A2 Further Practical Skills
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Transcript of A2 Further Practical Skills
8/3/2019 A2 Further Practical Skills
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The practical work in the second year of your A level
course builds on what you have covered in the first year.
The examination will test you on two areas: planning
experimerıts, and analysis and evaluation of your results.
PlanningAlrhough you should be thinking of experimental
procedures from as soon as you embark on an A
level course, if not before, by the time you complere
the course you should be competent in planning. experimental procedures.
There are different stages in planning an experiment.
D efin in g th e p ro ble m
it may seem obvious, but the first thing is to iderıtify
the problem. To do that you must idenrify:
• the independent variable in the experiment
• the dependent variable in the experiment
• the variables that are to be controlled.
For instance you may be investigating the response
time of a thermocouple when it is used to measure
changing temperatutes, as shown in Figure Al. I.
What are the variables here?
probe juction
water bath
i=---I~~ins
fixed junction
melting ice
FigureA2.1 Investigating the response time of a thermocouple
thermometer.
I
The independent variable would be the temperature
which is to be measured.
The dependent variable would be the time it takes
for the reading on the thermocouple meter to reach a
steady value.
There maybe severalother variables that could affect
the result. One might be the starting temperature
of the therrnostat 'probe junction', arıother is the
Appendix A2 F u rth er p ra c ti ca l s k ill s
- - ~ - - - - - - - - - - - - - - - - - - - - '
(fixed) temperature of the second junction of the
therrnocouple.
D a ta c o U ee tio n
The next task is to think about how you are going to
carry the experiment out. Once you have a method in
mind you need to:
• Deseribe the method to be used to vary the
independent variable.
• Deseribe how the independent variable is to be measured.
• Deseribe how the dependent variable is to be measured.
• Deseribe how other variables are to be controlled.
• Describe, with the aid of a clear, labelled diagram, the
arrangement of apparatus for the experiment and theprocedures to be followed.
In the experiment to investigate the response time
of a thermocouple you may decide that the sirnplest
way of varying the independent variable is to have
a water bath and to vary its temperature. When the
temperature has reached the required level, plunge the
'probe junction' into the water.
The procedure would be:
1 Measure the temperature of the water with a
laboratory thermometer.2 The reading from the therrnocouple will be
displayed on a millivoltmeter and the dependent
variable, the time taken for the reading to reach its
steady value, measured with a Stop watch.
3 There are several different variables that need to
be controlled. In practice you will only need to
consider one or two.
• The second junctiorı of the thermocouple should
be kept at a constant temperature by keeping it
in corıtact with melting ice.• The probe junction should be cooled to the same
temperature each time, again by holding it in themelting ice.
Your description of the experiment should be clear
and concise. It sometimes helps to give the description
in the form of bullet points or a numbered list. A
labelled diagram will always clarify your description.
In the experiment deseribed here, the diagram shown
in Figure Al.ıwould be sufficien t.
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th od o f an alys isrequires that you deseribe how the data should be
d in order to reach a conclusion, including details
erived quantities to be cakulated. In our example,
could plot a graph of the temperature of the water(x-axis) against the response time (y-axis).
ow would we use the graph? If the graph gives
raight line through the origin, therı we see that
response time is linear across the temperatlire
e tested.
Figures A2.2a and A2.2b show two possiblesets of results for the therrnocouple experiment.
For each, state how the response time depends
on the temperature.
Time Time
Temperature Temperaturea b
FigureA2.2 Graphs showing how the response time
of a thermocouple might depend on the temperature
being measured.
need to assess the risks of the experiment anderibe precautions that should be taken to keep
s to aminimum.
our example it is sufficient to mention simple
s such as wearing goggles to protect the eyes when
ting liquids, or ensuring that the water bath is
e and cannot be easily knocked over. The use of
s to put the thermocouple junction into the bath
other example.
A d d itio n al d e ta ils
This tests your experience of doing practical work.
Have you had suflicient experience to see things which
will improve the experiment?
Here are some ideas that might be incorporated in
the general description of the experiment:
• Stir the water in the bath so that it all reaches a unitorm
temperature.
• Use an oil bath to give a wider range of temperatures.
• Replace the water bath with different substances at rheir
melting or boiling points so that the temperatures are
more repeatable.
• Check the reading on the therrnocouple volrrneter at
the melting or boiling points of the different substances
before carrying out the main experiment.• Use a digital voltıneter so that it is easy to spot when
the thermocouple reaches the steady temperature.
