HW Question Regarding 1B.27


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Mark 1D
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Joined: Fri Sep 28, 2018 12:18 am

HW Question Regarding 1B.27

Postby Mark 1D » Sun Oct 21, 2018 8:00 pm

Howdy! 1B.27 says "A bowling ball of masss 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?" Can one of ya'll explain to me how I use the uncertainty equation to obtain the minimum uncertainty?

Faith Fredlund 1H
Posts: 68
Joined: Fri Sep 28, 2018 12:18 am

Re: HW Question Regarding 1B.27

Postby Faith Fredlund 1H » Sun Oct 21, 2018 8:49 pm

Since you know the two equations

ΔP=(m)(ΔV) uncertainty in momentum=(mass in kg)(uncertainty in velocity in m/s)

(ΔP)(Δx)> or = h/4π (uncertainty in momentum)(uncertainty in distance)=(plank's constant)/4π

you can put them together to create the equation

Δx=(6.626X10^-34)/4π(m)(ΔV) you set the equation equal because this will give you the minimum uncertainty

You then can plug in all of the values you are given where m=8.00kg and ΔV=10m/s because it is + or - 5m/s which gives a total uncertainty of 10.
This will give you the minimum uncertainty of distance Δx in meters.

LedaKnowles2E
Posts: 62
Joined: Fri Sep 28, 2018 12:27 am

Re: HW Question Regarding 1B.27

Postby LedaKnowles2E » Sun Oct 21, 2018 9:12 pm

Here's my work, hope it helps
Chem 14A 1B.27.JPG

Alyssa Bryan 3F
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Re: HW Question Regarding 1B.27

Postby Alyssa Bryan 3F » Sun Oct 21, 2018 10:38 pm

I think for this question the answer in the back of the book is wrong. On the chemistry 14A webpage in the solution manual errors page it shows this problem is one of them. So instead of the answer being 1.3 x 10^-36 it is actually 6.7 x 10^-37. This is because the Δv should be 10, not 5 in the equation as the given velocity is 5.00 ± 5.0

gwynlu1L
Posts: 62
Joined: Fri Sep 28, 2018 12:19 am

Re: HW Question Regarding 1B.27

Postby gwynlu1L » Tue Oct 23, 2018 1:53 am

I agree with you, since it says the velocity was 5+/- 5 m/s, the range of velocities is from 0-10, making change in velocity = 10m/s. I was confused by the solutions manual too.

Abby De La Merced 3F
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Re: HW Question Regarding 1B.27

Postby Abby De La Merced 3F » Tue Oct 23, 2018 6:41 pm

Wait, can someone actually explain why the change in velocity would be 10 and not 5 like it says in the solutions manual?

Adrian C 1D
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Joined: Fri Sep 28, 2018 12:19 am

Re: HW Question Regarding 1B.27

Postby Adrian C 1D » Tue Oct 23, 2018 9:54 pm

It would be 10 m/s instead of 5 m/s because you take the (highest value - lowest value) of 5 plus or minus 5 m/s. The highest value would be 5+5= 10 m/s and the lowest value would be 5-5=0 m/s. If you subtract 10 - 0, you would get 10 m/s which you would use as delta v.


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