# Homework Help: Precision in Physics problems

1. Jan 16, 2005

### haki

Hi.

I was wondering how do you calcultate physics problems when you must also note the error of measurement.

I mean if you have a weight-scale that has a percison of 20 g, and lets say that the scale shows 0,7 kg. Does that mean you have 0,7 kg +- 0,02 kg or 0,7 +- 0,01 kg. Or what. And if you have in a test. Something like this
mass of object is 20(1 +- 0,05)g what does that mean? Or if you need to calculate the speed of a bullet from kinetical energy. Let say the data is weigh 6,00 g and Wk = 3,12 J. The first data is of 2 or 3? significant figure the other is of 3 significant numbers. What is the precision of the resoult. You have v = sqrt((2*Wk)/m)) = 1,01980 do you write 1,019 or 1,02 +- xx? I am a bit puzzled by this +- precision and uncertanty. Can somebody explain this to me in simple language or give a link to a web site. I tried to ask my physics profesor but he just said "Look it up". I looked at the NIST referance for Constants, Units and Uncertanty but that make me more puzzled than before.

Thanks

2. Jan 16, 2005

### HallsofIvy

If the scale has a "precision of 20 g" or 0,2 kg, then that is +- 0,2 kg. A reading of 0,7 kg might mean the actual mass is as high as 0,9 kg or as low as 0,5 kg.

6,00 g has 3 significant figures and the true mass might be as high as 6,01 g or as low as 5,99 g. If the mass had been given as 6,0 g, that would have been 2 significant figures and the true mass might have been as high as 6,1 g or as low as 5,9 g.

Assuming the kinetic energy is 3,12 J and the mass is 6,00 g= 6,00 x 10-3 kg. (1/2)(6,00 x 10-3v2= 3,12.
v2= 2(3,12)/(6,00 x 10-3)= 1040 (m/s)2.
v= 32.249 m/s except that we haven't taken the precision into account.

One way to do that is to calculate the largest and smallest the numbers could be. Any fraction is larger if the numerator is larger and/or the denominator is smaller.

The energy could be as large as 3,13 J and the mass could be as small as 5,99 g. The calculation for v2 could be as large as 2(3,13)/(5,99 x 10-3)= 1045 so v could be as large as 32.3276.

The energy could be as low as 3,11 J and the mass could be as large as 6,01 g.
The calculation for v2 could be as small as 2(3,11)/(6,01 x 10-3= 1034.94 so v could be as small as 32,1705 m/s.

Well, you see what happened: the true velocity lies between 32,1 and 32,3, exactly what writing 3 significant figures gives: 32,2 m/s.

As a "rule of thumb", retaining the minimum number of significant figures through the calculations gives the correct precision for the result.

3. Jan 16, 2005

### haki

Thank you. It is a bit clearer now. Soo if you have something wirtten as 3,12 the value could be as low as 3,115 or as high as 3,124 the diff would be 0,09 = 0,0045. You could write that as 3,12 +- 0,0045 and lets say the bullet would be 20(1 +- 0,5)g, that means that bullet would weight as low as 19g or as high as 21g. To calculate with the precision we have then sqrt((2 * 3,12 +- 0,0045)/(20(1 +-0,5))) That would be max = 18,13 min = 17,22 avg = 17,68 diff = 0,91 => +- diff/2 = 0,455 = 0,46. Soo we can write this as 17,7 +- 0,46 m/s? Is there a way to calculate seperetely with the precision given (in this case +-0,0045 and +-0,5)?

Last edited: Jan 16, 2005