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Physics HUP help

  1. Mar 16, 2005 #1
    A ball of mass 50g moves with a speed of 30 m/s. If its speed is measured to an accuracy of 0.1%, what is the minimum uncertainty in its position? What does this answer indicate about the relevance of quantum mechanics to macroscopic objects?

    I am completely confused. My prof just gave us HUP but didn't explain how to use it.

    I know: (delta p)(delta x) > h /2, where h is actually h bar i just don't know how to type it.

    I think i need to sove for delta x, but i have tried similar problems that i have the answers to and i'm not even close.

    please, if anyone can help me understand this i would greatly appreciate it.
  2. jcsd
  3. Mar 16, 2005 #2
    Indeed solving for [itex]\Delta x[/itex] is correct. Why don't you show us the work you've done.
  4. Mar 16, 2005 #3
    honestly i haven't figured out where to start.
    i can get momentum by multiplying the mass and the speed. i wanted to multilpy that by 0.1% to get a delta p value and then solve for delta x. this is the method i used for other questions, and i did not get the right answer.

    also i am not sure what is significant about the fact that it asks for the minimum uncertainty.
  5. Mar 16, 2005 #4
    That sounds right. Post your calculations.

    Since you know the uncertainty in momentum, and [itex] \Delta p \Delta x \geq \frac{\hbar}{2}[/itex], the minimum uncertainty in position is achieved when there is equality.
  6. Mar 16, 2005 #5
    p=0.05kg * 30 m/s = 1.5 kgm/s.

    1.5 kgm/s * 0.1% =0.0015 kgm/s

    (delta x) > h/(2Delta p) where h is h bar ie) h/2pi

    delta x > 6.626*10^-34/ (4 *pi* 0.0015)

    delta x > 3.52*10^-32

    like i said, i tried this method with another question, but i gave the kinetic energy and a 5% accuracy in the momentum. for that, i calculated the velocity from 1/2mv^2 and then multiplied by the mass. then i multiplied that value by the 5% and used the same method above for determining delta x, and i was way off. i'm not too sure about this whole HUP thing.

  7. Mar 16, 2005 #6
    Yes, that looks right.

    Can you post the other question (that your answer was wrong for) in its entirety (preferably verbatim)?
  8. Mar 16, 2005 #7
    i just realized that i kept repeatedly using the same wrong number to figure out the other question. now i get it exactly right. thanks for your input on the my original question. guess i'm not as lost as i thought.
  9. Mar 16, 2005 #8
    Good :)

    I really would like to be able to make one word replies. asdf
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