Measurement.

The Scene at this step

Solar System clock

20 days/sdefaulthidden

Zoom out

Atom

Choices

Shown in a random order.

• Correct Kilogram, kg

• Wrong Gram, g

  Misconception: Takes the gram to be the SI base unit of mass, as the metre is for length.

  Feedback from the catalogue

  The base unit of mass is the kilogram, the only base unit with a prefix in its name. A gram is a thousandth of it: 1 g = 0.001 kg. For other units the name without a prefix is the base unit; mass is the exception.

• Wrong Newton, N

  Misconception: Confuses mass with weight: picks the newton as the unit of mass.

  Feedback from the catalogue

  The newton is a unit of force, and weight is a force: the pull of the Earth on a body. Mass is the amount of matter in the body, measured in kilograms. On the Moon the weight is six times smaller, and the mass stays the same.

• Wrong Litre, L

  Feedback

  A litre is a unit of volume, how much room a thing takes up, not of how much matter is in it. A litre of water is one kilogram, but a litre of oil is less than that.

Hint

Worked solution

The Scene at this step

Solar System clock

20 days/sdefaulthidden

Zoom out

DNA

Accepted within

± 0.01

A fixed value in the question

Wrong answers it recognises

• 2 × 10⁻⁹ ± 1 × 10⁻¹⁰

  Misconception: Leaves the number as it was in the old unit: 2 × 10⁻⁹ m written as 2 × 10⁻⁹ nm.

  Feedback from the catalogue

  When the unit changes, the number changes too. A nanometre is $10^9$⁠ times smaller than a metre, so the same length is $10^9$⁠ times as many nanometres: $2 \times 10^{-9}$⁠ metres is $2$⁠ nanometres. The same goes for micrometres: $0.01$⁠ millimetres is $10$⁠ micrometres.

Hint

Worked solution

The Scene at this step

Solar System clock

20 days/sdefaulthidden

Zoom out

Cell

Accepted within

± 0.01

A fixed value in the question

Wrong answers it recognises

• 0.00001 ± 0.000001

  Misconception: Converts in the wrong direction, dividing by the factor instead of multiplying: 0.01 mm = 0.00001 µm.

  Feedback from the catalogue

  Ask first: is the new unit smaller or bigger? A micrometre is smaller than a millimetre, so the same length is more micrometres: $0.01 \times 1000 = 10$$0.01 \times 1000 =$$10$⁠. To a smaller unit you multiply, to a bigger one you divide.

• 0.01 ± 0.001

  Misconception: Leaves the number as it was in the old unit: 2 × 10⁻⁹ m written as 2 × 10⁻⁹ nm.

  Feedback from the catalogue

  When the unit changes, the number changes too. A nanometre is $10^9$⁠ times smaller than a metre, so the same length is $10^9$⁠ times as many nanometres: $2 \times 10^{-9}$⁠ metres is $2$⁠ nanometres. The same goes for micrometres: $0.01$⁠ millimetres is $10$⁠ micrometres.

Hint

Worked solution

The Scene at this step

Solar System clock

20 days/sdefaulthidden

Zoom out

Sun

Accepted within

± 0.01

A fixed value in the question

Wrong answers it recognises

• 0.0014 ± 0.0001

  Misconception: Applies a prefix to a number still in kilometres: 1.4 × 10⁶ km ÷ 10⁹ = 0.0014 Gm.

  Feedback from the catalogue

  A gigametre is $10^9$⁠ metres, not kilometres. So go to the base unit first: $1.4 \times 10^6$⁠ kilometres is $1.4 \times 10^9$⁠ metres, and only then divide by $10^9$⁠.

• 1400 ± 1

  Misconception: Uses the wrong power of ten for a prefix, such as giga for 10⁶.

  Feedback from the catalogue

  Each prefix is a different power of $10$⁠: kilo $10^3$⁠, mega $10^6$⁠, giga $10^9$⁠. The other way, milli $10^{-3}$⁠, micro $10^{-6}$⁠ and nano $10^{-9}$⁠. Each step in the table changes the power by $3$⁠.

