Module Question #23


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Jamie2002
Posts: 75
Joined: Wed Sep 30, 2020 9:55 pm

Module Question #23

Postby Jamie2002 » Fri Oct 23, 2020 9:03 pm

You are caught in a radar trap and hope to show that the speed measured by the radar gun is in error due to the Heisenberg uncertainty principle. If you assume that the uncertainty in your position was +/- 5 m when your speed was measured, and that the car has a mass of 2150 kg, what is your calculated uncertainty in the speed of your car?

My work was:
2150 kg (delta v) (5m) >= h/4pi
delta v = 5 x 10^-39

This is wrong, where did I make a mistake?

Kaitlyn Hernandez 3I
Posts: 101
Joined: Wed Sep 30, 2020 9:37 pm
Been upvoted: 1 time

Re: Module Question #23

Postby Kaitlyn Hernandez 3I » Fri Oct 23, 2020 9:17 pm

The uncertainty in position should be 10m, not 5m.

So you should use the following equation to solve for the uncertainty in momentum first:
delta p >= h/(4pi)(delta x)
when you plug everything in, you'd get delta p >= (6.626x10^-34 Js)/(4pi)(10m)
and then delta p >= 5.2728 x 10^-36kgms^-1

After, you have to solve for the uncertainty in the speed using the following equation:
delta v = delta p / mass
when you plug anything in, you'd get delta v = (5.2728 x 10^-36 kgms^-1)/2150 kg
and then delta v = 2 x 10^-39 ms^-1

Hope this was easy to follow !

Jamie2002
Posts: 75
Joined: Wed Sep 30, 2020 9:55 pm

Re: Module Question #23

Postby Jamie2002 » Fri Oct 23, 2020 9:22 pm

Kaitlyn Hernandez 2E wrote:The uncertainty in position should be 10m, not 5m.

So you should use the following equation to solve for the uncertainty in momentum first:
delta p >= h/(4pi)(delta x)
when you plug everything in, you'd get delta p >= (6.626x10^-34 Js)/(4pi)(10m)
and then delta p >= 5.2728 x 10^-36kgms^-1

After, you have to solve for the uncertainty in the speed using the following equation:
delta v = delta p / mass
when you plug anything in, you'd get delta v = (5.2728 x 10^-36 kgms^-1)/2150 kg
and then delta v = 2 x 10^-39 ms^-1

Hope this was easy to follow !

Thank you so much! So when we see an uncertainty that is +/- x, the uncertainty is 2|x|?

Kaitlyn Hernandez 3I
Posts: 101
Joined: Wed Sep 30, 2020 9:37 pm
Been upvoted: 1 time

Re: Module Question #23

Postby Kaitlyn Hernandez 3I » Thu Oct 29, 2020 2:43 pm

Jamie2002 wrote:
Kaitlyn Hernandez 2E wrote:The uncertainty in position should be 10m, not 5m.

So you should use the following equation to solve for the uncertainty in momentum first:
delta p >= h/(4pi)(delta x)
when you plug everything in, you'd get delta p >= (6.626x10^-34 Js)/(4pi)(10m)
and then delta p >= 5.2728 x 10^-36kgms^-1

After, you have to solve for the uncertainty in the speed using the following equation:
delta v = delta p / mass
when you plug anything in, you'd get delta v = (5.2728 x 10^-36 kgms^-1)/2150 kg
and then delta v = 2 x 10^-39 ms^-1

Hope this was easy to follow !

Thank you so much! So when we see an uncertainty that is +/- x, the uncertainty is 2|x|?


Hi! I'm so sorry I just saw this. But the uncertainty is 10 because it is the range in the uncertainty since its + 5 and - 5. So for example, if the position was 15 +-5 m, the uncertainty would be +5 (15-20) from 15 and -5 (10-15) from 15 and the total range was 10. I hope this made sense !!

Sophia Spungin 2E
Posts: 60
Joined: Wed Sep 30, 2020 9:41 pm

Re: Module Question #23

Postby Sophia Spungin 2E » Thu Oct 29, 2020 7:06 pm

I did the same thing originally, and when a question uses +/- for uncertainty, I find it easier to apply this to a number. For example, if you apply +/-5 to 10, you get 15 and 5. Here, it is easier to see that the range in position is 10 units, and not 5.


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