7 concepts18 worked examples29 exercises4 exam-level7 figures
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01Introduction, measurement and estimating: uncertainty, significant figures, units and dimensions
You measure a sheet of paper with a 30 cm ruler: 21.6 cm along one edge, 27.9 cm along the other. The calculator returns 602.64 for the area and your lab partner copies all five digits into the report. The demonstrator crosses out the last two without touching the ruler, and takes a mark off for the missing unit as well.
By the end of this section you can say which digits of that 602.64 survive, why the demonstrator knew it without remeasuring, what the answer looks like with its unit and its attached, and how to get the same answer in square metres without slipping a factor of ten thousand.
In 60 seconds
Every number in physics is a measurement, so it comes with three things a bare number does not have: a unit, a limited count of trustworthy digits, and an uncertainty. This section is the grammar for all three.
every unit change, including squared and cubed units where the whole fraction is raised to the power
Dimensional consistency
$$[\text{every term}]=[M]^{a}[L]^{b}[T]^{c}\ \text{with the same }a,b,c$$
checking a half remembered formula, or reading the units of an unknown constant
Three most common mistakes
Copying the calculator. A ruler that reads to the nearest millimetre cannot support an area written to 0.01 cm squared; the extra digits were produced by the arithmetic, not by the measurement.
Converting a squared or cubed unit with the plain factor, so 1 cm squared becomes 0.01 m squared instead of 0.0001 m squared. The factor must be raised to the same power as the unit it carries.
Treating a dimensional check as a proof. Dimensions cannot see a pure number, so a wrong formula with the right dimensions passes the test unharmed.
The published weights for this course are Midterm 1 20%, Midterm 2 20%, quizzes 10%, homework 5%, final 25% and lab 20%. Nothing here is a topic that gets its own exam question later in the term; it is the notation every later answer is written in, and the lab component is where units, significant figures and uncertainties are written down by hand every week.
How much time do you have?
10 minutes
The two rules that cost marks in every single paper: how many digits to write, and how to change a unit without losing a power of ten.
card, c-significant-figures, c-unit-conversion, formula card
45 minutes
Everything that turns into a number on a lab sheet: the plus or minus, the digit count, the conversion chain, and the dimension check that catches a wrong formula before you spend five minutes on it.
card, c-measurement-uncertainty, c-significant-figures, c-unit-conversion, c-dimensional-analysis, faded ladder, practice B
full read
Adds the standards behind the units, order of magnitude estimating, and the one thing the first lecture is really about: what it means for a physical law to be tested by a number that is never exact.
hook, c-measurement-uncertainty, c-significant-figures, c-si-units-standards, c-unit-conversion, c-dimensional-analysis, c-estimating, c-models-laws, faded ladder, exam example, practice A to D, self audit
By the end of this section
Report a measured quantity as a best value with an uncertainty, and convert between absolute and percent uncertainty.
Apply the significant figure rules to a chain of multiplications, divisions and additions, rounding once at the end.
Write any quantity in with the correct SI base unit and prefix, and read a prefixed unit back into metres, kilograms and seconds.
Convert a quantity between units by a chain of factors equal to one, including units raised to a power.
Test an equation for dimensional consistency and deduce the SI units of an unknown constant inside it.
Estimate a quantity to the nearest power of ten by naming the model, bracketing each factor and multiplying one digit numbers.
Decide whether two measured results agree, and state the range of validity that any physical law carries with it.
Syllabus coverage
Introduction
The nature of physics: models, theories and laws, and the range of validity a law carries with it
Placed at the end rather than the start on purpose. The question of what it means for a law to be tested only becomes concrete once you can say what a measurement is and how far it can be trusted, which is the rest of the section.
covered
Measurement
Uncertainty, significant figures, SI units and standards, unit conversion,
Five concepts share this token: what an uncertainty is, how many digits survive, the SI base units and their prefixes, conversion chains, and dimensions.
covered
Estimating
Order of magnitude estimates and the modelling step that makes them possible
covered
as a worked context
Mass over volume used as the running example for conversion, uncertainty and plausibility checks
The week line does not name density. It appears here only as a vehicle: it is the shortest realistic quantity that forces a cubed unit conversion and a plausibility check at the same time, and every value used is either given in the question or looked up as a table value. You are not responsible for material properties.
off_syllabus
Recall first
Powers of ten
$10^{a}\times 10^{b}=10^{a+b}$, $\ 10^{a}/10^{b}=10^{a-b}$, and $(10^{a})^{b}=10^{ab}$. A negative exponent is a reciprocal: $10^{-3}=1/1000$.
Every prefix change and every conversion in this section is one of these three lines in disguise, and the cubed conversions are the third one.
Percentage of a quantity
A fraction $f$ written as a percentage is $100f$, so $p\%$ of $x$ is $(p/100)\,x$.
Uncertainties are quoted both ways and the exam switches between them inside a single question.
Areas and volumes from school geometry
Rectangle $A=\ell w$; circle $A=\pi r^{2}$; cylinder $V=\pi r^{2}h$; sphere $V=\tfrac{4}{3}\pi r^{3}$; box $V=\ell w h$.
These are the shapes every estimate and every lab volume is built from, and they are also the cleanest place to see a cubed unit appear.
Speed and density as definitions, not results
Speed is a distance divided by the time taken, $\mathrm{m/s}$; density is a mass divided by the volume it occupies, $\rho=m/V$ in $\mathrm{kg/m^{3}}$.
Both are used here purely as carriers of units and dimensions, never as physical results. Nothing in this section depends on how motion actually works; that question opens next week.
Rounding a decimal
To round to a given digit, look at the first digit dropped: $5$ or more rounds the kept digit up, less than $5$ leaves it alone. So $2.346\to 2.35$ and $2.344\to 2.34$.
Significant figures are a rounding rule with a stopping place chosen by the measurement rather than by taste.
Try it yourself first (3 questions)
1§01.2 — counting digits before the rules arrive●●○○○
Before any rule is stated, try the question the rest of the section answers. A balance prints the mass of a sample as 0.00250 kg. The zeros are doing two different jobs in that number.
Given
A printed reading of 0.00250 kg
Find
(a) How many of those digits carry information about the measurement?
Hint 1/4
Ask what each zero is for: is it holding the decimal point in place, or reporting something the balance actually resolved?
Hint 2/4
A zero that only fixes the position of the decimal point is a placeholder. A zero written after the last non zero digit, to the right of the point, is a measured digit.
Hint 3/4
Write the reading as $2.50\times 10^{-3}\ \mathrm{kg}$. The digits in front of the power of ten are exactly the ones the balance resolved.
Hint 4/4
So the reading carries three digits of information: $2$, $5$ and the final $0$.
Show solution
Rewriting in scientific notation is shorter than arguing about zeros in place, because the notation separates the digits from the decimal point by construction.
the power of ten now carries the position of the point, so nothing left in front of it can be a placeholder
$$2.50\ \Rightarrow\ \{2,\,5,\,0\}$$
the trailing zero was printed although it was not needed to place the point, which is the balance saying it resolved that digit
Answer $$\boxed{3\ \text{significant figures}}$$
Check
Independent check by contradiction: if that last zero were meaningless, the balance would have printed 0.0025 kg, which is a different claim about its resolution.
Scientific notation is not decoration; it is the only form in which the digit count of a number is unambiguous.
2§01.4 — the trap that costs a factor of ten thousand●●○○○
This one is here because almost everybody gets it wrong the first time, and getting it wrong now is free. A microscope slide has an area of 1.0 square centimetres and you have to hand it in in square metres.
Given
$A=1.0\ \mathrm{cm^{2}}$
$1\ \mathrm{m}=100\ \mathrm{cm}$
Find
(a) Write $A$ in $\mathrm{m^{2}}$.
Hint 1/4
Ask how many little squares of side one centimetre it takes to tile one square metre. That count is the whole answer.
Hint 2/4
Converting a unit raised to a power raises the whole conversion fraction to that power: $\left(\frac{1\ \mathrm{m}}{100\ \mathrm{cm}}\right)^{2}$.
Hint 3/4
Substitute: $1.0\ \mathrm{cm^{2}}\times\left(\frac{1\ \mathrm{m}}{10^{2}\ \mathrm{cm}}\right)^{2}=1.0\times 10^{-4}\ \mathrm{m^{2}}$, with $1\ \mathrm{m}=100\ \mathrm{cm}$ as given.
Hint 4/4
So the slide has an area of $1.0\times 10^{-4}\ \mathrm{m^{2}}$.
Show solution
Tiling the square is quicker to picture than the algebra, and the algebra is written underneath so the two agree.
Square the whole conversion fraction
$$\frac{1\ \mathrm{m}}{10^{2}\ \mathrm{cm}}=1$$
a fraction whose top and bottom are the same length is the number one, so multiplying by it changes nothing physical
Count instead of convert: a square metre is a hundred rows of a hundred one centimetre squares, so it holds ten thousand of them, and one of them is a ten thousandth of it.
The rule that follows for the rest of the course: whatever power sits on the unit sits on the conversion factor too.
3§01.4 — a speed in the unit the exam wants●○○○○
Every mechanics answer this term is expected in SI units, and speeds arrive in kilometres per hour from everyday life. A car on a city road is doing 72 kilometres per hour.
You need metres upstairs and seconds downstairs, so two separate replacements are needed, one on each part of the unit.
Hint 2/4
Multiply by fractions equal to one, arranged so the unwanted unit appears once on the top and once on the bottom.
Hint 3/4
Substitute, with $1\ \mathrm{km}=10^{3}\ \mathrm{m}$ and $1\ \mathrm{h}=3600\ \mathrm{s}$ as given: $72\times\frac{10^{3}}{3600}$.
Hint 4/4
So $v=20\ \mathrm{m/s}$.
Show solution
Two separate factors rather than the remembered shortcut of dividing by 3.6, because the shortcut hides which unit went where and is the thing that gets inverted under exam pressure.
hours now cancel; the fraction is written with the hour on top because the hour to be removed sits on the bottom of the speed
Answer $$\boxed{v=20\ \mathrm{m/s}}$$
Check
Order of magnitude check the other way: 20 metres every second is 1200 metres a minute, and 72 kilometres in an hour is 1200 metres a minute.
A useful anchor for the whole term: 36 km/h is 10 m/s, so dividing a km/h figure by 3.6 gives m/s.
Notation
symbol
reads as
means
watch out
$\pm$
plus or minus
the half width of the band a measurement lives in: $x\pm\delta x$ means the value is somewhere between $x-\delta x$ and $x+\delta x$
it is not a second answer and not a range of possible experiments; it is one measurement quoted honestly
$\delta x$
the uncertainty in x
the absolute uncertainty, carrying the same unit as the quantity itself
quote it to one significant figure, and never write more digits in $x$ than the position of $\delta x$ can support
$[\,Q\,]$
the dimensions of Q
which powers of mass, length and time the quantity is built from, written with $[M]$, $[L]$ and $[T]$
a dimension is not a unit: $[L]$ is length in general, while $\mathrm{m}$ and $\mathrm{cm}$ are two units for it
$\sim$
is of the order of
equal to within about a factor of ten, the an order of magnitude estimate claims
never write $\sim$ next to three digits; an estimate that deserves three digits is not an estimate
$\approx$
is approximately equal to
equal after rounding, or after an approximation stated in words
stronger than $\sim$ and weaker than $=$; if you can write $=$, write $=$
$\mathrm{m},\ \mathrm{kg},\ \mathrm{s}$
metre, kilogram, second
the three SI base units all of mechanics is written in
unit symbols are upright, never italic, never pluralised and never followed by a full stop
Conventions used here
How many digits an answer keeps.
Answers in these notes are rounded to three significant figures unless the data given deserve fewer, in which case the data win. Intermediate results carry one extra guard digit and are never rounded on the way through; rounding happens once, at the end.
Rounding twice moves the last digit by more than the measurement does, and the marker cannot tell that from a genuine error. One rounding at the end is a rule that leaves no room for argument.
What a quoted number means when no uncertainty is written.
A value written without an explicit plus or minus is taken to be uncertain by one unit in its last significant digit. So 4.50 s means an uncertainty of about 0.01 s, while 4.5 s means about 0.1 s. The two are different claims about the stopwatch.