Hint

Worked solution

The Scene at this step

Solar System clock

20 days/sdefault

Zoom out

Solar System

Choices

Shown in a random order.

• Correct m/s²

• Wrong m/s

  Misconception: Gives acceleration the unit of speed, m/s.

  Feedback from the catalogue

  m/s is the unit of speed. Acceleration is how fast the speed changes: speed divided by time, $\dfrac{\mathrm{m/s}}{\mathrm{s}} = \mathrm{m/s^2}$$\dfrac{\mathrm{m/s}}{\mathrm{s}} =$$\mathrm{m/s^2}$⁠. A car standing still and a car cruising at 30 m/s in a straight line both have zero acceleration.

• Wrong kg·m/s²

  Feedback

  That is the newton, the unit of force: mass times acceleration. Acceleration itself has no kilograms in it.

• Wrong m·s

  Feedback

  Metres times seconds grows the more time passes. Acceleration divides by time; it does not multiply.

Hint

Worked solution

The Scene at this step

Solar System clock

20 days/sdefault

Zoom out

Solar System

Choices

Shown in a random order.

• Correct $d = v \times t$⁠

• Wrong $d = \dfrac{v}{t}$⁠

  Misconception: Picks a formula without checking its units: distance = speed ÷ time.

  Feedback from the catalogue

  Check the units: $\dfrac{v}{t}$⁠ gives $\dfrac{\mathrm{m/s}}{\mathrm{s}} = \mathrm{m/s^2}$$\dfrac{\mathrm{m/s}}{\mathrm{s}} =$$\mathrm{m/s^2}$⁠, a unit of acceleration, not of distance. And by sense: the longer you drive, the farther you go, so time should multiply, not divide.

• Wrong $d = \dfrac{t}{v}$⁠

  Feedback

  The units: $\dfrac{\mathrm{s}}{\mathrm{m/s}} = \dfrac{\mathrm{s^2}}{\mathrm{m}}$$\dfrac{\mathrm{s}}{\mathrm{m/s}} =$$\dfrac{\mathrm{s^2}}{\mathrm{m}}$⁠, not metres.

• Wrong $d = v + t$⁠

  Misconception: Adds quantities with different units, such as a speed and a time.

  Feedback from the catalogue

  You can add only quantities of the same kind: metres to metres, seconds to seconds. $v + t$⁠ adds metres per second to seconds, like adding apples to hours. A formula like that is wrong before you put a number into it.

Hint

Worked solution

The Scene at this step

End of the bar

3.43 cmhidden

Smallest division

1 cmfixed

Before reading / Read

Before reading

Accepted within

± 0.01 cm

Computed from the Scene at this step’s setting

Ruler: the length, endReading − startReading/nodes/board/outputs/reading

Any unit of the same quantity is accepted; it is converted before checking.

Wrong answers it recognises

• 4 cm ± 0.01 cm

  Misconception: Reads the mark after the end of the object instead of the nearest one.

  Feedback from the catalogue

  Read the mark nearest the end, even when it comes before the end. The end of the bar is a little past the mark for $3$⁠, but far from $4$⁠, so the reading is 3 cm.

Hint

Worked solution

The Scene at this step

End of the bar

3.43 cmhidden

Smallest division

1 cmfixed

Before reading / Read

Before reading

Accepted within

± 0.001 cm

Computed from the Scene at this step’s setting

Ruler: the length's uncertainty, half the smallest division for each reading taken (one from 0, two otherwise)/nodes/board/outputs/uncertainty

Any unit of the same quantity is accepted; it is converted before checking.

Wrong answers it recognises

• 1 cm ± 0.001 cm

  Misconception: Takes the uncertainty of a reading to be a whole division instead of half of one.

