## Problem 1B.27

$\Delta p \Delta x\geq \frac{h}{4\pi }$

Gabriella Bates 2L
Posts: 113
Joined: Thu Jul 11, 2019 12:15 am

### Problem 1B.27

A bowling ball of mass 8.00 kg is rolled down a bowling alley lane at 5.00 +/- 5.0 m/s. What is the minimum uncertainty in its position?

In the answer key, they used 5.00 m/s as the uncertainty in velocity (delta v). Why wouldn't the uncertainty be 10.00 m/s? I thought it would be that because 10.00 m/s is the entire interval. Thanks in advance!

Posts: 87
Joined: Sat Aug 24, 2019 12:15 am

### Re: Problem 1B.27

If I am correct, the reasoning behind this is that the value can only be over or below it. The value could never be both larger and smaller simultaneously. Therefore the uncertainty or indeterminacy would only take in the values that we are unsure of. This may be incorrect as I was slightly confused by this too, but I can imagine this is the reasoning. Hope it helps!

Chris Tai 1B
Posts: 102
Joined: Sat Aug 24, 2019 12:16 am

### Re: Problem 1B.27

I also used delta v = 10, which got me a solution of delta x >= 6.6 x 10^-37. I think the answer key may be incorrect here since the change in velocity could vary from 0 m/s to 10 m/s, which has a clear difference of 10-0 = 10

sarahforman_Dis2I
Posts: 109
Joined: Sat Aug 17, 2019 12:18 am

### Re: Problem 1B.27

Gabriella Bates 1C wrote:A bowling ball of mass 8.00 kg is rolled down a bowling alley lane at 5.00 +/- 5.0 m/s. What is the minimum uncertainty in its position?

In the answer key, they used 5.00 m/s as the uncertainty in velocity (delta v). Why wouldn't the uncertainty be 10.00 m/s? I thought it would be that because 10.00 m/s is the entire interval. Thanks in advance!

Yes, you are correct. The uncertainty in position should be 10 m/s. If you look on the class website, there is a document of mistakes in the textbook and their corrections.