Otherwise the digit count is decoration. This convention is what makes the significant figure rules a statement about the apparatus rather than a style guide.
Which value of g is used.
Where the magnitude of the free fall constant is needed for a plausibility check, these notes use $g=9.80\ \mathrm{m/s^{2}}$ throughout the course, never 9.8 in one place and 10 in another.
A constant that changes value between sections quietly changes every third digit of every answer, and the reader has no way to tell which version a given number came from.
How an order of magnitude answer is written.
An estimate is reported as a single power of ten, or one digit times a power of ten, with the tilde symbol: $N\sim 10^{9}$. The model behind it is stated in one sentence before the arithmetic, and the answer is never dressed up with extra digits.
An estimate written as 3.47 times ten to the ninth claims a the method cannot deliver, and it hides the fact that the reader was supposed to check the model, not the arithmetic.
Signs and directions in this section.
Every quantity here is a magnitude with a unit: a length, a mass, a time, a density, a count. No direction and no sign convention is needed anywhere in this section, so none is declared. Negative signs appear only inside exponents.
Stating this now means that when a sign convention does get declared later, it is a genuine choice being made and not a piece of background noise.
How agreement between two results is judged.
Two measured values agree when their uncertainty bands overlap, and disagree when the bands are disjoint. The comparison is between intervals, never between the two central numbers on their own.
Central values essentially never coincide, so comparing them alone makes every experiment a failure; the width of the band is the part that decides.
1.1Every measurement is a band, not a point
Fixes what a measured number actually claims: a best value, a unit, and a half width either side of it.
Physics begins with a number that came off an instrument, and is therefore not exact.
Solvable with what we have
Read a ruler: the paper edge falls on 21.6 cm.
Read the other edge: 27.9 cm.
Multiply the two numbers and copy what the calculator shows: 602.64.
Not solvable yet
Say how many of those five digits the ruler is entitled to.
Say how far the area could be out if each edge is out by a millimetre.
Decide whether a group reporting 599 disagrees with us.
The arithmetic is perfect and the claim is not. Written like that, the number says the area is known to a hundredth of a square centimetre, which is a claim about the ruler.
Why it fails
The smallest division is 1 mm and a length needs two edges read, so each reading is good to about 0.1 cm. Push both edges up and the area is near 607.6, both down and it is near 597.7. Everything after the third digit came from the multiplication.
$\delta x$ carries the same unit as $x$, quoted to one significant figure
the last digit of $x$ sits in the same decimal place as $\delta x$
the band describes this apparatus and this reading
$$\boxed{\;x_{\text{measured}} = x \pm \delta x,\qquad \text{percent uncertainty}=\frac{\delta x}{|x|}\times 100\%\;}$$
A measurement is not one number but an interval: a best value, and a half width saying how far the reading could be out. The percent form divides that half width by the value, turning two lengths into one pure number.
Two groups measure the same edge; the bands are disjoint, so extra repeats will not fix it.
Looks like this, but is not
This is a precise measurement: a caliper reads the same rod five times as $12.31$, $12.30$, $12.32$, $12.31$, $12.30$ mm, a spread of two hundredths of a millimetre.
This looks like an accurate measurement and is not one: the caliper was zeroed on a speck of dust, so every reading is $0.24$ mm too small. Repeating shrinks the scatter and leaves the offset untouched: precision and accuracy are separate words.
Percent uncertainty of a length read from a millimetre ruler
A rod lines up with the 21.6 cm mark on a ruler whose smallest division is 1 mm. Quote the reading with an absolute uncertainty, then as a percent uncertainty.
Given
Reading 21.6 cm
Smallest division on the ruler 1 mm, that is 0.1 cm
Find
The reading written as a band, and the same band written as a percentage.
Solution
We charge the full smallest division rather than half of it, because a length is the difference of two readings and each end can be off by half a division; taking half would quietly assume one end is exact.
Turn the instrument into an uncertainty
$$\delta x = 0.1\ \mathrm{cm}$$
half a division at each of the two ends, and half plus half is one whole division
$$x = (21.6 \pm 0.1)\ \mathrm{cm}$$
the last digit of the value and the digit of the uncertainty now sit in the same decimal place, which is what makes the pair readable
Convert the band into a percentage
$$\frac{\delta x}{x}=\frac{0.1}{21.6}=0.00463$$
the units cancel, which is the whole point: the result no longer depends on whether we worked in centimetres or metres
$$0.00463\times 100\% = 0.463\% \approx 0.5\%$$
an uncertainty gets one significant figure, so the extra digits are dropped here and not carried forward
A route with no division in it: one millimetre inside about twenty centimetres is one part in two hundred, and one part in two hundred is half a percent.
The percent form is the one that travels. When measurements get multiplied together, it is the percentages that combine, and the centimetres would have nothing to combine with.
Area of the paper sheet, with the uncertainty the ruler allows
The same ruler gives the two edges of a sheet as 21.6 cm and 27.9 cm. Report the area with its uncertainty, and say how many digits survive.
Given
$a=(21.6\pm 0.1)\ \mathrm{cm}$
$b=(27.9\pm 0.1)\ \mathrm{cm}$
Find
The area as a value plus or minus an uncertainty.
Solution
Adding percent uncertainties is shorter than working out the largest and the smallest possible area, and for a product the two routes agree; the long route is kept below as the check.
Multiply the best values
$$A = 21.6\times 27.9 = 602.64\ \mathrm{cm^{2}}$$
the best estimate of a product is the product of the best estimates; every digit here is still provisional
for a product the fractional slops add, since both factors can sit at the top of their bands at the same time
Come back to square centimetres and round once
$$\delta A = 0.0082\times 602.64 = 4.9\ \mathrm{cm^{2}}\approx 5\ \mathrm{cm^{2}}$$
the uncertainty is rounded to one significant figure, which puts it in the units place
$$A = (603 \pm 5)\ \mathrm{cm^{2}}$$
the value is rounded to the same decimal place as the uncertainty; writing 602.64 next to a slop of 5 would contradict itself
Answer $$\boxed{A=(603\pm 5)\ \mathrm{cm^{2}}}$$
Check
The long route, which shares no algebra with the short one: the largest area the readings allow is $21.7\times 28.0=607.6$ and the smallest is $21.5\times 27.8=597.7$. Half the difference is $5.0\ \mathrm{cm^{2}}$, the same slop, obtained without ever writing a percentage.
Two readings, one multiplication and one addition of percentages: the uncertainty costs less work than the area did.
This closes the question the section opened with. The calculator was never wrong; 602.64 was a claim about a ruler, and the ruler could not back it.
Checkpoint
§01.1 — percent uncertainty of a stopwatch reading●○○○○
Thirty seconds on the same move with a different instrument. A hand stopwatch is used to time a swing and the reader quotes 2.34 s, with a reaction time slop of 0.02 s.
Given
$t=(2.34\pm 0.02)\ \mathrm{s}$
Find
(a) Give the percent uncertainty in the timing.
Hint 1/4
The question is not asking for a new measurement. It asks for the same band expressed as a fraction of the value.
Hint 2/4
Percent uncertainty is $\dfrac{\delta t}{t}\times 100\%$, and the seconds cancel.
Hint 3/4
Substitute the reading given above, $t=2.34\ \mathrm{s}$ and $\delta t=0.02\ \mathrm{s}$: $\dfrac{0.02}{2.34}\times 100\%$.
Hint 4/4
That comes to 0.855 percent, which is 0.9 percent once an uncertainty is cut to one significant figure.
Show solution
Straight into the definition; there is no shorter route and no approximation worth making at this size.
1.2Significant figures: how much of the calculator to write down
Turns the digit count of your data into the digit count of your answer, and nothing more.
The band told us the area was uncertain by about five square centimetres. Significant figures are the shorthand that carries that information without writing the band out every time.
RuleRule: significant figures in arithmetic
Conditions
counted objects and defined conversion factors are exact and never limit the answer
carry one guard digit through the middle of a calculation and round once, at the end
a zero to the right of the decimal point and after the last non zero digit is significant; a leading zero never is
a bare integer such as $2500$ is ambiguous, so scientific notation is the only safe way to write it
$$\boxed{\;\begin{aligned}\times,\ \div\ &:\ \text{keep the fewest significant figures among the inputs}\\[2pt] +,\ -\ &:\ \text{keep the fewest decimal places among the inputs}\end{aligned}\;}$$
A chain is as strong as its weakest link, and the two operations disagree about which link is weakest. When you multiply, it is the factor with the fewest significant digits. When you add, it is the term whose last trustworthy digit sits furthest to the left, because a column you cannot fill in cannot be added.
The same three digits survive in the first two numbers; the third is unreadable as written, which is the whole argument for scientific notation.
Looks like this, but is not
This is the rule working: $2.3\times 3.14159 = 7.2$, cut to two significant figures because the weakest factor has two.
This looks like the same rule and is a different one: $12.3+3.14159 = 15.4$. Addition counts decimal places, not significant figures, and here the answer has three significant figures although one input had six. Change the first number and the count changes with it: $2.3+3.14159 = 5.4$, two significant figures, from exactly the same decimal place rule.
Counting the significant figures in 0.00250, 100.0, 2500 and 1.20 times ten cubed
For each number, say how many digits are significant and why.
Given
$0.00250$
$100.0$
$2500$
$1.20\times 10^{3}$
Find
A digit count for each, with the reason.
Solution
Rewriting each number in scientific notation settles every case in one move, because that form has no room for a placeholder.
The unambiguous ones
$$0.00250 = 2.50\times 10^{-3}\ \Rightarrow\ 3$$
the leading zeros only position the point; the final zero was written although it was not needed for that, so it is a measured digit
$$100.0 = 1.000\times 10^{2}\ \Rightarrow\ 4$$
the decimal point followed by a zero says the instrument resolved the tenths, and that forces the zeros in front of it to be significant too
$$1.20\times 10^{3}\ \Rightarrow\ 3$$
the power of ten carries the size, so every digit in front of it was written on purpose
The ambiguous one
$$2500 \Rightarrow 2,\ 3\ \text{or}\ 4$$
the two zeros could be measured digits or could be holding the place, and the notation gives the reader no way to tell
Independent check on the second one: if the trailing zeros of 100.0 were placeholders, the writer would have written 100, so the extra point and zero are doing work that only a measured digit can do.
Whenever you find yourself arguing about whether a zero counts, you have already found the answer: write the number in scientific notation and the argument disappears.
Area and perimeter of a plot measured as 23.5 m by 8.1 m
The same two measurements feed a multiplication and an addition. Report both results with the right number of digits.
Given
$\ell = 23.5\ \mathrm{m}$ (3 significant figures)
$w = 8.1\ \mathrm{m}$ (2 significant figures)
Find
The area and the perimeter, each correctly rounded.
Solution
Both are computed from the raw numbers first and rounded at the very end, because rounding the sum before doubling it would throw away the guard digit the doubling needs.
The area, decided by significant figures
$$A = 23.5\times 8.1 = 190.35\ \mathrm{m^{2}}$$
the raw product, carrying every digit so that the rounding happens once
$$A = 1.9\times 10^{2}\ \mathrm{m^{2}}$$
the weaker factor has two significant figures; writing $190$ instead would look like three, so the power of ten is not optional here
The perimeter, decided by decimal places
$$\ell + w = 23.5+8.1 = 31.6\ \mathrm{m}$$
both terms are known to the same decimal place, so the sum keeps that place and nothing is lost
$$P = 2\times 31.6 = 63.2\ \mathrm{m}$$
the 2 counts sides, so it is exact and cannot weaken the answer; three significant figures survive from two numbers, one of which had only two
Bracket the area instead of trusting the rule: the smallest the sides allow is $23.45\times 8.05=188.8$ and the largest is $23.55\times 8.15=191.9$. The spread already moves the third digit, so only two digits can be defended, which is what the rule said in one line.
Two numbers, two operations, two different digit counts: this is the pair that catches people who learned only one of the rules.
Do not memorise the two rules as a pair of unrelated facts. Both say the same thing: an answer may not claim a column that none of its inputs could fill.