M ath em atic al an a ly s is o f d ata
In Appendix Al, we saw how to rearrange simple
equations into the form y= mx+ c and how to use
a straight-lirıe graph to hnd the constants m and c.
However, you also need to be able to deal with
quantities related by equations of the form y= ax"
and y = ae", For these, you need to be able to uselogarithms (logs).
In handling data, our aim is usually to process
the data to obtain a straight line graph. Then we
can deduce quantities from the gradient and the
intercepts. Table A2. ishows the graph which should
be plotted in each case, and the quantities which can
be deduced from the graph.
Relationship Graph Gradient Intercept because ...on j-axis
r==« y against x m c
y=axnInyagainst
n InaIny=
Inx nlnx+lna
y=aekx Iny against x k Ina
Iny=kx+
Ina
Table A2.1 Choice ofaxes for straight-line graphs.
IILppendix A2 Fu rth e r p ra c tic a l s kil ls
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D
A rela tionsh ip of th e form v = a x "
Abali falls under graviry in the absence of air
resistance. It falls a distance s in time t.The results
are given in the firs t two columns ofTable Al.2.Notice that the unit for the logarithm is written as
In ( s / m) and not In ( s ) / m or In ( s ) / In (m).
Time tlsDistanee
in (tl s) in (sım)fallen sım
0.20 0.20 -L.61 -L.61
0.40 0.78 -0.92 -0.25
0.60 L.76 -0.51 0.57
0.80 3.14 -0.22 1.14
L.OO 4.90 0.00 L.59
1.20 7.05 0.18 L.95
Table A2.2 Results for Example 1.
8.0
6 . 0
E
- -c~~ 4.0
< L Iuc
2.~o
2.0
o~~--,-----,------r------o 0.80 1.20
Time/s
FigureA2.3 A distance-time graph plotted using the data
in Table A2.2.
A graph of the distance fallen against time gives the
curve shown in Figure Al.3. This, being a curve, tells
us little about the relationship between the variables.
If, however, we suspect that the relationship is of the
Appendix A2 Fu rt he r p ra c tic a l s k ills
- - ~ - - - - - - - - - - - - - - - - - - -
form y= ax", we can test this idea by plotting a graph
of In s against In t (a 'log-log plot'). Table Al.2 shows
the values for I n s and Int,
and the resulting graph isshown in Figure Al.4. (Notice that here we are using
'natural logs', bur we could equally well use logs to
base 10,written as 'lg'.)
In (sım)
< i . s In (t/s )
2.0
1.0
-15 -1 -0·5
-1.0
-2.0
lRgure A2.4 A log-log plot for the data shown in Table A2.2.
From this graph the gradient is equal to the valueof n, the power of t.
. (-1.55-0.5)n = gradıent = -----------
(- 1.50 - 2.55)
= 1.98 ",,2.0
So the equation is of the form s = a t? The intercept
on the y-ax:is is equal to In a, so:
Ina = 1.6
By taking the antilogarithm we ger:
a= 4.95 ms-2"" 5.0 rns?
If we think of the equation for free fall 5= 1 - g t ? - , the
constant a= 1 - g. But g = 9.8 m S-2which is consistent
with the value we get for our constant.
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A re la t ionsh ip o f th e fo rm y= ae"
Acurrent flows from a charged capacitor when it is
connected in a circuit with a resistor. The current
decreases exponentially with time (the same pattemwe see in radioactive decay).
Figure A2.S shows the circuit and Table A2.3
shows typical values of current iand time t from
such an experiment.
Current I/mA Time t/s ln(I/mA)
ıo.OO 0.00 2.303
6.70 0.20 1.902
4.49 0.40 1.5023.01 0.60 1.102
2.02 0.80 0.703
1.35 1.00 0.300
Table A2.3 Results from a capacitor discharge experiment.
c =10IlF
L o =10 m A
R= 20.0 kQ
Figure A2.5 A circuit for investigating the discharge of a
capacitor.
The graph obtained from these results (Figure
A2.6) shows a typical decay curve, but we cannot
be sure that it is exponential. To show that the
curve is of the form 1= 10 e k " we plor Iniagainstt (a 'log-linear plor'). Values of ını are included
in TableA2.3. (Here, we must use logs to base e
rather than to base 10.)
The graph oflnI against t is a straight line (Figure
A2.?), confirming that the decrease in current
followsan exponential pattern. The negative gradient
showing exponential decay, rather than growth.
r ı / m A
5
10
0+------,-----,,-----,------
o 0.4 0.8
Time / s
1.2
FigureA2.6
(
2.0
InI
1.0
o+------,------,---~,------o 0.8
Time/s
1.2
. J
Figure Az.z
The gradient of the graph gives us the value of the
constant k.
