## 1B27

$\lambda=\frac{h}{p}$

Natallie K 3B
Posts: 61
Joined: Wed Sep 30, 2020 9:51 pm

### 1B27

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

On the solution manual error list, it says that the answer is 6.7x10^-37m. For some reason, I keep on getting 6.6x10^-37. I know its not that big of a different, but it was wondering if I was doing something wrong.

First I solved for delta(p) by multiplying the mass 8 kg by delta(v), which is 10m.s^-1, which got me 80.
Then I solved for delta(x) by inputing in the equation h/4pi(80), and I got the answer 6.5983x10^-37

Am I doing something wrong or is this difference simply from rounding?

Ryan Laureano 3I
Posts: 92
Joined: Wed Sep 30, 2020 10:09 pm
Been upvoted: 1 time

### Re: 1B27

I did this problem too, and I ended up getting 6.59x10^-37. I would assume the difference is in how much of planck's constant you use. In this case, I wouldn't worry about the small difference as long as the sig figs are correct. I noticed sapling allows some room for error in rounding.

Posts: 159
Joined: Wed Sep 30, 2020 9:32 pm
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### Re: 1B27

I got the same answer for that question. I would assume that a difference that small simply comes from rounding (maybe for Planck's constant or pi).

Hope this helps!

Jenny Lee 2L
Posts: 94
Joined: Thu Oct 08, 2020 12:15 am

### Re: 1B27

Hello! It's most likely from Planck's constant. Planck's constant is not the exact calculation we use (we usually use 6.626x10^-34 Js), and there is more after the 6.626. We can probably assume that because Planck's constant is actually slightly larger than what we use, it's probably making that small difference between our answers and the textbook's answer. (As for pi, if we use the calculator function, it should still be more accurate than the Planck's constant that we use, thus leading me more to think that it's Planck's constant.)