Checkpoint
§01.2 — rounding a mixed product and quotient●●○○○
Half a minute on the rule in the form the exam uses it: a short chain with a deliberately weak link in the middle. The reading 4.00 came from a digital meter, 2.0 from an analogue scale.
Given
$\dfrac{8.4\times 2.0}{4.00}$
Find
(a) Which is the correctly reported value?
Hint 1/4
The arithmetic is not the question. The question is which of the three inputs decides how much of the result you may keep.
Hint 2/4
For a chain of multiplications and divisions, the answer keeps as many significant figures as the weakest input has, and no more.
Hint 3/4
Substitute, with the data from the question, $8.4$ (two), $2.0$ (two) and $4.00$ (three): $\dfrac{8.4\times 2.0}{4.00}=4.2$.
Hint 4/4
So the correct report is 4.2, with two significant figures.
Show solution
Count first and compute second: knowing the answer will carry two digits stops you from copying five of them off the display.
Find the weakest link
$$8.4\ (2),\quad 2.0\ (2),\quad 4.00\ (3)$$
the digit counts, before any arithmetic, so the target is fixed in advance
$$\frac{8.4\times 2.0}{4.00} = 4.2$$
the arithmetic happens to be exact here, which is why the question tests the rule and not the calculator
Answer $$\boxed{4.2}$$
Check
Order of magnitude check: 8 times 2 is 16, and 16 over 4 is 4, so a first digit of 4 is right and only the second digit was ever in doubt.
⚠ Rounding at every step instead of once at the end
each intermediate line looks like an answer, so the significant figure rule gets applied to it, and the error that introduces is then carried forward
1.3SI base units, standards, and the prefixes that ride on them
Gives every number a unit others can reproduce, and a prefix so you do not carry the exponent by hand.
We now know how many digits a measurement supports. The next question is what the number is a measurement of, and the answer has to be something a stranger in another country can rebuild.
DefinitionDefinition: the three SI base units of mechanics
Conditions
every other mechanical unit is a product of powers of these three
modern standards fix a constant of nature rather than name an object: caesium 133 for the second, the speed of light for the metre, and the Planck constant for the kilogram since 2019
unit symbols are upright, never pluralised and never followed by a full stop
Mechanics needs only three independent units. Once a metre, a kilogram and a second are fixed, every area, volume, speed and density in the course is fixed with them, because each of those is nothing but these three multiplied and divided together.
The reason prefixes and powers of ten exist: the same three units have to describe a proton and a galaxy, and the distance between them is thirty six factors of ten.
Looks like this, but is not
This is a base unit: the kilogram, written $\mathrm{kg}$, the SI unit of mass.
This looks like a unit you may prefix again, and is not: the prefix is already inside the name. A million grams is a megagram, $\mathrm{Mg}$, never a kilokilogram, and a thousandth of a kilogram is a gram, not a millikilogram. Any prefix you add replaces the one already there.
prefix
symbol
factor
a length that size
giga
G
10⁹
1 Gm is about two and a half times the distance to the Moon
mega
M
10⁶
1 Mm is roughly a sixth of the Earth's radius
kilo
k
10³
1 km is a ten minute walk
centi
c
10⁻²
1 cm is the width of a fingernail
milli
m
10⁻³
1 mm is the smallest division on your ruler
micro
µ
10⁻⁶
1 µm is a bacterium
nano
n
10⁻⁹
1 nm is about ten atoms side by side
Read the whole table as a single instruction: a prefix is a power of ten wearing a letter, so it can always be traded back for that power and then cancelled like any other factor. Nothing else in the table has to be memorised, because the right hand column rebuilds the left hand one from things you can see.
Writing 0.000 000 45 m with a prefix
A wavelength is measured as 0.000 000 45 m. Write it in scientific notation and then with the most natural prefix.
Given
$\lambda = 0.000\,000\,45\ \mathrm{m}$
Find
The same length in scientific notation and with a prefix.
Solution
Scientific notation first and the prefix second, because the exponent is what tells you which prefix to reach for; guessing the prefix first is how a factor of a thousand goes missing.
Cross check against the scale figure: this length sits between an atom and a virus, which is where visible light belongs, so the exponent is in the right region.
Prefixes step in thousands. If your exponent is not a multiple of three, move digits until it is, and then read the prefix straight off.
A density quoted as 7.85 g/cm³ in SI units
A handbook gives the density of a mild steel sample as 7.85 g/cm³. Convert it to the SI unit of density.
An independent anchor: water is 1.00 g/cm³, and a cubic metre of water is a tonne, that is 10³ kg, so the factor between the two unit systems really is a thousand. Steel comes out about eight times denser than water, which is why a steel bolt sinks.
Remember the single fact and you never need the conversion again: 1 g/cm³ is exactly 10³ kg/m³.
Checkpoint
§01.3 — a prefix traded for a power of ten●○○○○
Thirty seconds on a prefix in the direction that catches people out. A cell in a laboratory photograph is 35 micrometres across and the report wants SI base units.
Given
$d = 35\ \mathrm{\mu m}$
$1\ \mathrm{\mu m}=10^{-6}\ \mathrm{m}$
Find
(a) Write the diameter in metres, in scientific notation.
Hint 1/4
Nothing is being measured again here. The prefix is a factor that has to be written out and then folded into the number.
Hint 2/4
Replace the prefix by its power: $1\ \mathrm{\mu m}=10^{-6}\ \mathrm{m}$, so the whole quantity is multiplied by that factor.
Hint 3/4
Substitute, with $d=35\ \mathrm{\mu m}$ from above: $35\times 10^{-6}\ \mathrm{m}$, then move one digit to normalise.
Hint 4/4
So $d = 3.5\times 10^{-5}\ \mathrm{m}$.
Show solution
Write the prefix out as a power before touching the digits, so the normalising step at the end cannot be confused with the conversion itself.
Changes the name of a quantity without changing the quantity, which is why the units alone can check the work.
The prefix table already contains every conversion in this section. What is missing is a way of using it that cannot silently drop a power of ten, and that is what the chain provides.
MethodMethod: the conversion chain
Conditions
the top and the bottom of each fraction must describe the same physical amount, so the fraction is the number one
a unit raised to a power carries the whole fraction to that power
defined factors such as $1\ \mathrm{h}=3600\ \mathrm{s}$ or $1\ \mathrm{in}=2.54\ \mathrm{cm}$ are exact and never limit the significant figures
write each fraction so the unwanted unit appears once above and once below, and strike it out on the page
Multiplying by one never changes a quantity, and a fraction whose top and bottom describe the same amount is one. A conversion is therefore a string of multiplications by one, arranged so that the units you do not want appear once above and once below and cancel in writing. If they do not cancel, the chain is wrong and the units say so before the number does.
Every unit that appears once above and once below is gone by the end of the line; the ones left standing are the answer's units, and they are chosen before any arithmetic happens.
Looks like this, but is not
This conversion is right: $2.5\ \mathrm{km} = 2.5\times 10^{3}\ \mathrm{m}$.
This one reuses the same factor and is wrong by a thousand: $2.5\ \mathrm{km^{2}} = 2.5\times 10^{3}\ \mathrm{m^{2}}$. The unit is squared, so the fraction is squared with it and the true value is $2.5\times 10^{6}\ \mathrm{m^{2}}$. Picture it instead of trusting the habit: a square kilometre is a thousand rows of a thousand square metres.
55.0 mi/h in kilometres per hour and in metres per second
A speed limit is posted as 55.0 mi/h. Express it in km/h and then in the SI unit of speed.
One chain, two stopping points: converting all the way to metres per second in a single line and reading off the intermediate value costs nothing and gives a free consistency check.
Go backwards along a different route: 24.6 m/s times 3.6 gives 88.6 km/h, which differs from 88.5 by one in the last digit, exactly the rounding that was applied. Any larger discrepancy would mean a factor was inverted.
Three fractions, three cancellations, one rounding at the end. Every fraction was written down before any number was multiplied.
Worth keeping for the whole term: 36 km/h is 10 m/s, so a km/h figure divided by 3.6 is the same speed in m/s.
A tank holding 1.50 times ten to the fourth cubic centimetres, in cubic metres and litres
A tank is filled with $1.50\times 10^{4}\ \mathrm{cm^{3}}$ of water. Give the volume in cubic metres and in litres.
Given
$V = 1.50\times 10^{4}\ \mathrm{cm^{3}}$
$1\ \mathrm{cm}=10^{-2}\ \mathrm{m}$
$1\ \mathrm{L}=10^{3}\ \mathrm{cm^{3}}$
Find
The volume in SI units and in litres.
Solution
The cube is applied to the whole fraction rather than remembered as a number, because the remembered number is the one that comes out as ten to the minus two under exam pressure.
Picture it: a litre is a cube of side 10 cm, and a cubic metre holds a thousand of them, so 15 litres has to be 0.015 of a cubic metre. Fifteen one litre bottles is also a volume you can hold in your head.
Two conversions from one starting value cost almost nothing extra, and each one checks the other.
Checkpoint
§01.4 — an area conversion with the power attached●●○○○
Thirty seconds on the trap the whole concept exists for. A microscope field of view has an area of 45 square centimetres and the report is in SI units.
Given
$A = 45\ \mathrm{cm^{2}}$
$1\ \mathrm{cm}=10^{-2}\ \mathrm{m}$
Find
(a) Write $A$ in $\mathrm{m^{2}}$.
Hint 1/4
Before converting anything, ask what power sits on the unit, because that same power has to sit on the factor.
Hint 2/4
Use the whole fraction raised to the power: $\Big(\dfrac{1\ \mathrm{m}}{10^{2}\ \mathrm{cm}}\Big)^{2}=\dfrac{1\ \mathrm{m^{2}}}{10^{4}\ \mathrm{cm^{2}}}$.
Hint 3/4
Substitute the value given above, $A=45\ \mathrm{cm^{2}}$: $45\times 10^{-4}\ \mathrm{m^{2}}$.
Hint 4/4
So $A = 4.5\times 10^{-3}\ \mathrm{m^{2}}$.
Show solution
Squaring the fraction rather than the number keeps the unit visible at every step, so a wrong exponent shows up as a unit that fails to cancel.
Strips a quantity down to powers of mass, length and time, which is enough to kill a wrong formula in seconds.
Converting units treats metres and centimetres as different names for the same thing. Dimensions go one step further and forget the names entirely, keeping only what kind of thing the quantity is.
RuleRule: dimensional consistency
Conditions
only quantities with identical dimensions may be added, subtracted or set equal to one another
the argument of $\sin$, $\cos$, $\exp$ or $\log$ must be dimensionless
a pure number is invisible to the test, so consistency is necessary and never sufficient
dimensions do not care which unit is chosen: $[L]$ is satisfied by a metre and by a light year alike
$$\boxed{\;[\,Q\,] = [M]^{a}[L]^{b}[T]^{c};\qquad \text{every term of an equation carries the same } a,\ b,\ c\;}$$
Write down what each term is made of and ignore how big it is: so many powers of mass, so many of length, so many of time. If two terms disagree in that bookkeeping, no choice of units will ever rescue the equation, because the disagreement is about what kind of quantity each side is, not about how it is measured.
The chart has a dead end on the left and an open door on the right, which is the honest shape of this test: a failed check ends the matter, a passed check only lets you continue.
Looks like this, but is not
This passes the test and is a real formula: the volume of a cylinder, $V=\pi r^{2}h$, since $[L]^{2}\times[L]=[L]^{3}$.
This passes exactly the same test and is not the volume of anything: $V=\pi r h^{2}$, since $[L]\times[L]^{2}=[L]^{3}$ too. Dimensions cannot tell you which of the two lengths gets squared. Passing the check is permission to carry on, not a result, and the choice between these two has to be made by looking at the shape: a cylinder is a circle of area $\pi r^{2}$ stacked to a height $h$.
Reading the units of the constants in a laboratory fit
A group hangs masses on a rubber band and fits their data with $x = A + Bm$, where $x$ is the extension in centimetres and $m$ is the hanging mass in grams. What are the units of $A$ and of $B$, and what are they in SI?
Given
$x = A + Bm$
$x$ in cm, $m$ in g
Find
The units of $A$ and $B$, first as written and then in SI.
