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Students reinventing the general law of energy conservation
Logman, P.S.W.M.
Publication date
2014
Link to publication
Citation for published version (APA):
Logman, P. S. W. M. (2014). Students reinventing the general law of energy conservation.
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Appendix A: Upcoming Dutch exam program
The exam requirements relevant to the concept of energy conservation in the upcoming Dutch exam program are the following (van Weert et al., 2012): A1 The student is capable of purposefully looking for, judging, selecting and
processing information on the subject.
A2 The student is capable of adequately communicating in the public domain on subjects on physics orally, in writing and digitally.
A4 The student can tell in which ways knowledge about physics is used in study and profession and can express his interest in related studies and professions.
A5 The student is capable of analyzing problems in contexts making use of relevant concepts and theory, translating these problems into subject-specific research, performing such research, and drawing conclusions from the research results making use of consistent reasoning and relevant computational and mathematical skills.
A6 The student is capable of preparing, implementing, testing, and evaluating a technological design based on a given problem in a certain context, while using relevant concepts, theory, skills, and valid and consistent reasoning.
A10 The student is capable of analyzing in which ways physical and technological knowledge is being developed and applied in their contexts.
A12 The student is capable of applying a number of computational and mathematical skills relevant to physics correctly and routinely in physics-specific problems.
A15 The student is capable of quantifying physical quantities and relate mathematical expressions to physical concepts.
C2 The student is capable of using the concepts of conservation of energy, efficiency, work, and heat to describe and analyze energy transformations in their contexts.
I1 The student is capable of performing experiments and analyzing and interpreting the results from those experiments in contexts covered by the central exam.
I3 The student is capable of preparing, implementing, testing, and evaluating a design based on a given problem in contexts covered by the central exam.
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Appendix B: Exam-like question concerning near transfer
The first example for the “near transfer” type of test questions, taken from domain parts that were addressed in the teaching-learning sequence, is the following:
Rock, iron and water
Rock has a specific heat of 750 J/(kg∙K). We put a hot block of iron in a container filled with cold water and rock. Iron has a specific heat of 460 J/(kg∙K). The rock has been in the water for a while and thus has assumed the same temperature as the water. The specific heat of water is 4180 J/(kg·K).
Calculate the temperature once it has settled after putting a 237 g hot block of iron with a temperature of 652 °C in a container filled with 420 g of cold water and 125 g of rock both with a temperature of 20 °C.
The answer should be something like this:
mrock·crock·Trock + mwater·cwater·Twater + miron·ciron·Tiron = mrock·crock·Tafter +
mwater·cwater·Tafter + miron·ciron·Tafter
0.125·750·20 + 0.420·4180·20 + 0.237·460·652 = 0.125·750·Tafter +
0.420·4180·Tafter + 0.237·460·Tafter
1875 + 35112 + 71081.04 = 93.75·Tafter + 1755.6·Tafter + 109.02·Tafter
108068.04 = 93.75·Tafter + 1755.6·Tafter + 109.02·Tafter
108068.04 = 1958.37·Tafter
Tafter = 108068.04/1958.37 = 55.182646792996 °C ≈ 55 °C.
The second example for the first type of test questions is the following:
Diver
A diver dives from a diving board 10 m above the water. The highest point he reaches is 12 m above the water, see figure. The mass of the diver, including shorts, is 75 kg.
Consider the diver to be a point mass, ignore friction, and assume the diver jumps nearly vertically upwards.
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The diver
a. Calculate the minimum velocity with which the diver jumps off the diving board.
b. Calculate the minimum velocity with which the diver hits the water.
The answer should be something like this:
a. (m∙g∙h + ½∙m∙v2)before = (m∙g∙h + ½∙m∙v2)after 75∙9.81∙10 + 0.5∙75∙vbefore2 = 75∙9.81∙12 + 0.5∙75∙02 7357.5 + 37.5∙vbefore2 = 8829 + 0 37.5∙vbefore2 = 8829 – 7357.5 = 1471.5 vbefore2 = 1471.5 / 37.5 = 39.24 vbefore = √39.24 = 6.2641839 m/s ≈ 6.3 m/s. b. (m∙g∙h + ½∙m∙v2)before = (m∙g∙h + ½∙m∙v2)after 75∙9.81∙12 + 0.5∙75∙02 = 75∙9.81∙0 + 0.5∙75∙vafter2 8829 + 0 = 0 + 37.5∙vafter2 37.5∙vafter2 = 8829 vafter2 = 8829 / 37.5 = 235.44 vafter = √235.44 = 15.344054223 m/s ≈ 15 m/s.
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Appendix C: Exam-like question concerning far transfer
An example for the “far transfer” type of test questions, taken from domain parts that were not addressed in the teaching-learning sequence is the following:
Short airstrips
To be able to reach some remote places very short airstrips are constructed like Soldier Bar in Idaho (see figure).
Soldier Bar in Idaho over Big Creek canyon
Source: http://www.canyonflying.com/soldierbar.jpg
Because the airstrips are shorter than normal and the brakes should not get overheated these airstrips are often constructed on a slope. By this construction airplanes may land uphill and take off downhill.
A rich man wants to go fishing in the Big Creek canyon and wonders whether it is possible to land his Cessna on Soldier Bay airstrip. His Cessna including everything weighs 2500 kg and has a minimum air velocity of 80 km/h for landing. Of course it has to come to a standstill before the end of the airstrip.
Cessna airplane
Source: http://upload.wikimedia.org/wikipedia/commons/c/c8/Cessna172-CatalinaTakeOff.JPG
The airstrip starts at a height of 155 m above the river and ends at a height of 175 m above the river. The Cessna’s two steel brakes (the front wheel does not have a brake) weigh 1200 g a piece and lose their braking qualities at temperatures above 500 °C.
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Brake from a Cessna airplane
Source: http://www.aircraftspruce.com/catalog/graphics/RAPCOBRAKE.jpg The specific heat of steel is 460 J/(kg∙K). It is a beautiful day with a temperature of 25 °C.
Use a clear calculation to hand the rich fisherman a sound advice on landing at Soldier Bay.
The answer should be something like this:
Ek,before + Eg,before + Eth,before = Ek,after + Eg,after + Eth,after
½∙mairplane∙vbefore2 + mairplane∙g∙hbefore + mbrakes∙csteel∙Tbefore = ½∙mairplane∙vafter2 +
mairplane∙g∙hafter + mbrakes∙csteel∙Tafter
0.5∙2500∙(80/3.6)2 + 2500∙9.81∙155 + 2.4∙460∙25 = 0.5∙2500∙02 + 2500∙9.81∙175 + 2.4∙460∙Tafter 617284 + 3801375 + 27600 = 0 + 4291875 + 1104∙Tafter 4446259 = 4291875 + 1104∙Tafter 1104∙Tafter = 4446259 – 4291875 = 154384 J Tafter = 154384 / 1104 = 139.84057971 ≈ 1.4∙102 °C No problem.