. (O- ı. 16)k=gradıent= =-0.S04çl::::-0.SOçl
(2.30 - O)
From the graph, we can also see that the intercept
on the y-axis has the value 2.30 and hence (taking
the inverse log) we have 10= 9.97 = 10.OmA.Hence we can write an equation to represent the
deereasing current as follows:
1= 10.Oe-o .50'
We could use this equation to calculate the current
at any time t.
11lppendix A2 Fu rth e r p ra c tic a l s kills
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2 In the expressions that follow x and y are
variables in an experiment. All the other
quantities in the expressions are constants.
In each case, state the graph you would plot to
produce a straight line. Give the gradient of each
line in terms of the constants in the expression.
a y= k X 3/2
b v = c . x 'l
c m= 8x
B i
dy= yoekx
e R - (Y- Y o )- 2
x
3 The period of oscillation T of a small spherical
mass supported by a length i of thread is given
by the expression
T=2n f - fwhere g is the acceleration due to gravity.
Design a laboratory experiment using this
expression to determine the acceleration due
to gravity. You should draw a diagram showingthe arrangement of your equipment. In your
account, you should pay particular attention to:
a the procedure to be followed
b the measurements to be taken
c how to analyse the data to determine g
d any safety precautions that you would take.
T re atm e n t o f u n c erta in tie sAll results should indude an estimate of the absolute
uncertainty. For example, when measuring the time for
a runner to cornplete the 100 m you may express this as
12.1 ± 0.2 s. This can also be expressed as a percentage
uncertainty (see Appendix Al); the percentage
uncertairıry is equal to 0.2 x 100% = 1.65%. so we12.1
write the value as 12.1 s ± ı. 7%.
- Appendix A2 Fu rt he r p ra c ti ca l s kil ls
- - - - . - - - - - - - - - - - - ~
C o m b in in g u n c e r ta in t ie s
When quantities are added or subtracted, their
absolute uncertainties are added. A simple example
is that when measuring the length of a stick using a
millimetre scale there is likely to be an uncertainty
of 0.5 mm at both ends, giying a total uncerta irıtyof 1.0 mm.
To combine uncertainties when quantities are
rnultiplied or divided is a little more complex. You
can only add or subtract quantities if theyare in the
same units; you cannot subtract current from voltage!
However, you can multiply or divide them. To find
the combined uncertainry in a cas e like this we add
the percentage uncertainties.
Wo rk ed e xample
1 The poterıtial difference across aresistor is
measured as 6.0 ± 0.2 V, whilst the current is
measured as 2.4 ± O . 1A.
Calculate the resistance of the resistor and the
absolute uncertainty in its measurement.
Step 1 Find the percentage uncertainty in each
of the quantities:
percentage uncertainty in p.d. = 0.2 x 100%6.0
=3.3%
percentage uncertainty in current
= 0.1 x 100%=4.2%2.4
Step 2 Add the percentage uncertainties
= (3.3 + 4.2)% = 7.5%
Step 3 Calculate the resistance value and findthe absolute uncertairıry
R= V = 6.0 =2.50i 2.4
7.5% of2.5=0.1875===0.20
The resistarıce of the resistor is 2.5 ± 0.2 O.
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en you calculate the uncertainty in the square
a quantity rherı, since this is an example of
ltiplicatiorı you should double the percentage
. For example if A =2.0 + 0.2 cm
n A has a percentage uncertainty of 10% so
=4.0 m ' ± 20%; or using the absolute uncertai nty=4.0 ±0.8 crn".
4 You measure the following quantities:
A= 1.0m±0.4m B=2.0m±0.2m
C=2.0ms-l±O.5mçl D=0.20s±0.01s.
Calculate the answer and the uncertainry forthe following expressions. You may express
your uncertainty either as an absolute value or
as a percentage.
aA+B
bB-A
c CxD
d l LD
e A2
f 2xA
g the square root of (A xB)
5 A rifie bullet is photographed in Hight using
rwo Hashes of light separated by a time interval
of 1.00 ms ± 0.02 ms. The nrst image of the
bullet on the photograph appears to be at a
position of 22.5 ±0.5 cm on a scale underneath
the Highr path. The positiorı of the second
image is 37.5 ±0.7 cm on the same scale.
Find the speed of the bullet and the absolute
uncertainty.
ta in tie s a n d lo g ar ith m s
hen a log graph is used and we need to include
r bars (see uncertainties and graphs) we must nnd
logarithm of the measured value and either the
ithm of the largest or smallest possible value. The
inty will be the difference between the two.