Solution
Work term by term rather than trying to interpret the fit physically; the units are fixed by the equation alone, and no knowledge of rubber is needed or wanted here.
Every term must match the left hand side
$$[\,x\,] = [\,A\,] = [\,Bm\,]$$
the three are added or set equal, and only quantities of the same kind may be added
$$[\,A\,] = [L] \Rightarrow A\ \text{in cm}$$
$A$ stands alone next to a length, so it is a length; it is the extension at zero hanging mass
Put the units back into the equation: centimetres per gram times grams gives centimetres, which matches the centimetres on the left, so no term is left standing on its own with the wrong kind of quantity.
Any constant in any fitted equation gets its units this way. You never have to be told them, and being told them is not a reason to skip the check.
Which combination of a mass and a density is a volume?
You have a mass $M$ and a density $\rho$ and you want a volume, but you cannot remember how they go together. Test the four combinations $M\rho$, $M/\rho$, $\rho/M$ and $M/\rho^{2}$.
Given
$[\,M\,]=[M]$
$[\,\rho\,]=[M][L]^{-3}$
target: $[\,V\,]=[L]^{3}$
Find
The only combination that can be a volume.
Solution
Testing all four costs four lines and removes the guesswork completely, which is cheaper than getting one of them wrong later inside a longer calculation.
both leave a power of length or of mass that a volume cannot have
Answer $$\boxed{V = \frac{M}{\rho}}$$
Check
Test it on a number instead of a symbol: a kilogram of water has a density of 10³ kg/m³, and dividing gives 10⁻³ m³, which is one litre. A kilogram of water is a litre, so the combination survives the numerical test as well as the dimensional one.
When two quantities can only go together in one dimensionally legal way, the dimensions have done more than check the formula: they have found it.
Three half remembered formulas for the volume of a cylinder
Under exam pressure three expressions come to mind for the volume of a cylinder of radius $r$ and height $h$: $2\pi r h$, $\pi r^{2} h$ and $\pi r h^{2}$. Use dimensions to narrow the field, and then finish the job.
Given
$r$ and $h$ are lengths
$V$ is a volume, $[\,V\,]=[L]^{3}$
Find
Which candidates survive the dimensional test, and how to choose between the survivors.
Solution
Dimensions first because they are free, geometry second because it costs a sentence of thought; doing it in the other order means thinking hard about a candidate that could have been dropped for nothing.
Run the test
$$[\,2\pi rh\,] = [L]\cdot[L] = [L]^{2}$$
two powers of length, so this is an area; it is in fact the curved surface of the cylinder, which is why it feels familiar
both survive, because the test counts powers of length and cannot see which symbol carries them
Break the tie with geometry, not with dimensions
$$V = (\text{area of the circular face})\times(\text{height}) = \pi r^{2}h$$
the cross section is the circle, so the radius is the length that gets squared, and the height enters once because it is the direction the circle is stacked along
Try an extreme case: make the cylinder a thin disc by letting $h$ become very small. The volume should fall off in proportion to $h$, and $\pi r^{2}h$ does, while $\pi rh^{2}$ falls off far faster than the shape does.
Three candidates, three one line tests, one of them killed for free. The remaining decision needed a picture, not algebra.
This is the honest summary of the whole concept: the test eliminates, and what survives still has to be earned.
Checkpoint
§01.5 — units of a constant inside a given relation●●○○○
Thirty seconds on the move that appears in every exam that mentions dimensions. A model of a dust cloud writes its density as $\rho = k/r^{3}$, where $r$ is a distance from the centre.
Given
$\rho = \dfrac{k}{r^{3}}$
$\rho$ is a density, $r$ is a length
Find
(a) What are the SI units of the constant $k$?
Hint 1/4
Do not try to understand the model. Ask only what $k$ has to be so that the two sides of the equation are the same kind of quantity.
Hint 2/4
Rearrange first, then read the dimensions off: $k = \rho r^{3}$.
Hint 3/4
Substitute the dimensions given above, $[\,\rho\,]=[M][L]^{-3}$ and $[\,r\,]=[L]$: $[\,k\,]=[M][L]^{-3}\cdot[L]^{3}$.
Hint 4/4
The lengths cancel, so $k$ is a mass and its SI unit is the kilogram.
Show solution
Rearranging for the unknown constant first turns the problem into a multiplication of two known dimensions, which removes the chance of dividing the wrong way round.
Isolate the constant and count
$$k = \rho\, r^{3}$$
multiplying both sides by the cube removes the only place the unknown was hiding
$$[\,k\,] = [M][L]^{-3}\cdot[L]^{3} = [M]$$
the three powers of length cancel exactly, which is the whole content of the answer
Check by substitution in units: kilograms divided by metres cubed is a density, and that is what the left hand side had to be.
⚠ Adding two quantities that are not the same kind of thing
both symbols came from the same figure and both look like lengths on the page, so the sum feels harmless until the units are written in
wrong$$V = \pi r^{2} + h$$
right$$V = \pi r^{2} h$$
⚠ Treating a passed dimension check as a proof
the check feels like work, and work that produces no objection feels like a confirmation
wrong$$[\,\pi r h^{2}\,] = [L]^{3}\ \Rightarrow\ V = \pi r h^{2}$$
right$$[\,\pi r h^{2}\,] = [L]^{3}\ \Rightarrow\ \text{possible, and still wrong}$$
1.6Order of magnitude estimates
Buys the exponent of an answer for one minute of work, which is often the only part of the answer that matters.
Every tool so far tightened a number that was already there. This one produces a number where there was none, and deliberately refuses to make it precise.
MethodMethod: an order of magnitude estimate in four moves
Conditions
state the model in one sentence before any number appears: what shape, what typical value, what is being ignored
round every input to one significant figure or to a bare power of ten
report the result as a power of ten with the symbol $\sim$, never with extra digits
name the input the answer is most sensitive to, because that is the one worth improving
$$\boxed{\;Q \sim q_{1}\times q_{2}\times\cdots\times q_{n},\qquad \text{each } q_{i}\ \text{rounded to one digit}\;}$$
Break the unknown into a product of things you can bracket, round each of them savagely, multiply, and report the power of ten. The claim being made is not that the answer is right, but that it is right to within roughly a factor of ten, which is enough to decide whether something is worth doing.
The modelling step, drawn: the real pool has rounded corners and a sloping floor, and the estimate throws all of that away in exchange for three numbers that can be multiplied.
Looks like this, but is not
This is an estimate: the pool holds about $5\times 10^{5}$ litres of water.
This looks like a better estimate and is a worse one: the pool holds $486\,320$ litres. Those six digits came out of the arithmetic, not out of the pool, whose corners are rounded and whose floor slopes. Writing them claims a survey that was never done, and it hides the only thing the reader needed to check, which is whether the box was a fair stand in for the pool.
quantity
order of magnitude
how you rebuild it
seconds in a year
3 × 10⁷ s
365 × 24 × 3600, which lands on 3.15 × 10⁷
height of a person
2 m
a doorway is about 2 m and people fit through it
mass of a person
10² kg
a bathroom scale reads between 50 and 100 kg
density of water
10³ kg/m³
a litre of water is a kilogram and a cubic metre is 1000 litres
thickness of a sheet of paper
10⁻⁴ m
a 500 sheet ream is about 5 cm thick
one heartbeat
1 s
about 60 to 80 beats in a minute
The right hand column is the important one. An estimate built from remembered numbers is only as good as your memory, but an estimate built from numbers you can rederive in ten seconds survives being forgotten.
How much water is in a 25 metre swimming pool?
Estimate the volume of water in a standard 25 m pool, in cubic metres and in litres, and say how much that water weighs on a lorry scale.
Given
Length about 25 m
Width about 10 m
Depth about 2 m
Find
The volume as a power of ten, and the corresponding mass.
Solution
The model is one sentence: treat the pool as a rectangular box. That is worth less than a factor of two, and the answer is only claimed to a factor of ten.
Sanity by comparison rather than by arithmetic: 500 tonnes is roughly a dozen fully loaded lorries, which is a believable amount of water for a building to hold and an unbelievable amount for a bathtub, so the exponent is in the right place.
Notice that the mass came out numerically equal to the number of litres. That is not luck, it is the density of water being 1 kg per litre, and it is worth carrying for the rest of the term.
How thick is one sheet of paper?
Estimate the thickness of a single sheet of printer paper without measuring one, then check the estimate by a completely different route.
Given
A ream of 500 sheets is about 5.0 cm thick
Find
The thickness of one sheet, to the nearest power of ten.
Solution
Measuring 500 sheets and dividing beats measuring one sheet, because the ruler's uncertainty is divided by 500 as well; this is the standard trick for anything too small to measure directly.
A second route with nothing in common with the first: printer paper is sold as 80 grams per square metre, and paper is a pressed fibre mat with a density of roughly 8 × 10² kg/m³. Thickness is then mass per area divided by density, that is 0.080 divided by 800, which gives 1 × 10⁻⁴ m. Two independent routes landing on the same power of ten is the strongest evidence an estimate can have.
One division and one conversion for the estimate; the check cost one more division and used no measurement at all.
This is the shape of a good estimate: a model in one sentence, arithmetic you can do in your head, and a second route that could have disagreed and did not.
Checkpoint
§01.6 — the most reused estimate in physics●○○○○
Thirty seconds on the number that turns up in every estimate involving time. You are not allowed a calculator and you are not allowed more than one digit in the answer.
Given
A year is about 365 days
A day is 24 hours, an hour is 3600 s
Find
(a) Estimate the number of seconds in a year to one significant figure.
Hint 1/4
This is a product of three numbers you already know; the work is arranging them, not remembering anything new.
Hint 2/4
Round each factor to one digit or to a power of ten, multiply, and read off the exponent.
Independent memory hook: the number is very close to π times ten to the seventh, which is a coincidence but a useful one for checking that the exponent has not slipped.
⚠ Reporting an estimate with three digits
the calculator produced them and deleting digits feels like losing information, when the information was never there
wrong$$V = 4.86\times 10^{5}\ \mathrm{L}$$
right$$V \sim 5\times 10^{5}\ \mathrm{L}$$
⚠ Guessing the answer instead of guessing the ingredients
the unknown is the thing being asked about, so it feels like the thing to guess, but a guess at the answer cannot be argued with while a guess at each ingredient can
wrong$$N_{\text{breaths in a lifetime}} \approx 10^{6}$$
1.7Models, theories, laws, and the range they come with
Says what a physical law actually claims, which is a prediction with a range attached, never a statement about everything.
Now that a measurement is an interval rather than a number, the first lecture's real question can finally be asked properly: what does it mean to test a physical law with something that is never exact?
NoteNote: how a law meets a measurement
Conditions
a model is a picture used for thinking, a theory is a quantitative account with a scope, and a law is a compact relation that has survived testing inside that scope
a law is never proved, only tested; each successful test narrows the region where it could still fail
outside its stated range a law is not wrong, it is simply not the right tool for the job
Testing a law means comparing two intervals, not two numbers. If the band the theory predicts and the band the experiment measures share even one value, the test has been passed. If the two bands are disjoint, something in one of them has not been accounted for, and finding out which is the whole job.
The inner box is where this course has been tested; the rest of the picture is why every law in it is quoted with a range rather than as a universal truth.
Looks like this, but is not
This is a scientific claim: in a vacuum, a heavy object and a light object fall at the same rate, a statement that has been tested to a few parts in a million and could have failed at any of those tests.
This looks like a scientific claim and is not testable as written: heavier objects fall faster. Faster than what, by how much, and under what conditions? A claim with no number and no range gives a measurement nothing to disagree with, so no experiment can ever count against it, and that is exactly what makes it useless rather than safe.
Two laboratory groups measure the same constant and get different numbers
Group A reports $(9.76\pm 0.05)\ \mathrm{m/s^{2}}$ and group B reports $(9.85\pm 0.03)\ \mathrm{m/s^{2}}$ for the same constant, using different apparatus. Do the two results agree, and what should each group do next?
Given
A: $(9.76\pm 0.05)\ \mathrm{m/s^{2}}$
B: $(9.85\pm 0.03)\ \mathrm{m/s^{2}}$
Reference value used in this course: $9.80\ \mathrm{m/s^{2}}$
Find
Whether the bands overlap, and what the answer implies about the two experiments.
