# Physics Question (1 Viewer)

#### theKingPin

##### Member
Hi All,
I was practicing some physics questions from Australian Physics Olympiads Online and came across a question that I had not seen before. The answer is given however I do not completely understand it. Could someone explain it.
Thanks

Question:

Lucy is measuring the acceleration due to gravity in Melbourne by dropping a ball through a vertical distance 1.00 m and timing how long it takes.

The ball starts at rest, and Lucy times its fall four times. The results are: 0.47 s, 0.42 s, 0.48 s and 0.41 s. The uncertainty in her distance measurement is 1 cm and the uncertainty in the timer is 0.01 s. What is the uncertainty in the value of g that Lucy calculates?

Select one:
a. at least 0.01 ms−2 and at most 0.03ms−2.
b. more than 0.03 ms−2but at most ms−2.
c. more than 0.1 ms−2 but at most 0.4 ms−2.
d. more than 0.4 ms−2 but at most 0.6 ms−2.
e. more than 0.6ms−2 but at most 2ms−2.

Correct Answer: E

A value of t = 0.45 ± 0.04 s just covers the range of measured values.

Alternative 1: Use this value and the range in possible distance measurements to calculate the largest and smallest possible values of the acceleration due to gravity.

Alternative 2 (using fractional errors): t has approximately 9% fractional error. The fractional error in the distance is 1%, so the fractional error in g is (2 × 9 + 1)% = 19% which gives an absolute uncertainty of around 2 ms−2.

http://csma31.csm.jmu.edu/physics/Courses/P140L/appendices/Combining Uncertainties in Experimental Results.htm

#### InteGrand

##### Well-Known Member
Hi All,
I was practicing some physics questions from Australian Physics Olympiads Online and came across a question that I had not seen before. The answer is given however I do not completely understand it. Could someone explain it.
Thanks

Question:

Lucy is measuring the acceleration due to gravity in Melbourne by dropping a ball through a vertical distance 1.00 m and timing how long it takes.

The ball starts at rest, and Lucy times its fall four times. The results are: 0.47 s, 0.42 s, 0.48 s and 0.41 s. The uncertainty in her distance measurement is 1 cm and the uncertainty in the timer is 0.01 s. What is the uncertainty in the value of g that Lucy calculates?

Select one:
a. at least 0.01 ms−2 and at most 0.03ms−2.
b. more than 0.03 ms−2but at most ms−2.
c. more than 0.1 ms−2 but at most 0.4 ms−2.
d. more than 0.4 ms−2 but at most 0.6 ms−2.
e. more than 0.6ms−2 but at most 2ms−2.

Correct Answer: E

A value of t = 0.45 ± 0.04 s just covers the range of measured values.

Alternative 1: Use this value and the range in possible distance measurements to calculate the largest and smallest possible values of the acceleration due to gravity.

Alternative 2 (using fractional errors): t has approximately 9% fractional error. The fractional error in the distance is 1%, so the fractional error in g is (2 × 9 + 1)% = 19% which gives an absolute uncertainty of around 2 ms−2.

http://csma31.csm.jmu.edu/physics/Courses/P140L/appendices/Combining Uncertainties in Experimental Results.htm
Well the answer is definitely greater than 0.6 m s-2, so I guess that mades E the answer.

$The formula for g is g=\frac{2s}{t^2}, where s is the height dropped from.$

$Therefore, the standard deviation, or uncertainty, in g is given by$

$\sigma _g = \sqrt{\left( \frac{\partial g}{\partial s}\right)^2 \sigma _s ^2 + \left( \frac{\partial g}{\partial t}\right)^2 \sigma _t ^2 }=\sqrt{\left( \frac{2}{t^2}\right)^2 \sigma _s ^2 + \left( \frac{-4s}{t^3}\right)^2 \sigma _t ^2 } \ \ (*)$

(see the propagation of uncertainty article on wiki: https://en.wikipedia.org/wiki/Propagation_of_uncertainty#Simplification (you should familiarise yourself with propagation of error concepts)).

$Uncertainty in s is \sigma_s = 0.01 (units being m (metres)).$

$Uncertainty in time measurements is calculated to be roughly \sigma _t = 0.0351 (units being seconds). You can calculate this on your calculator using the statistics mode (on the Casio one, you should use the sx'' button for standard deviation after inputting the time values, as this gives sample standard deviation (if you don't know about the STAT mode of the calculator, consult its manual. For example, the manual for the CASIO fx-82 AU PLUS is linked below, and start reading from the lower half of page E-17, which is page 18 of the PDF, to see how to use STAT mode, as well as how to obtain standard deviation etc.)).$

http://support.casio.com/storage/en/manual/pdf/EN/004/fx-82AU_PLUS_II_EN.pdf

$The value to use for s in Equation (*) is 1 (unit being m). The value to use for t is the average \bar{t}=0.445 (units being seconds), i.e. average of the t measurements. (There will be uncertainty in this and \sigma _t too actually, so we should be using different values, but these uncertainty is small enough to neglect for this MCQ.)$

$Then by Equation (*), \sigma _g = \sqrt{\left( \frac{2}{(0.445)^2}\right)^2 \cdot (0.01)^2 + \left( \frac{-4\times 1}{(0.445)^3}\right)^2 \cdot (0.0351)^2 }\approx 1.59 (slightly different if you use more exact values for \sigma _t), units being \text{m s}^{-2}. This is between 0.6 and 2, so (E) is the answer.$

$If you look at the highest and lowest possible values of g(s,t)=\frac{2s}{t^2} based on the given data and uncertainties, you'll find that they are g_\text{max.}=g(1.01,0.4)\approx 12.6 and g_\text{min.}=g(0.99,0.49)\approx 8.2, which would suggest that uncertainty lies roughly between the following: 9.81-8.2=1.61, and 12.6-9.81=2.79 (this was their Alternative 1 solution). Option (E) best fits these.$

#### theKingPin

##### Member
Thanks InteGrand, for once again solving a physics question on BOS.