Wo rk ed e xample
2 The resistance of aresistor is given as 47 ±5o.The value of In (R i O) is to be plotred on a
graph. Calculate the value and uncertainty in
In (RIO).
Step ıCalculate the logarithm of the given value:
In (RIO) =ln47 =3.85
Step 2 Calculate the logarithm of the
maximum value:
maximum value =47 + 5=520
In 52 =3.95
Step 3 The uncertainty= 3.95 - 3.85 =O.10
Thus In (RIO) =3.85 ± O.10
U n c e rta in tie s a n d g ra p h s
We can use error bars to show uncertainties on
graphs. Table A2.4 shows results for an experiment on
stretching a spring.
Load/NLength of
Extension icmspring/cm
O 12.4±0.2 0.0
1.0 14.0±0.2 1.6±0.4
2.0 15.8±0.2 3.4±0.4
3.0 17.6±0.2 5.2±0.4
4.0 18.8±0.2 6.4±0.4
5.0 20.4±0.2 8.0±0.4
Table A2.4 Results from an experiment on stretching a spring.
When plotting the graph the points are plotted as usual,
and then theyare extended to show the extreme values,
as shown in Figure A2.8. Then the best nt line is drawn.
To estimate the error in the gradient we draw not
only the best nt line but also the 'worst acceptable
line'. This line is the worst line which goes through all
the error bars, and is shown in Figure A2.9.
EDppendix A2 Fu rth er p ra ctic al s kills
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- : : , - . : , : ~ ~:~, ' - -
- -
Extension i cm
8.0
6.0
2. 0
o~------r------.------.-------
o 2 64
Load i N
FigureA2.8 A graph representing the data in Table A2.4, with
error bars and a line of best fit drawn.
Extension i cm
8.0
6.0
2. 0
o~-----,r------.------.-------
o 4 6Load i N
FigureA2.9 The same graph as in Figure A2.8, with a 'worst
acceptable' line drawn (dashed).
The gradients for both best fit and worst fir lines are
calculated and the error is the difference in their gradients:
D
error = (gradient of best fi t line) - (gradient of
worst acceptable line)
Appendix A2 Further pract ica l skills
~ - - - - - - - - - - - - - - - - - - - - '
In our experiment, the gradients are:
(8 . 2 - O Jline ofbest fit: gradient = cmN-I
5.0 - °= 1.58 cm N-I", 1.6 N cm N-i
(8.4-0J
line of worst fi t: gradient = cm N-i5.0 - °
= 1.4cmN-I
So the uncertain ty in the gradient = ı.6 - ı.7=±O.l cm N:'
The gradient is therefore: 1.6 ± O.1cm N-I.
6 Suggest why there are no uncertainties
induded in the measurements of the load.
7 it İs suggested that R and r are related by the
equation:
R = al' where a and b are constants.
a A graph is plotted with In R on the y-axis and
In r on the x-axis. Express the gradient and
y-intercept in terrns of a and b.
b Values of R and r are given in Table A2.5.
r ı mm RıO.
2.0±0.1 175.0
3.0±0.1 77.8
4.0±0.1 43.8
5.0±0.1 28.0
6.0±0.1 19.4
Table A2.5 Results for Test yourself Q 7.
Calculate and record values of In (R iQ) and
In (rı mm) in the table and indude the absolute
uncertainties in In (rı mm).
c Plot a graph ofln (RI Q) and In (rı mm).
Indude error bars for In (rı mm).
cont inued ...~ .
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d Draw the line of best nt and a worst
acceptable straight line on your graph.
e Determine the gradient of the line of best nt.
Include the uncertainty in your answer.
f Using your answer to e, determine the value
of b .
g Determine the value of a and its uncertainty.
c lu s io ns a nd e va lu atio nre s u ltsthe previous experiment we can conclude that
extension/load for the spring in this example
.6 ± 0.2 cm N-I. If a hypothesis is rnade that the
extensiorı is proportional to the load therı there
is enough evidence here for the conclusion to be
supported, as a straight line can be drawn from the
origin through all the error bars. If this is not possible
therı the hypothesis is not validated.
Now, suppose that the hypothesis is that the springobeys Hooke's lawand stretches by 5.8 cm when a
load of 3.5 N is applied. The firsr part is validated for
the reasons given above. However, an extension of
5.0 cm for a load of 2.5 N gives a value of 2.0 cm N-I
for the gradient. This is clearly outside the range
allowed for by the uncertainty in our measurements,
and therefore the hypothesis is not supported .