Yr 12 EMPA Section B
You are in the EMPA and you have completed measurements of how a time measured
with a stop clock precise to 0.01s varies with a length, x measured with a tape
measure with a precision of 1mm.
T_1
T_2
T_3
18.46
21.27
23.00
22.14
23.32
25.23
27.34
27.26
31.60
35.99
35.99
37.65
47.37
43.74
45.32
58.58
57.08
56.1
68.63
66.27
68.21
85.15
85.57
83.93
T_av
x (m)
20.91
23.56
28.73
36.54
45.48
57.25
67.70
84.88
Uncertainty
1
2
3
4
5
6
7
8
2.27
1.55
2.17
0.83
1.82
1.24
1.18
0.82
% Unc.
10.9
6.6
7.6
2.3
4.0
2.2
1.7
1.0
1. Correct the table for any mistakes that would cost you marks in the exam.
T_1 through T_av are lacking units and should be of the
form T_1 /s
All timings should be to 2 d.p. 56.1 is not
X should have the units specified x /m rather than x(m)
Values of x should be precise to mm so should read 1.000 etc
Uncertainty, (Calculated from half range) should have
units /s
2. Fill in the data in table for the columns marked T_av,.
3. Estimate the uncertainty in the average time for each data point.
Note uncertainty is half the range
4. Calculate the percentage uncertainty in the average time for each data point.
5. Plot a graph of T_av against x.
Graph shown below, note the quality of the curve is low due
to the need to draw in pen rather than pencil so that it
would scan well.
6. Evaluate the gradient of the line when x= 4 m. Show your working below.
Gradient = Change in y / Change in x
Gradient = (70.0-22.5) / (8.0-2.3)
Gradient = 47.5 / 5.7
Gradient = 8.33 s/m
7. Describe the method you used to determine the gradient of the line at x=4 m.
The mirror/protractor method. The mirror is placed
approximately perpendicular to the tangent to the curve at
the x=4 point. Looking along the curve the continuation of
the curve can be seen reflected in the mirror. The angle of
the mirror is adjusted until the reflected image of the curve
follows the remainder of the actual curve. A normal line is
then drawn along the plane of the mirror. The protractor is
then use to draw the tangent at right angles to the normal
line drawn. Gradient construction lines were made as large
as possible, (at least 8x8)
8. Below is an image of a chessboard, make suitable measurements to determine
the width of the uniformly-sized squares.
Not done since image very reduced on my printout. We’ll
assume I measured them as 2.9cm
9. Show with calculations how the procedure you used reduced the percentage
uncertainty in measuring the width of each square.
The correct approach is to “measure across n identical
squares and divide by n”.
If we assume that each square is around 2.9cm and we use a
normal 30cm ruler marked in mm. The absolute uncertainty
for a single reading will be 0.1cm
So our reading will be 2.9.0 + 0.1cm As a percentage
uncertainty this is 3.4%
If we measure across 8 squares we can expect the reading to
be around 23.2cm, the uncertainty for this single reading
remains the same at 0.1cm
So our reading will be 23.2 + 0.1cm
Since we measured 8 square, we divide each by 8 so the
reading for each square is now
2.9 + 0.0125cm (Note I’ve shown more dp for the
uncertainty than usual to make the working out clearer)
As a percentage uncertainty this becomes 0.0125/2.9 = 0.43%
which is a massive improvement over the reading for a
single square.
Note : since we are told the board is square, one could take
repeat readings horizontally and vertically and find the
average.
10. The chessboard box states that the sides of the squares are 3cm long. What is
the percentage difference between the value you measured and the quoted
value?
Actual accurate value is 3.0, measured value is 2.9.
%difference = difference / actual value
%difference = 0.1/3
%difference = 3.3%