Solution
Compare intervals rather than central values, because two central values are essentially never equal and comparing them alone would condemn every experiment ever done.
repeating shrinks the scatter and leaves an offset untouched, so B needs to hunt for a systematic effect rather than take more readings
Answer $$\boxed{\text{The bands are disjoint: the results disagree.}}$$
Check
The same conclusion by subtraction: the centres differ by 0.09 while the two half widths add to only 0.08. A difference bigger than the combined slop is precisely what disjoint bands mean, reached without drawing a single interval.
Two honest groups can disagree, and when they do, the interesting quantity is not the average of their answers but the difference between their setups.
Checkpoint
§01.7 — what an overlap does and does not establish●●○○○
Half a minute on the sentence that gets written at the end of a laboratory report. A group finds that their band overlaps the accepted value and writes a conclusion.
Given
Claim: two results whose bands overlap have been shown to be the same.
Find
(a) True or false, with one sentence of justification.
Hint 1/4
Ask what would happen to the overlap if a group simply worked less carefully and reported a much wider band.
Hint 2/4
Overlapping bands mean the two results are consistent: there is at least one value both experiments allow. Consistency is not equality.
Hint 3/4
Substitute an extreme case into the claim: a band of plus or minus 5 overlaps almost everything, and no one would say it has shown agreement with all of it.
Hint 4/4
So the claim is false. Overlap establishes consistency only, and a wide band makes consistency cheap.
Show solution
Attack the claim with an extreme case rather than with a definition; the extreme case is what makes the weakness visible.
Push the claim to its breaking point
$$9.8 \pm 5.0\ \Rightarrow\ [\,4.8,\ 14.8\,]$$
a careless experiment produces a wide band that overlaps nearly every plausible answer
the wide band has not learned anything, so overlap on its own cannot be evidence of agreement
Answer $$\boxed{\text{False}}$$
Check
The converse case checks the same point from the other side: disjoint bands really do establish a disagreement, so the two directions are genuinely different in strength.
⚠ Saying a law has been proved
the word is borrowed from mathematics, where a proof settles the matter for ever, and physics has no operation of that kind
wrong$$\text{many successful tests}\ \Rightarrow\ \text{proved for all time}$$
right$$\text{many successful tests}\ \Rightarrow\ \text{no failure found inside the tested range}$$
⚠ Comparing central values instead of bands
the central values are the numbers that got written in bold, and the uncertainties look like a formality at the end of the line
Reporting a computed number: the order the steps have to happen in
Any time a number leaves your page: a laboratory result, an exam answer, a line in a table.
Count before you compute
Look at the data and write down, in the margin, how many significant figures the weakest input has. That number is the target and it does not change later.
Carry a guard digit
Do the whole calculation with one digit more than the target and never round in the middle. Rounding twice moves the last digit more than the measurement does.
Apply the right rule at each operation
Multiplications and divisions are decided by significant figures; additions and subtractions by decimal places. A chain containing both is decided step by step, on the raw numbers.
Round once, at the end
Cut to the target you wrote in step one. If the target is ambiguous in ordinary notation, as 190 or 2500 are, switch to scientific notation instead of hoping the reader guesses.
Attach the unit, then the uncertainty
A bare number is not an answer. If an uncertainty was given or can be worked out, quote it to one significant figure and place the value's last digit in the same column.
Check the size out loud
Say what the answer is comparable to. A density near that of water, a speed near a walking pace, a time near a heartbeat. A number with no comparison is a number you cannot defend.
Where it goes wrong
Rounding the intermediate result and then rounding again.
Taking the digit count from the most precise input instead of the least precise one.
Writing 190 when two significant figures were meant, so the reader reads three.
Letting an exact number, such as a count of sides or a defined conversion factor, cut the digits down.
Building a conversion chain that cannot go wrong
Whenever the units you were given are not the units the answer is wanted in, which in this course is most of the time.
Write the target units first
Put the units you want on the right hand side of the page before writing any number. Everything after this is arranging cancellations to reach them.
Write each factor as a fraction that equals one
Never write a bare multiplier. Write $\frac{10^{3}\ \mathrm{m}}{1\ \mathrm{km}}$, where the top and the bottom are the same length, so that multiplying by it is provably harmless.
Orient each fraction to cancel
Put the unit you are removing on the opposite side of the bar from where it currently sits. If the unwanted unit is downstairs in the quantity, it goes upstairs in the fraction.
Raise the whole fraction to the power on the unit
For $\mathrm{cm^{2}}$ or $\mathrm{cm^{3}}$, square or cube the entire fraction, not just the unit. This single step is where most of the lost powers of ten happen.
Cancel on the page and read what is left
Strike out the units that appear above and below. What survives must be exactly the target you wrote in step one; if it is not, a fraction is upside down and the number is wrong.
Then, and only then, multiply
Do the arithmetic last, with a guard digit, and round once. The units have already told you whether the structure is right.
Where it goes wrong
Remembering the factor as a number so that there is nothing to raise to a power.
Writing the fraction upside down, which shows up as a unit that refuses to cancel.
Letting an exact defined factor limit the significant figures.
Converting the prefix but forgetting that it sits inside a bracket that is squared or cubed.
Making an order of magnitude estimate you can defend
When no data are given, when you want to know whether a calculation is worth starting, or when you need to check an answer you already have.
State the model in one sentence
Say what you are replacing the real thing with, out loud: the pool is a box, the person is a cylinder of water, the city is a square of houses. The model is the part that can be argued with, so it goes on the page.
Break the unknown into factors you can bracket
Each factor should be something you can name an upper and a lower bound for without looking anything up. If you cannot bracket it, split it again.
Round each factor to one digit
One significant figure, or a bare power of ten. Precision in an input you guessed is a waste of effort and gives the answer a false air of authority.
Multiply the powers separately from the digits
Add the exponents in your head, multiply the leading digits, then normalise. This keeps the whole calculation off paper, which is the point.
Report a power of ten and name your weakest link
Write the result with $\sim$ and one digit at most, then say which input the answer is most sensitive to. That last sentence is what turns a guess into an estimate.
Where it goes wrong
Guessing the answer directly instead of guessing the ingredients.
Keeping three digits at the end, which claims a precision the method never had.
Leaving the model unstated, so the reader cannot tell whether the answer is wrong or the picture is.
Bracketing nothing, so a factor that could be off by a hundred passes unnoticed.
Five caliper readings that agree with each other and not with the truth
A rod whose true length is $12.55\ \mathrm{mm}$ is measured five times with a digital caliper: $12.31$, $12.30$, $12.32$, $12.31$, $12.30\ \mathrm{mm}$. Report the result and judge it.
The reported value with its scatter, and whether it is trustworthy.
Solution
The scatter of the readings is the only uncertainty the data themselves can reveal, so it is quoted first and then compared with the truth, which the data could never have supplied.
half the full spread is a fair half width, and it is tiny, so the instrument is repeatable
Compare with the truth
$$[\,12.30,\ 12.32\,] \not\ni 12.55$$
the true value is nowhere near the band, so the small scatter was measuring the instrument and not the rod
Answer $$\boxed{(12.31\pm 0.01)\ \mathrm{mm},\ \text{precise and wrong by }0.24\ \mathrm{mm}}$$
Check
Repeating a sixth time would land inside the same narrow band, which is the signature of an offset rather than of noise.
Five ruler readings that scatter widely and straddle the truth
The same rod is measured five times with a plastic ruler read to the nearest half millimetre: $12.5$, $13.0$, $12.5$, $12.5$, $12.0\ \mathrm{mm}$. Report the result and judge it.
fifty times the caliper's scatter, which is what reading by eye between divisions costs
Compare with the truth
$$[\,12.0,\ 13.0\,] \ni 12.55$$
the true value sits comfortably inside the band, so this cruder instrument is telling the truth, just not very sharply
Answer $$\boxed{(12.5\pm 0.5)\ \mathrm{mm},\ \text{imprecise and correct}}$$
Check
The band is wide enough to contain the caliper's answer too, which is the honest statement: this ruler cannot distinguish the two claims.
The caliper gives a band fifty times narrower than the ruler and misses the true value completely, while the ruler gives a wide band that contains it.
How to tell them apart
Scatter is what repeating a measurement shows you, and repeating is the only cure for it. An offset is invisible to repetition and is cured only by calibrating against something you already trust. If the readings agree with each other and disagree with a known standard, the problem is an offset and taking more readings will make it worse, not better, by shrinking a band that is centred in the wrong place.
Scaffolding comes off
The common skeleton
Write the units the answer has to be in, before touching the numbers.
Write the given quantity with its units, and count its significant figures.
Write each conversion as a fraction equal to one, raised to the power that sits on its unit.
Orient every fraction so the unwanted unit appears once above and once below, and cancel on the page.
Check that only the target units survive, then do the arithmetic with one guard digit.
Round once to the counted significant figures, and ask whether the number moved in the direction you expected.
1 · fully worked
A density of 13.6 g/cm³ converted to kilograms per cubic metre
Mercury has a density of $13.6\ \mathrm{g/cm^{3}}$. Express it in SI units.
Given
$\rho = 13.6\ \mathrm{g/cm^{3}}$
$1\ \mathrm{g}=10^{-3}\ \mathrm{kg}$
$1\ \mathrm{cm}=10^{-2}\ \mathrm{m}$
Find
The density in $\mathrm{kg/m^{3}}$.
Solution
Both fractions are written before anything is multiplied, because the cube on the second one is the step that decides whether the answer is out by a factor of ten thousand.
Independent anchor rather than a repeat of the algebra: water is 1.00 g/cm³ and 1.00 × 10³ kg/m³, so the factor between the two ways of writing any density is exactly one thousand. Mercury is 13.6 times denser than water in both systems, which is why a steel ball floats on it.
Two fractions, one cube, one rounding. Nothing was multiplied until every unit had been cancelled on paper.
This is the whole skeleton in one example. The next three rungs use it on the same kind of problem with less and less of the work done for you.
2 · you write the reasoning
Easier than the one above, because only one power of ten is in play and no unit is raised to a power. Convert a speed of $55.0\ \mathrm{km/h}$ into metres per second, and write, in the empty column, why each line is allowed.
A kilometre and a thousand metres are the same length, so the fraction is the number one and multiplying by it cannot change the speed. Kilometres appear once above and once below, so they cancel and metres per hour is what is left.
The hour has to be removed from downstairs, so it is written upstairs in the fraction. The three thousand six hundred is a defined value, not a measurement, so it cannot reduce the three significant figures the data carry.
The arithmetic is done last and rounded once, to the three significant figures that 55.0 supplied. As a check on direction, metres per second should be a much smaller number than kilometres per hour, and it is.
3 · find the buried error
Harder than the rung above, and the work has been done for you, badly. A laboratory pump delivers $2.5\ \mathrm{L/s}$ and the report needs cubic metres per hour. Exactly two of the four steps below are wrong. Find both.
Step 4. The 3600 is exact, so keep every digit: $6.944\times 10^{-7}\ \mathrm{m^{3}/h}$.
the two buried errors (2)
⚠ step 2
The fraction is upside down. The seconds to be removed already sit downstairs in $\mathrm{m^{3}/s}$, so the factor must be $\dfrac{3600\ \mathrm{s}}{1\ \mathrm{h}}$; as written, seconds appear twice below and the units come out as $\mathrm{m^{3}\,h/s^{2}}$, which is not a flow rate at all.
The step is remembered as the instruction divide by 3600 rather than being built to cancel the unit that is actually present, and the memory is reinforced by the fact that going the other way really does need a division.
An exact conversion factor does not license extra digits. The digit count is set by the weakest measured input, and the only measured number here is 2.5, which carries two significant figures.
The rule that exact numbers do not limit the answer gets over generalised into the belief that they improve it, so a defined 3600 is taken as permission to keep everything the calculator shows.
right
Two significant figures survive, so the answer is $9.0\ \mathrm{m^{3}/h}$, not $9.000\ \mathrm{m^{3}/h}$.
4 · the bare problem
§01.4 — the same skeleton with nothing filled in●●●○○
No scaffolding this time. A block of ice is weighed and measured in a cold room, and the handbook value it is being compared with is in SI units, so the measurement has to be converted before the comparison means anything.
