About paper
Czech originalApproximation in physics and life
Introduction
Evolving the ability to estimate the values of different quantities like mass and length and also the amount, that could be useful in everyday life, should be a part of physics education. The subjects of this paper are a few suggestions for creating a sense for size and amount of things, together with practical ideas for the classroom.
Limber up your brain - rough estimations based on mental calculations
1. How high would a barometer be containing water instead of mercury?
Answer: Approximately 10 metres.
Solution: Consider that the working principle of a barometer is based on the balance between the atmospheric pressure (equivalent to the normal atmospheric pressure pn = 105 Pa) and hydrostatic pressure ph = h ρ g of the liquid (water). If you introduced the density of water and the Earth’s gravitational acceleration to the equation pn = ph, the calculated height of the water column (difference in liquid level) would be 10 m.
2. Imagine that somebody would like to balance you out with gold. How much of gold would you need?
Answer: A few litres.
Solution: Let us imagine a man with about 70 kilos of mass. The human body is mostly composed of water and because water density is approximately ρ = 1000 kg·m-3, 1 kg of water occupies a volume of 1 litre; thus the man weighing 70 kg occupies a volume of 70 litres. Even if we do not remember the density of gold, we can guess that it is about 1 order of magnitude higher than the density of water (to be precise, the nominal value is ρAu = 19290 kg·m-3[1]). With respect to our rough estimation, the volume of gold which is necessary for balancing out the weight of our body would be ten times lower than the volume of the water we are composed of, i.e. a few litres.
3. Estimate the mass of one million steel balls with a diameter of 1 mm.
Answer: A few kilograms
Solution: Let us use the relation between the mass, density and volume \( m = \rho V \). The density of steel is estimated to be \(\rho\approx 5000\,\mathrm{kg\cdot m^{-3}}\) (higher than the density of water but lower than the density of gold). The values of real density of different types of steel range from 7500 kg·m-3 to 8300 kg·m-3 for different kinds of steel, but our simple estimations are accurate enough for our purposes. The volume of one ball is \[ V= \frac{4}{3} \pi r^3 \approx \frac{4}{3} \cdot 3 \cdot \left(\frac{1}{2}\right)^3\,\mathrm{mm^3} \approx\frac{1}{2}\,\mathrm{mm^3}\]
(One student estimated it considering the simple idea of a ball trapped in the cube with 1mm edge. Because the ball did not fill up the space completely, we considered just half of the volume – \(\frac{1}{2}\,\mathrm{mm^3}\) ). One million steel balls would occupy a volume that is million times bigger, \( \approx\frac{1}{2}\,\mathrm{dm^3}\); multiplying it by our estimated steel density, the resulting mass is a few kilograms.
4. How many moles of water do you carry on your back?
Answer: A few kilomoles.
Solution: The mass of one mole or the molar weight of water is equal to \[ M_m = M_r (\mathrm{H_2O}) \, \mathrm{g\, mol^{-1}} = [2 A_r (\mathrm{H}) + A_r (\mathrm{O})] \, \mathrm{g\, mol^{-1}} \approx (2\cdot1 + 16) \, \mathrm{g\, mol^{-1}} \approx 18\, \mathrm{g\, mol^{-1}} .\] One thousand moles have a mass of about 18 kg which is the mass of a heavier bag. We are able to carry approximately thousands moles of water, i.e. a few kilomoles.
5. Which one contains more money- one wagon of 10-crown coins or ten wagons of 1-crown coins?
Answer: Ten wagons of 1-crown coins
Solution: We can start with the following idea: Ten wagons of 1-crown coins give the same amount of money as one wagon of 10-cown coins, assuming both coins have the same volume. A 1-crown coin is smaller than a 10-crown coin (we mean Czech coins), thus more of them can be put into a wagon, so 10 wagons filled with 1-crown coins contain a bigger financial deposit than 1 wagon filled with 10-crown coins.
Have you ever seen one million or do you have a sense about this huge amount?
This activity could help the students to create an impression of huge amounts. It is suitable for children of different ages. It is appropriate to separate children into several groups containing 6 to 8 members, because they will stick together a huge number of millimetre paper sheets. The activity has two variants but the beginning of both is still the same.
Task: Stick together so many sheets of paper that the number of millimetre squares in the resulting structure approaches 1 million. Mark then so many squares that the amount
a) is equal to the number of the citizens in your regional capital, local town or the village you live in;
b) is the number of hours that a 100-year-old person is alive, or the number of hours you spend in an elementary school.
Accessories: a lot of millimetre sheets (20 sheets per group), scissors, glue, ruler, calculator, stationery
Procedure: Using a simple calculation, students will discover that to reach 1 million of millimetre squares, approximately 20 (16) sheets of A4 paper are needed (the size of final sheet will be equal to 1m2).
a) To estimate the number of citizens that live in the regional capital, students at first estimate the number and then find more precise data on the internet. Then they mark the area which contains the relevant number of millimetre squares by line. It is appropriate to suggest choosing a square area-it is the clearest way of demonstration. If it is possible, have them mark the area representing the town (village) where they live. If the students from different groups live in different places, each of them can mark their own town (village). The procedure is similar when emphasizing the number of classmates.
b) Students will calculate how many hours would fit into 100 years. Considering that the year has 365 days on average and the day contains 24 hours, the result would be n = 100·365·24 = 876000. A 100-year-old person will be alive for 900 000 hours which could seem to be not enough! Marking of the appropriate values (hours in that case) could be done the same manner as in the previous case.
