
 

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