# what happens to absolute uncertainty when you multiply two quantities

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#1
for example: mass 25 Kg with absolute uncertainty of 0.1 kg and acceleration of 2 m/s^(2) with absolute uncertainty of 0.01 m/s^(2)

Can anyone tell me what is the absolute uncertainty of force here?

Would appreciate any help !
Last edited by tahmidbro; 11 months ago
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11 months ago
#2
Usually the safest way to work out absolute uncertainty is to try to find the minimum and maximum from the given bounds - e.g. the minimum would be 24.9 kg * 1.99 ms-2 and then find the maximum the same way to manually find the uncertainty. Not the quickest way but you probably won't go wrong.
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#3
Well thanks for your reply But I want to be sure whether this method would really work or not if a question like that appears in my exam.

If it does not work I may lose a some marks and I don't want that 0
11 months ago
#4
(Original post by tahmidbro)
for example: mass 25 Kg with absolute uncertainty of 0.1 kg and acceleration of 2 m/s^(2) with absolute uncertainty of 0.01 m/s^(2)

Can anyone tell me what is the absolute uncertainty of force here?

Would appreciate any help !
The usual way to do this at A-Level is to add the % uncertainties in the 2 values to find the % uncertainty in the final result.
Covert the % uncertainty in the final result back to absolute.

eg. % uncertainty in mass is 100 x 0.1 / 25
% uncertainty in acceleration is 100 x 0.01 / 2
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#5
(Original post by Stonebridge)
The usual way to do this at A-Level is to add the % uncertainties in the 2 values to find the % uncertainty in the final result.
Covert the % uncertainty in the final result back to absolute.

eg. % uncertainty in mass is 100 x 0.1 / 25
% uncertainty in acceleration is 100 x 0.01 / 2
Using your theory, %U in mass+%U in acceleration = 0.4%+0.5% = 0.9%

then to convert the % uncertainty in the final result back to absolute, (U/50) x 100= 0.9

U= 0.9/100 x50 = 0.45

Is this correct?

And should I do the same thing for the case of dividing two quantities as well? for example, density= mass/volume
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11 months ago
#6
(Original post by tahmidbro)
Using your theory, %U in mass+%U in acceleration = 0.4%+0.5% = 0.9%

then to convert the % uncertainty in the final result back to absolute, (U/50) x 100= 0.9

U= 0.9/100 x50 = 0.45

Is this correct?

And should I do the same thing for the case of dividing two quantities as well? for example, density= mass/volume
Yes Using F=ma to get the force you have
F = 25 kg x 2 m/s = 50N - so that is correct
The % error is also correct 0.4% + 0.5%
The absolute uncertainty is then 50N x 0.9/100 - so that is also correct

This comes to 0.45N which is correct

so from those values the absolute uncertainty in the final value of 50N is plus or minus 0.45N

And yes, you also add the % uncertainties when you divide two values.
Last edited by Stonebridge; 11 months ago
1
#7
(Original post by Stonebridge)
Yes Using F=ma to get the force you have
F = 25 kg x 2 m/s = 50N - so that is correct
The % error is also correct 0.4% + 0.5%
The absolute uncertainty is then 50N x 0.9/100 - so that is also correct

This comes to 0.45N which is correct

so from those values the absolute uncertainty in the final value of 50N is plus or minus 0.45N

And yes, you also add the % uncertainties when you divide two values.
Many many thanks :-) for the help!
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