(a) Express the density in $\mathrm{kg/m^{3}}$ with the right number of significant figures.
(b) Say in one sentence what the number tells you about ice placed in water.
Hint 1/4
The target units are kilograms per cubic metre. Everything that follows is arranging two fractions so that only those units are left standing.
Hint 2/4
One fraction trades grams for kilograms; the other trades cubic centimetres for cubic metres, and it must be cubed as a whole.
Hint 3/4
Substitute the value given above, $\rho = 0.917\ \mathrm{g/cm^{3}}$, with $1\ \mathrm{g}=10^{-3}\ \mathrm{kg}$ and $1\ \mathrm{cm^{3}}=10^{-6}\ \mathrm{m^{3}}$: the two powers combine into a single factor of $10^{3}$.
Hint 4/4
So $\rho = 9.17\times 10^{2}\ \mathrm{kg/m^{3}}$, which is below the $1.00\times 10^{3}\ \mathrm{kg/m^{3}}$ of water.
Show solution
The two prefix trades are combined into one factor of a thousand only after both have been written out, because writing the single factor from memory is exactly how it ends up as a hundred.
The one thousand rule as an independent check: any density in g/cm³ is the same number times 10³ in kg/m³, so 0.917 must become 917 without any arithmetic at all.
About nine tenths of an ice cube sits below the waterline, which is the same ratio as the two densities, and is the kind of everyday fact a converted number should be checked against.
Full exam-style question
A measured steel rod: volume, density, uncertainty and a comparison with the handbookexam format
This is what a full question on this material looks like on a paper: four parts, each worth a few marks, and every part failing if the units or the digits are wrong.
A solid cylindrical steel rod is measured with a caliper and a balance: diameter $d = (12.4\pm 0.1)\ \mathrm{mm}$, length $\ell = (85.0\pm 0.5)\ \mathrm{mm}$, mass $m = (81.5\pm 0.1)\ \mathrm{g}$.
(a) Find the volume in $\mathrm{m^{3}}$. (b) Find the density in $\mathrm{kg/m^{3}}$. (c) Find the percent uncertainty in the density and quote the density as a band. (d) A handbook gives $7.85\times 10^{3}\ \mathrm{kg/m^{3}}$ for this steel. Does the measurement agree?
Given
$d = (12.4\pm 0.1)\ \mathrm{mm}$
$\ell = (85.0\pm 0.5)\ \mathrm{mm}$
$m = (81.5\pm 0.1)\ \mathrm{g}$
$V = \pi r^{2}\ell$ for a cylinder, and $\rho = m/V$
Handbook value $7.85\times 10^{3}\ \mathrm{kg/m^{3}}$
Find
The volume, the density, the percent uncertainty in the density, and a verdict on the comparison.
Solution
Everything is converted to SI before any formula is used. Converting at the end would mean carrying a mixed unit through a squared quantity, which is where the factor of ten thousand goes missing in the mm to m step.
The uncertainty by the long route, which shares no algebra with the percentage rule: the largest density the data allow uses the biggest mass with the smallest volume, giving 8.13 × 10³, and the smallest uses the smallest mass with the largest volume, giving 7.76 × 10³. Half the difference is 1.85 × 10², the same slop as the percentages gave. As a separate plausibility check, the answer is about eight times the density of water, which is where every common steel sits.
Three conversions, one squared quantity, three percentages and one comparison of intervals. The single most expensive mistake available here is the millimetre that never became a metre before being squared.
Notice which measurement dominated: the caliper reading on the diameter, because it enters squared and its percentage is doubled. If this experiment were to be improved, a better caliper would buy more than a better balance, and knowing that took one line of the uncertainty calculation.
Practice
A · concept 4 questions
1§01.2 — do extra digits buy anything●○○○○
A classmate defends a laboratory report by saying that copying more digits from the calculator can only help, because throwing digits away is throwing information away.
Given
Claim: writing more digits from the calculator makes a reported result more accurate.
Find
(a) True or false, with one sentence of justification.
Hint 1/4
Ask what the extra digits are a measurement of. If nothing in the apparatus produced them, they cannot be reporting anything about it.
Hint 2/4
An answer may not claim a precision that no input had; the digit count comes from the weakest input, not from the display.
Hint 3/4
Substitute the claim into the case from earlier in this section: 21.6 cm times 27.9 cm gives 602.64, and the ruler could not resolve a hundredth of a square centimetre.
Hint 4/4
So the claim is false: the extra digits are generated by the multiplication and are not information about the sheet of paper.
Show solution
Argue from a concrete case rather than from the rule, because the rule is exactly what is in dispute.
Track one digit back to its source
$$21.6\times 27.9 = 602.64$$
the two hundredths in the answer came from multiplying two tenths together, not from any reading
$$\delta A \approx 5\ \mathrm{cm^{2}}\ \Rightarrow\ A = 603\ \mathrm{cm^{2}}$$
the uncertainty already covers the units place, so the two digits after it describe nothing
Answer $$\boxed{\text{False}}$$
Check
Test the claim in reverse: if extra digits helped, measuring with a worse ruler and then multiplying by 1.000000 would improve the result, which is plainly absurd.
2§01.1 — what repeating a measurement can and cannot fix●●○○○
A caliper was zeroed while a speck of dust sat between its jaws, so every reading it gives is 0.24 mm too small. The group notices that their five readings agree with each other to within 0.01 mm and decide to take twenty more.
Given
Every reading is 0.24 mm low
The five readings agree to within 0.01 mm
Find
(a) What will the twenty extra readings achieve?
Hint 1/4
Split the error into the part that changes between readings and the part that is the same every time, and ask which one averaging can touch.
Hint 2/4
Averaging reduces random scatter roughly as the number of readings grows; a systematic offset is identical in every reading and survives untouched.
Hint 3/4
Substitute the numbers from the question: the scatter is 0.01 mm and the offset is 0.24 mm, so the thing averaging can attack is already twenty times smaller than the thing it cannot.
Hint 4/4
So the extra readings will tighten a band that is centred in the wrong place, and the offset will still be there.
Show solution
Separating the two kinds of error before touching any statistics is what makes the answer obvious; treating the total error as one number makes it impossible.
Consistency check with the earlier contrast pair: the precise caliper missed the true value and the crude ruler contained it, which is the same conclusion arrived at from data rather than from algebra.
3§01.7 — what a disagreement between two groups means●●○○○
Two groups measure the same quantity with different apparatus and their uncertainty bands do not overlap. One student concludes that the theory is in trouble.
Given
Claim: if two measured bands are disjoint, at least one experiment has an effect it has not accounted for.
Find
(a) True or false, with one sentence of justification.
Hint 1/4
Ask what a band claims. If both bands are honest, what should they both contain?
Hint 2/4
An uncertainty band claims to contain the true value. Two bands that both contain it must overlap, since they both contain at least that one point.
Hint 3/4
Substitute the earlier case: A gives the interval from 9.71 to 9.81 and B gives 9.82 to 9.88, and no single value lies in both.
Hint 4/4
So the claim is true: at least one of the two bands is either too narrow or centred wrongly, and something unaccounted for is causing it.
Show solution
Argue from what a band claims rather than from what the theory predicts, because the two experiments contradict each other whatever any theory says.
a shared point is exactly what an overlap is, so honest bands cannot be disjoint
$$A\cap B = \varnothing\ \Rightarrow\ \text{one claim fails}$$
the contrapositive: something in at least one experiment is not what it was reported to be
Answer $$\boxed{\text{True}}$$
Check
A second route: if only random scatter were present, extra repeats would widen nothing and eventually the two means would have to converge; they do not, which is the signature of a systematic effect.
4§01.5 — what a passed dimension check establishes●●●○○
In an exam a student checks a half remembered formula, finds that both sides have the same dimensions, and writes underneath that the formula is therefore correct.
Given
A candidate formula whose two sides have identical dimensions
Find
(a) What has actually been established?
Hint 1/4
Ask what the check would have said about a formula that is right, and about a wrong formula built from the same symbols.
Hint 2/4
Dimensional consistency is necessary and not sufficient: it can convict, never acquit.
Hint 3/4
Substitute the cylinder case from this section: $\pi r^{2}h$ and $\pi r h^{2}$ both carry three powers of length, and only one of them is a volume.
Hint 4/4
So all that has been established is that dimensions have not ruled the formula out.
Show solution
One counterexample settles the question faster than any general argument, and this section already supplied one.
Produce a survivor that is wrong
$$[\,\pi r^{2}h\,] = [\,\pi r h^{2}\,] = [L]^{3}$$
two different expressions pass the same test, so passing cannot single one out
$$V_{\text{cyl}} = \pi r^{2}h\ \text{only}$$
the choice between survivors was made by geometry, which is information the dimensions never carried
Check the logic in the other direction: a failed test really does settle the matter, and that asymmetry is exactly what necessary but not sufficient means.
B · computation 7 questions
1§01.2 — counting significant figures in four written forms●●○○○
Four numbers copied from four different instrument displays. Each one has its zeros doing a different job, and that is the whole question.
Given
$0.0304$
$4.500\times 10^{2}$
$6000.0$
$0.0900$
Find
(a) Give the number of significant figures in each.
Hint 1/4
Sort the zeros in each number into two classes: those holding the decimal point in place and those reporting a resolved digit.
Hint 2/4
Leading zeros are never significant, zeros between non zero digits always are, and trailing zeros count when there is a decimal point in the number.
Hint 3/4
Substitute the four values above one at a time; rewriting each in scientific notation, as $3.04\times 10^{-2}$ and $9.00\times 10^{-2}$, settles the two awkward cases.
Hint 4/4
The counts are 3, 4, 5 and 3.
Show solution
Scientific notation is used only where the plain form is ambiguous, which keeps the answer short and shows exactly which cases needed it.
The interior and leading zeros
$$0.0304 = 3.04\times 10^{-2}\ \Rightarrow\ 3$$
the two leading zeros place the point, and the interior zero sits between resolved digits so it must be resolved too
$$4.500\times 10^{2}\ \Rightarrow\ 4$$
already in scientific notation, so every digit written in front of the power was written deliberately
The trailing zeros
$$6000.0\ \Rightarrow\ 5$$
the decimal point followed by a zero reports the tenths, which forces the three zeros before it to be significant as well
$$0.0900 = 9.00\times 10^{-2}\ \Rightarrow\ 3$$
the two trailing zeros are after the point and after the nine, so they are not placeholders
Answer $$\boxed{3,\quad 4,\quad 5,\quad 3}$$
Check
Cross check on the third: written without the point, 6000 would be ambiguous, so the point and the extra zero are doing work that only measured digits can do.
2§01.3 — scientific notation and the nearest prefix●●○○○
An oscilloscope reports a pulse length as 0.000 072 5 s and the report wants both the scientific form and a prefixed unit that a reader can picture.
Given
$t = 0.000\,072\,5\ \mathrm{s}$
Find
(a) Write the time in scientific notation.
(b) Write it with the most suitable SI prefix.
Hint 1/4
Two separate jobs: first put the number into the form one digit, point, the rest, times a power of ten; only then look for a prefix.
Hint 2/4
Prefixes step in thousands, so the exponent has to be pushed to a multiple of three before a prefix will fit.
Hint 3/4
Substitute the reading given above: moving the point five places gives $7.25\times 10^{-5}\ \mathrm{s}$, and shifting to a multiple of three gives $72.5\times 10^{-6}\ \mathrm{s}$.
Hint 4/4
So $t = 7.25\times 10^{-5}\ \mathrm{s} = 72.5\ \mathrm{\mu s}$.
Show solution
Normalise first and choose the prefix second, because the exponent is what names the prefix; guessing the prefix first is how a factor of a thousand disappears.
Multiply back: 72.5 times ten to the minus six is 0.0000725, which is the number the instrument printed.
3§01.4 — one speed in three unit systems●●○○○
A motorway sign abroad gives a limit of 95 km/h. Your calculation needs SI units, and a friend with an old car wants it in miles per hour.
