Textbook Problem 1B27


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Constance Newell
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Textbook Problem 1B27

Postby Constance Newell » Tue Oct 27, 2020 1:36 pm

1B.27 A bowling ball of mass 8.00 kg is rolled down a bowling alley lane at
5.00±5.0m⋅s−1. What is the minimum uncertainty in its position?

Could someone explain why we don't convert 8.00kg to grams in this problem

Thank you!

Sabina House 2K
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Re: Textbook Problem 1B27

Postby Sabina House 2K » Tue Oct 27, 2020 1:46 pm

SI units are always in kg, not g. Because of this, the constant h used in the Heisenberg's indeterminacy equation is in units of J/s, and 1 J = 1 kg m^2 s^-2. Because the constant h uses kg, you must use units of kg for the mass.

Isaac Wen
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Re: Textbook Problem 1B27

Postby Isaac Wen » Thu Oct 29, 2020 9:44 pm

We've been using grams in the previous chapter because molar mass is in g/mol. Now, since we're no longer dealing with molar mass, we use the standard SI unit, kg!

DanielHong2L
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Re: Textbook Problem 1B27

Postby DanielHong2L » Thu Oct 29, 2020 9:52 pm

Constance Newell wrote:1B.27 A bowling ball of mass 8.00 kg is rolled down a bowling alley lane at
5.00±5.0m⋅s−1. What is the minimum uncertainty in its position?

Could someone explain why we don't convert 8.00kg to grams in this problem

Thank you!


I'm also confused in the uncertainty in position. I thought Dr. Lavelle explained that if it was +/- 5.0 m/s then the uncertainty would be 5*2; so delta v = 10. However, the txtbook answer simply uses 5 as the delta v to calculate it.

If anyone / mods can clarify this that'll be amazing!

DanielHong2L
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Joined: Wed Sep 30, 2020 9:51 pm

Re: Textbook Problem 1B27

Postby DanielHong2L » Thu Oct 29, 2020 9:55 pm

DanielHong2L wrote:
Constance Newell wrote:1B.27 A bowling ball of mass 8.00 kg is rolled down a bowling alley lane at
5.00±5.0m⋅s−1. What is the minimum uncertainty in its position?

Could someone explain why we don't convert 8.00kg to grams in this problem

Thank you!


I'm also confused in the uncertainty in position. I thought Dr. Lavelle explained that if it was +/- 5.0 m/s then the uncertainty would be 5*2; so delta v = 10. However, the txtbook answer simply uses 5 as the delta v to calculate it.

If anyone / mods can clarify this that'll be amazing!



Oh oops nevermind just checked other post on this same problem. They said it's a txtbook error (check on dr lavell's website answer key thing) so

tl;dr:

The uncertainty in poisition is *2 the +/- so it'd be 10 m/s because that includes both the possibility of being +5 and -5.

905290504
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Re: Textbook Problem 1B27

Postby 905290504 » Fri Oct 30, 2020 10:00 am

for uncertainty you want to keep it in kg because planck's constant's units are J*s which is a simplification of (kg*m^2*s^-2)(s)
*** 1 J = kg*m^2*s^-2

905290504
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Re: Textbook Problem 1B27

Postby 905290504 » Fri Oct 30, 2020 10:01 am

for uncertainty you want to keep it in kg because planck's constant's units are J*s which is a simplification of (kg*m^2*s^-2)(s)
*** 1 J = kg*m^2*s^-2


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