How many hours will we spend in for example an elementary school? Each lesson has 45 min, but adding the breaks and the time for lunch in a canteen, we can count 60 min per lesson, because we are interested only in rough estimation. Assume that we spend in an elementary shool 9 years. Every school year has for example 33 weeks and every week contains 25 lessons. The number of hours we spend at school is then p = 9·33·25 = 7425. It is approximately 1 % of 100 years and more than 1 % of the average age which is not so much. The students would probably not agree.
Methodological remarks: At first, the teacher should figure out whether to let the students use a calculator or not. If you allow a calculator, I suggest that they use those in their mobile phones, even if it is not commonly allowed at school. They will practise the work with the machine they carry all the time. In variant b) it is possible to count and mark how many hours they usually spend at work. It is good to discuss all the acquired data together with the students and it is appropriate to let them subjectively sum up if the particular numbers seem to be huge or not.
How long is our corridor or measurements leading to more precise estimations
This activity is suitable as a slightly nontraditional introduction to measurements, especially for younger students. It uses no special equipment and is suitable for work in pairs.
Task: Try to estimate the length of a school corridor.
a) Estimate the length standing at one of its ends
b) Estimate the length walking along the corridor as fast as you usually walk, without stopping
c) Measure the length using objects which do not have a scale in units of length (you are not allowed to use a ruler)
Tools: stationery, all the instruments that fulfil c).
Procedure:
a) Students will either guess how long the corridor is or they will base their estimation of its length on the distance between two neighbouring class-doors, width of the window, distances between the lines on the floor etc. The estimated values will be marked carefully.
b) Working on this problem, they probably will take into account the possibility of counting the number of steps and multiplying it by the length of their foot (or the step itself). The resulting values should be precisely marked again.
c) This task forces the students to take some object with known length, such as a notebook, or different parts of the human body (such as the arms). Some students will enthusiastically lie on the floor to measure the length of the corridor using their own body.
Methodological remarks:
If we compare the methods described in a, b and c, we can guess that the least precise is method a, more precise is method b and the most precise one is c. It happens sometimes, that some pairs reach more accurate results from estimations than from the measurements. It could happen by good fortune or be caused by some systematic error, as when the length of the notebook is measured incorrectly. If we calculate the arithmetic average of the results of all groups separately for the methods in a, b and c it is highly probable, that c would be more accurate than a and b.
Which one is heavier or “how the objects tell lies”.
Fig. 1: Which is heavier? Do not be misled by the size and the material!
This activity should directly explain, that the measuring of weight of objects by pure balancing in the hand is unreliable, because we are misled by our own experience (like assuming that big or metal objects must be heavy).
Task: Compare two objects a) the bigger one which is lighter than the smaller one, b) the bigger one is really heavier than the smaller one, keeping each in a different hand at the same time.
Tools: heavy small object and light bigger object – a chemical stand and an aluminum ladder, heavy big object and small lighter object – plastic bottle filled with water and metal weights, weights (personal and kitchen)
Procedure: Prepare at first the couple of objects where the bigger one is heavier than the smaller one. The frame or the ladder is appropriate.
We will let the students hold both objects in their hand and ask them to not describe or somehow comment the situation, only to mark personally which one seems to be heavier. After finishing this experiment with all students, we will compare all estimations and weigh both objects precisely. After the discussion, we will start comparing the second couple of objects.
Methodological remarks: A lot of students will consider the ladder to be heavier than the stand. To persuade them of the real truth, weigh both of them on a personal scale. Asking the students why they judged the bigger object to be heavier, the resulting answer would be that it was due to its size. Having the new experience that the bigger object can be lighter than the smaller one now, some of the students will use it in comparing the mass of heavier and bigger objects. Discussing afterwards why most of them have been misled, it is good to point out the fact that we usually consider the metal object to be the heavy one. Thus the main result is that the comparison of the weight of different objects only by balancing it in a hand is deceptive.
Conclusion: Without any calculator and even any stationery, it is possible to estimate the values of different qualities only using simple estimations. The idea of huge quantities can be visualised by using the millimeter paper. A measurement is usually more precise than a rough estimation, but when trying to guess approximate values, we can use a comparison with the value of a known quantity; thus the difference between the measurement and the estimation might not necessarily be so dramatic, and thus the judgement is not without value. On the other hand, it is important to keep in mind that estimation could be strongly influenced by conditions that could not be totally ignored (like size of the object or the material, estimating its weight and comparing it with another object).
References:
[1] Mikulčák J. a kol.: Matematické, fyzikální a chemické tabulky pro střední školy. Prometheus Praha, 1988. ISBN 80-85849-84-4.