Given
$v = 95\ \mathrm{km/h}$
$1\ \mathrm{mi} = 1.609\ \mathrm{km}$
$1\ \mathrm{h} = 3600\ \mathrm{s}$
Find
(a) Express $v$ in $\mathrm{m/s}$.
(b) Express $v$ in $\mathrm{mi/h}$.
Hint 1/4
Two independent chains starting from the same quantity. Write the target units of each before you start it.
Hint 2/4
For the first, replace kilometres by metres and hours by seconds. For the second, use $1\ \mathrm{mi}=1.609\ \mathrm{km}$ oriented so kilometres cancel.
Hint 3/4
Substitute the data above: $95\times\frac{10^{3}}{3600}$ for part (a), and $95\div 1.609$ for part (b), since one mile is more than one kilometre.
Hint 4/4
So $v = 26\ \mathrm{m/s}$ and $v = 59\ \mathrm{mi/h}$.
Show solution
The mile conversion is done from the original figure rather than from the metres per second answer, so that a rounding in part (a) cannot leak into part (b).
Direction check on both: metres per second must be far smaller than kilometres per hour, and miles per hour must be smaller than kilometres per hour because the mile is the longer unit. Both moved the right way.
4§01.1 — the area of a measured rectangle with its band●●●○○
A rectangular plate is measured with a ruler on both sides and the report needs an area that a marker can check, which means an area with an uncertainty on it.
Given
$a = (2.34\pm 0.02)\ \mathrm{m}$
$b = (1.06\pm 0.02)\ \mathrm{m}$
Find
(a) Find the area.
(b) Find its absolute uncertainty and quote the area as a band.
Hint 1/4
The area itself is one multiplication. The real question is how the two slops combine, and they do not combine by being added as they stand.
Hint 2/4
For a product the fractional uncertainties add: $\dfrac{\delta A}{A}=\dfrac{\delta a}{a}+\dfrac{\delta b}{b}$.
Hint 3/4
Substitute the data given above: $\dfrac{0.02}{2.34}=0.855\%$ and $\dfrac{0.02}{1.06}=1.887\%$, so the total is $2.74\%$ of $2.4804\ \mathrm{m^{2}}$.
Hint 4/4
So $A = (2.48\pm 0.07)\ \mathrm{m^{2}}$.
Show solution
Percentages first, because the same 0.02 m is more than twice as damaging on the shorter side, and that fact is invisible while the slops are still in metres.
The area itself
$$A = 2.34\times 1.06 = 2.4804\ \mathrm{m^{2}}$$
the raw product with guard digits, to be rounded only once the uncertainty is known
The extremes, which share no algebra with the percentage rule: the largest area is $2.36\times 1.08 = 2.5488$ and the smallest is $2.32\times 1.04 = 2.4128$. Half the difference is $0.068\ \mathrm{m^{2}}$, the same as before.
5§01.4 — a cubic metre in the units people can picture●●○○○
A tank is specified as holding one cubic metre. Nobody can picture a cubic metre, so the specification has to be translated before anyone can check whether the tank will fit through the door.
Given
$V = 1.00\ \mathrm{m^{3}}$
$1\ \mathrm{m}=10^{2}\ \mathrm{cm}$
$1\ \mathrm{L}=10^{3}\ \mathrm{cm^{3}}$
Find
(a) Express $V$ in $\mathrm{cm^{3}}$.
(b) Express it in litres.
Hint 1/4
Ask what power sits on the unit before writing any factor down, because that power has to go on the whole fraction.
Picture it: a cube of side one metre holds ten rows of ten columns of ten litre cubes, which is a thousand litres, and a cubic metre of water is therefore a tonne.
6§01.5 — units of a constant in front of a square root●●●○○
A group hangs masses on a spring and finds that their extension data are fitted well by an expression with a square root in it. The constant has to be reported with its units or the fit means nothing to anyone else.
Given
$x = C\sqrt{m}$
$x$ measured in cm, $m$ measured in g
Find
(a) What are the units of $C$?
Hint 1/4
Do not think about springs. Ask only what units the constant must carry so that the right hand side comes out as a length.
Hint 2/4
Rearrange for the unknown constant and read the units off: $C = x/\sqrt{m}$.
Hint 3/4
Substitute the units given above, $x$ in cm and $m$ in g: $C$ has units of cm divided by the square root of a gram.
Hint 4/4
So $C$ is in $\mathrm{cm\,g^{-1/2}}$.
Show solution
Rearranging for the constant first turns a puzzle about square roots into a division of two known units, which is a step no one gets wrong.
Isolate and count
$$C = \frac{x}{\sqrt{m}}$$
the constant is the only unknown, so the units follow from the two measured quantities alone
Put it back: centimetres per root gram, times root grams, gives centimetres, which is what the left hand side is. A fractional power in a unit is unusual to look at and perfectly legal.
7§01.6 — an estimate with no data supplied●●●○○
Nothing is given in this question except what you already know about people. That is the point: the whole answer has to be assembled from things you can bracket without looking anything up.
Given
A resting pulse is roughly one beat a second
A lifetime is of order 80 years
Find
(a) Estimate the number of heartbeats in a human lifetime, to the nearest power of ten.
Hint 1/4
Break it into two factors you can each bracket on their own: beats per unit time, and total time lived.
Hint 2/4
State the model in one sentence, round every factor to one digit, and multiply the powers of ten separately from the digits.
Hint 3/4
Substitute the anchors given above, roughly $70$ beats per minute and $5\times 10^{5}$ minutes in a year, over $80$ years.
Hint 4/4
That gives about $3\times 10^{9}$ beats.
Show solution
Working in minutes rather than seconds keeps every factor to one digit and a power of ten, which is what lets the whole thing be done without paper.
State the model, then bracket each factor
$$f \approx 70\ \mathrm{min^{-1}}$$
resting pulse sits between 60 and 80 for most people, so one digit is all this factor deserves
digits times digits, exponents added, then normalised, which is the whole arithmetic of an estimate
$$N \sim 3\times 10^{9}$$
one digit only, because the pulse rate was never known to better than about twenty percent
Answer $$\boxed{N \sim 3\times 10^{9}}$$
Check
A second route: a beat a second gives one times the seconds in a lifetime, that is 80 times 3 × 10⁷, which is 2.4 × 10⁹. The two routes agree to within the factor of ten the method claims, and the weakest link in both is the assumed lifetime.
C · exam level 4 questions
1§01.4 — density of a machined rod, exam style●●●○○
This is the shape a full marks question takes: a shape formula, a unit conversion and a digit count, all inside one line of arithmetic. A solid ceramic cylinder is machined to a diameter of 2.40 cm and a height of 5.00 cm, and the balance reads 68.0 g.
Given
$d = 2.40\ \mathrm{cm}$
$h = 5.00\ \mathrm{cm}$
$m = 68.0\ \mathrm{g}$
$V=\pi r^{2}h$
Find
(a) Which value is the density in SI units?
Hint 1/4
Three separate decisions hide in this question: which length goes into the formula, what power it carries, and which unit system the answer is wanted in.
Hint 2/4
Use $V=\pi r^{2}h$ with $r=d/2$, then $\rho = m/V$, then the fact that any density in $\mathrm{g/cm^{3}}$ is a thousand times as many $\mathrm{kg/m^{3}}$.
Hint 3/4
Substitute the data above, $d=2.40\ \mathrm{cm}$ so $r=1.20\ \mathrm{cm}$, $h=5.00\ \mathrm{cm}$ and $m=68.0\ \mathrm{g}$: $V=\pi(1.20)^{2}(5.00)=22.6\ \mathrm{cm^{3}}$.
The whole calculation is done in centimetres and grams and converted once at the end, because converting three quantities at the start is three chances to lose a power of ten instead of one.
Halve the diameter and find the volume
$$r = \frac{2.40}{2} = 1.20\ \mathrm{cm}$$
the 2 is exact, so three significant figures survive
Plausibility rather than repetition: this is three times the density of water, which is where ceramics and glasses sit, and well below the eight thousand of steel. A value of 3.01 kg/m³ would be lighter than air and a value of 7.5 × 10² would float.
The two wrong answers on offer both come from the same slip, using a diameter where a radius belongs, and one of them is out by exactly four. Whenever an answer is out by four in a problem with a circle in it, look at that step first.
2§01.1 — how an uncertainty grows when a length is cubed●●●○○
A cube of metal is measured on one side only, since the sides are known to be equal, and the side comes out with a percent uncertainty of 1.0 percent. The report needs the uncertainty in the volume.
Given
A cube of side $s$, with $\dfrac{\delta s}{s} = 1.0\%$
$V = s^{3}$
Find
(a) What is the percent uncertainty in the volume?
Hint 1/4
Write the volume as a product of measured quantities rather than as a power, and ask how many of the factors carry the same slop.
Hint 2/4
For a product the percent uncertainties add, so $s\times s\times s$ contributes its percentage three times.
Hint 3/4
Substitute the value given above, $\delta s/s = 1.0\%$, into $\delta V/V = 3\,\delta s/s$.
Hint 4/4
So the volume carries $3.0\%$.
Show solution
Writing the power out as a product makes the rule for products do all the work, so no new rule has to be remembered for powers.
Turn the power into a product
$$V = s\cdot s\cdot s$$
three factors, each carrying the same measurement and therefore the same percentage
the rule for products applied three times, which is exactly what a power is
Answer $$\boxed{\frac{\delta V}{V} = 3.0\%}$$
Check
Numerical check with a concrete cube: a side of 10.0 cm with 1.0 percent gives a side between 9.9 and 10.1, so a volume between 970 and 1030 cubic centimetres. That spread is 60 in 1000, that is six percent full width, which is three percent either side.
The same reasoning explains why the caliper dominated the steel rod question: a diameter that enters squared contributes twice its own percentage.
3§01.4 — rainfall on a roof, in cubic metres per hour●●●○○
A building services calculation, and a typical way conversion appears on an exam paper: a small unit multiplied by a large one. Rain falls at 4.0 mm per hour onto a flat roof of area 2.0 times ten to the fourth square metres.
Given
Rainfall rate $4.0\ \mathrm{mm/h}$
Roof area $A = 2.0\times 10^{4}\ \mathrm{m^{2}}$
Find
(a) At what rate does water arrive on the roof, in cubic metres per hour?
Hint 1/4
A rainfall rate is a depth per unit time, and a depth times an area is a volume, so the shape of the answer is fixed before any number is used.
Hint 2/4
Convert the depth into the same length unit the area is built from, then multiply; mixing millimetres with square metres is what the question is testing.
Hint 3/4
Substitute the data above: $4.0\ \mathrm{mm} = 4.0\times 10^{-3}\ \mathrm{m}$, and multiply by $2.0\times 10^{4}\ \mathrm{m^{2}}$.
Hint 4/4
So the rate is $80\ \mathrm{m^{3}/h}$.
Show solution
Convert the small quantity to match the large one rather than the other way round; turning the roof area into square millimetres would work too and would carry an exponent of ten that nobody wants to handle.
digits multiply and exponents add, and the units combine to cubic metres exactly as they should
Answer $$\boxed{\dot V = 80\ \mathrm{m^{3}/h}}$$
Check
Order of magnitude sanity: 80 cubic metres is 80 tonnes of water an hour off a roof the size of two football pitches, which is a believable load for a drainage system and would be absurd at 80 000 tonnes.
Whenever a small unit multiplies a big one, convert the small one. It is the one whose power of ten you are most likely to get wrong, and doing it first puts the risky step where you are still paying attention.
4§01.5 — building a density out of the three base dimensions●●○○○
A dimensional question in the form the paper likes: nothing physical to picture, only bookkeeping. You are given quantities whose dimensions are a mass, a length and a time, and nothing else.
Given
$M$ has dimension $[M]$, $L$ has $[L]$, $T$ has $[T]$
A density has dimension $[M][L]^{-3}$
Find
(a) Which combination has the dimensions of a density?
Hint 1/4
Write down the dimensions the answer must have before looking at the options, so that the options are being checked rather than compared.
Hint 2/4
A density is a mass divided by a volume, so its dimensions are $[M][L]^{-3}$, with no time in it at all.
Hint 3/4
Substitute the target given above and read each option: only one has a single power of mass on top and exactly three powers of length underneath.
Hint 4/4
So the answer is $M/L^{3}$.
