## Textbook Problem 1B27

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

SLai_1I
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Joined: Wed Sep 30, 2020 9:52 pm

### Textbook Problem 1B27

A bowling ball of mass 8.00 kg is rolled down a bowling alley lane at 5.00 plus/minus 5.0 m*s^-1. What is the minimum uncertainty in its position?

What is the delta V for this problem? I keep rereading the question, and it seems to be 10 m/s. However, when I use this value, I get the wrong answer. The textbook key solves the problem using delta V=5 m/s. Why is delta V 5 and not 10?

Crystal Yu 1D
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### Re: Textbook Problem 1B27

I think this is an error in the solution guide. If you check on the website, there is a link called "Solution Manual Errors".
Last edited by Crystal Yu 1D on Mon Oct 19, 2020 2:24 pm, edited 1 time in total.

Ria Nawathe 1C
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### Re: Textbook Problem 1B27

Delta v should be 10 m/s. I believe this is an error in the book's solution. Here's a full list of errors in the solution manual:

https://lavelle.chem.ucla.edu/wp-conten ... rs_7Ed.pdf

Lorraine Jiang 2C
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### Re: Textbook Problem 1B27

Hi!

Whenever we are given "+ -" the delta x is always 2 times the number after "+-". I believe there is a mistake in the solution and here is the link to all the mistakes that professor Lavelle found.

https://lavelle.chem.ucla.edu/wp-conten ... rs_7Ed.pdf

Hope it helps!

Alexa Pham 1D
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Joined: Wed Sep 30, 2020 10:03 pm

### Re: Textbook Problem 1B27

I was going to ask the same thing too. Thank you for pointing out the mistake, I thought I was misunderstanding what the "change in velocity" was and as a result, was quite confused.

Frankie Mele 3J
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### Re: Textbook Problem 1B27

I was also confused on this problem, but the +/- symbol in front of the value 5.0 m/s should be multiplied by 2 to give deltaV the value of 10 m/s.

AnjikaFriedman-Jha2D
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Joined: Wed Sep 30, 2020 9:55 pm

### Re: Textbook Problem 1B27

You use 10 m/s for the uncertainty because if the velocity is +/- 5 m/s then the values for velocity can vary by 10 m/s so that is the value you use in your calculation