Show solution
Fix the target dimensions first and test each candidate against it; comparing the candidates with each other instead invites the reciprocal to look as good as the answer.
Test each candidate against the target
$$[\,M/L^{3}\,] = [M][L]^{-3}$$
matches the target exactly, with no time dimension left over
each fails on a different count, which is why each is a plausible thing to write and none is filler
Answer $$\boxed{M/L^{3}}$$
Check
Substitute units instead of dimensions: kilograms over metres cubed is the SI unit of density, and none of the other three reads as a unit anyone quotes densities in.
D · interleaved 3 questions
1§01.4 — a small cube weighed and measured●●●○○
Mixed practice, and the type is not announced. A small cube of metal is measured with a caliper and weighed, and the result has to be handed in as a number a handbook could be compared with.
Given
Side $s = 1.25\ \mathrm{cm}$
Mass $m = 16.6\ \mathrm{g}$
Find
(a) Find the density in $\mathrm{kg/m^{3}}$, with the right number of significant figures.
(b) Say in one sentence how the result compares with water.
Hint 1/4
Decide first what the answer is: a mass divided by a volume, in units that are not the units of either measurement.
Hint 2/4
Use $V=s^{3}$ and $\rho = m/V$, then the factor of a thousand between $\mathrm{g/cm^{3}}$ and $\mathrm{kg/m^{3}}$; the digit count is set by the weaker of the two measurements.
Hint 3/4
Substitute the data above, $s=1.25\ \mathrm{cm}$ and $m=16.6\ \mathrm{g}$: $V = 1.953\ \mathrm{cm^{3}}$ and $\rho = 16.6/1.953$.
Hint 4/4
So $\rho = 8.50\ \mathrm{g/cm^{3}} = 8.50\times 10^{3}\ \mathrm{kg/m^{3}}$.
Show solution
Work in centimetres and grams and convert once at the end, because cubing a length that has already been converted invites the wrong power of ten at the riskiest moment.
Independent size check: the cube is about two cubic centimetres and weighs about seventeen grams, so roughly eight and a half grams per cubic centimetre can be read straight off without doing the division, and that matches.
Comparing a density with water costs nothing and catches an error of a factor of a thousand instantly, which is the most common size of error in this whole section.
2§01.4 — rain on a field, from millimetres to litres●●●●○
Mixed practice again, with a squared conversion buried in the middle of it. A weather station records 12 mm of rain overnight on a field of area 0.25 square kilometres.
Given
Rain depth $12\ \mathrm{mm}$
Field area $0.25\ \mathrm{km^{2}}$
Find
(a) How many litres of water fell on the field?
Hint 1/4
The answer is a depth multiplied by an area, so the only real work is getting both into units that can be multiplied together.
Hint 2/4
Convert the depth to metres and the area to square metres, remembering that the square on the kilometre squares the conversion factor as well.
Hint 3/4
Substitute the data above: $12\ \mathrm{mm}=1.2\times 10^{-2}\ \mathrm{m}$ and $0.25\ \mathrm{km^{2}} = 0.25\times 10^{6}\ \mathrm{m^{2}} = 2.5\times 10^{5}\ \mathrm{m^{2}}$.
Hint 4/4
So $V = 3.0\times 10^{3}\ \mathrm{m^{3}} = 3.0\times 10^{6}\ \mathrm{L}$.
Show solution
Both quantities are pushed into SI base units before anything is multiplied, so that the squared conversion happens on its own line where it can be checked.
Estimate it independently: a quarter of a square kilometre is 250 000 square metres, and a centimetre of rain on each square metre is about ten litres, so a few million litres is the right scale. The mass is three thousand tonnes, which is why a night of rain changes a river.
The squared kilometre is the step worth slowing down for. A factor of a million looks unreasonable until you notice that a square kilometre really does contain a million square metres.
3§01.2 — a perimeter reported with one digit too many●●●●○
Mixed practice with no announced type. A rectangle is measured as 12.25 cm on one side, with a caliper, and 3.1 cm on the other, with a ruler, and the perimeter is reported as 30.70 cm.
Given
$a = 12.25\ \mathrm{cm}$
$b = 3.1\ \mathrm{cm}$
Reported perimeter $30.70\ \mathrm{cm}$
Find
(a) What should the reported perimeter be?
Hint 1/4
Two rules are competing here and only one of them applies. Ask which operation is actually being carried out on the two measurements.
Hint 2/4
A perimeter is a sum, so it is decided by decimal places, and the 2 that multiplies the sum is a count and therefore exact.
Hint 3/4
Substitute the data above: $12.25 + 3.1 = 15.35$, where the ruler reading stops at the tenths, and $2\times 15.35 = 30.70$.
Hint 4/4
Round once, at the end, to the tenths that the ruler supports, giving 30.7 cm.
Show solution
Carry the full sum through the doubling and round at the very end, since rounding the intermediate value would move the answer by a whole tenth, and the two candidate answers on offer differ by exactly that.
Add, keeping the guard digit
$$a + b = 12.25 + 3.1 = 15.35\ \mathrm{cm}$$
the hundredths column exists in one term and not in the other, so it is not trustworthy, but it is kept until the end
Double, then round once
$$P = 2\times 15.35 = 30.70\ \mathrm{cm}$$
the 2 counts pairs of sides, so it is exact and cannot change the digit count
$$P = 30.7\ \mathrm{cm}$$
the coarser measurement stops at the tenths, so the answer does too; this is the single rounding of the whole calculation
Answer $$\boxed{P = 30.7\ \mathrm{cm}}$$
Check
Bracket it: the ruler reading lies between 3.05 and 3.15, so the perimeter lies between 30.6 and 30.8. The hundredths column really is undetermined, and the tenths column really is not.
The two wrong answers that survive to the end here differ only by where the rounding happened. That is why the rule is to round once, at the end, rather than whenever a line looks finished.
Mistake ledger (14 entries)
⚠ Quoting the uncertainty with as many digits as the value
the calculator produced 0.0234 and it feels dishonest to throw digits away, when in fact keeping them is the dishonest move
wrong$$t = 2.34 \pm 0.0234\ \mathrm{s}$$
right$$t = 2.34 \pm 0.02\ \mathrm{s}$$
⚠ Adding absolute uncertainties when the quantities are multiplied
adding is what uncertainties do when quantities are added, and the rule gets carried across to products where the units no longer even match
wrong$$\delta A = \delta a + \delta b = 0.1+0.1 = 0.2\ \mathrm{cm}\ \ (\text{a length, for an area})$$
⚠ Adding two quantities that are not the same kind of thing
both symbols came from the same figure and both look like lengths on the page, so the sum feels harmless until the units are written in
wrong$$V = \pi r^{2} + h$$
right$$V = \pi r^{2} h$$
⚠ Treating a passed dimension check as a proof
the check feels like work, and work that produces no objection feels like a confirmation
wrong$$[\,\pi r h^{2}\,] = [L]^{3}\ \Rightarrow\ V = \pi r h^{2}$$
right$$[\,\pi r h^{2}\,] = [L]^{3}\ \Rightarrow\ \text{possible, and still wrong}$$
⚠ Reporting an estimate with three digits
the calculator produced them and deleting digits feels like losing information, when the information was never there
wrong$$V = 4.86\times 10^{5}\ \mathrm{L}$$
right$$V \sim 5\times 10^{5}\ \mathrm{L}$$
⚠ Guessing the answer instead of guessing the ingredients
the unknown is the thing being asked about, so it feels like the thing to guess, but a guess at the answer cannot be argued with while a guess at each ingredient can
wrong$$N_{\text{breaths in a lifetime}} \approx 10^{6}$$
intervals are compared, never the two central values
Check yourself
Close the page and write, from memory: what the two parts of a measurement are; the two significant figure rules and which operation each belongs to; how percent uncertainties combine in a product; the three SI base units of mechanics; how a conversion fraction is built and what happens to it when the unit carries a power; what a dimensional check can and cannot establish; and the test for whether two results agree. Then reopen and compare. The gaps are your reread list and none of this is scored.
Quote a ruler reading as a value plus or minus an uncertainty, convert it to a percentage, and combine two of them into the uncertainty of an area?
c-measurement-uncertainty
Say how many significant figures 0.00250 and 6000.0 carry, and report a chain containing both a product and a sum without rounding twice?
c-significant-figures
Write a very small or very large quantity in scientific notation, attach the right prefix, and explain why the kilogram cannot take a second prefix?
c-si-units-standards
Build a conversion chain out of fractions equal to one, and convert a squared or cubed unit without losing a power of ten?
c-unit-conversion
Read the units of an unknown constant out of an equation, and say in one sentence why a passed dimensional check is not a proof?
c-dimensional-analysis
State a model in one sentence, bracket each factor, and produce an order of magnitude answer without writing more than one digit?
c-estimating
Decide whether two measured results agree, and say what a disagreement implies about the apparatus rather than about the theory?
c-models-laws
Glossary (19 terms)
measurementölçüm
A comparison of a physical quantity with an agreed standard, producing a value, a unit and an uncertainty.
uncertaintybelirsizlik
The half width of the band a measured value is quoted with, carrying the same unit as the value itself.
percent uncertaintyyüzde belirsizlik
The absolute uncertainty divided by the value and multiplied by one hundred, which turns a slop into a pure number that can be compared across quantities.
precisionkesinlik
How closely repeated readings of the same quantity agree with one another, which says nothing about whether they are right.
accuracydoğruluk
How close a reading is to the true value, which repetition cannot improve and only calibration can.
sistematik hata
An error that is the same in every reading, such as an instrument that was zeroed wrongly, and that averaging therefore cannot remove.
rastgele hata
The part of the error that changes from reading to reading, which shows up as scatter and shrinks when readings are averaged.
significant figuresanlamlı basamak
The digits of a number that carry information about the measurement, as opposed to the zeros that only place the decimal point.
scientific notationbilimsel gösterim
Writing a number as one digit, a decimal point, the remaining digits and a power of ten, which makes the digit count unambiguous.
SI base unitSI temel birim
One of the seven units all others are built from; mechanics uses three of them, the metre, the kilogram and the second.
birim öneki
A letter standing for a power of ten attached to a unit, such as the k in km, which can always be traded back for that power.
conversion factordönüşüm çarpanı
A fraction whose top and bottom describe the same physical amount, so that it equals one and multiplying by it changes only the name of a quantity.
dimensionboyut
The kind of quantity something is, written as powers of mass, length and time, independently of the units chosen to measure it.
dimensional analysisboyut analizi
Checking that every term of an equation carries the same powers of mass, length and time, which can rule an equation out but never confirm it.
order of magnitudebüyüklük mertebesi
The power of ten a quantity is closest to, and the accuracy claimed by an estimate written with a tilde.
estimatekestirim
A deliberately one digit calculation built from bracketed inputs and a stated model, used to find an exponent rather than a value.
densityyoğunluk
Mass divided by the volume it occupies, in kilograms per cubic metre, used throughout this section as the standard target of a unit conversion.
modelmodel
A deliberately simplified picture of a system, such as treating a swimming pool as a rectangular box, adopted so that a calculation becomes possible.
physical lawfizik yasası
A compact relation between measured quantities that has survived testing within a stated range, and that is never proved, only left uncontradicted.
What comes next
§02 · Describing Motion: Kinematics in one Dimension
Everything here was static: a length, a mass, a volume, each measured once and reported properly. Next week the quantity being measured starts to change while you are measuring it, and the questions become how far, how fast and how fast the fastness itself is changing. Every number in that section is still a measurement with a unit and an uncertainty, so nothing you have just learned gets put away.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition, the opening chapter of the same name as this week's syllabus line The required book for the course. The treatment of uncertainty, significant figures, SI standards, unit conversion, order of magnitude estimating and dimensional analysis follows this text; no section numbers are quoted because the week line in the syllabus gives none.
PHYS 101 syllabus: week 1 line and the published assessment weights The week line reads Introduction, Measurement, Estimating. The weights quoted in the sixty second card are the published ones and nothing beyond them is claimed.
The International System of Units, as revised in 2019 Used only for the statements about how the second, the metre and the kilogram are now defined through fixed values of constants rather than through physical artefacts.