  Feedback from the catalogue

  An end you read as $3$⁠ is closer to the mark for $3$⁠ than to the marks for $2$⁠ and $4$⁠. So it is at most halfway to the next mark: half a division, 0.5 cm. An end a whole division from $3$⁠ would already read as $2$⁠ or $4$⁠.

Hint

Worked solution

The Scene at this step

End of the bar

3.43 cmhidden

Smallest division

1 cm

Before reading / Read

Read

Choices

• Correct It will get smaller

• Wrong It will stay 0.5 cm

  Feedback

  The uncertainty comes from the marks, not the bar: closer marks narrow down where the end can be.

• Wrong It will get larger

  Feedback

  More marks do not leave more room for error. On the contrary: there is always a mark closer to the end, so the reading is more precise.

Once the prediction is locked in, the Scene changes

Smallest division

0.1 cm

Before reading / Read

Read

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Worked solution

The Scene at this step

End of the bar

3.43 cmhidden

Smallest division

0.1 cmfixed

Before reading / Read

Before reading

Accepted within

± 0.04 cm

Computed from the Scene at this step’s setting

Ruler: the length, endReading − startReading/nodes/board/outputs/reading

Any unit of the same quantity is accepted; it is converted before checking.

Wrong answers it recognises

• 3 cm ± 0.01 cm

  Misconception: Reads only the numbered marks and ignores the millimetre marks between them.

  Feedback from the catalogue

  The small marks are what this ruler adds. After the 3 whole centimetres, count millimetre marks up to the one nearest the end: four, that is 0.4 cm. Together 3.4 cm.

• 34 cm ± 0.5 cm

  Misconception: Counts millimetres but writes the count in centimetres: 34 cm for 34 mm.

  Feedback from the catalogue

  You counted $34$⁠ millimetre marks, which is 34 mm. A centimetre has $10$⁠ millimetres, so that is 3.4 cm. You can also pick millimetres in the unit list beside the answer.

Hint

Worked solution

The Scene at this step

End of the bar

4.2 cmhidden

Smallest division

0.1 cmfixed

Before reading / Read

Before reading

Other settings

Start/nodes/board/props/start = 1

Accepted within

± 0.04 cm

Computed from the Scene at this step’s setting

Ruler: the length, endReading − startReading/nodes/board/outputs/reading

Any unit of the same quantity is accepted; it is converted before checking.

Wrong answers it recognises

• 4.2 cm ± 0.04 cm

  Misconception: Reads where the object ends and takes it as the length, though it does not start at 0.

  Feedback from the catalogue

  A ruler shows a position, not a length. Only when the bar starts at zero is the position of its end also its length. Here it starts at 1 cm, so subtract: $4.2 - 1 = 3.2$⁠ centimetres.

Hint

Worked solution

The Scene at this step

End of the bar

3.43 cmhidden

Smallest division

0.1 cmfixed

Before reading / Read

Before readingfixed

Choices

Shown in a random order.

• Correct 3.4 cm ± 0.05 cm

• Wrong 3.4271 cm

  Misconception: Thinks more digits make a result more precise, even digits the instrument did not measure.

  Feedback from the catalogue

  A digit the instrument did not read is not precision, it is a guess. A millimetre ruler cannot tell 3.42 cm from 3.43 cm, so a digit beyond the tenths claims something nobody measured. Write what was measured, and no more.

• Wrong 3 cm ± 0.05 cm

  Misconception: Writes fewer digits than were measured: 3 cm from a millimetre ruler that read 3.4 cm.

  Feedback from the catalogue

  The ruler read the millimetres too, so $3.4$⁠, not $3$⁠. Dropping a measured digit throws information away: $3$⁠ says the end is somewhere between 2.5 cm and 3.5 cm, and we know more than that.

• Wrong $3.4$⁠

  Misconception: Gives a measurement as a bare number with no unit.

  Feedback from the catalogue

  $3.4$⁠ on its own says nothing: $3.4$⁠ centimetres, metres or kilometres? A measurement is always a number and a unit, with an uncertainty saying how far to trust the last digit.

Hint

Worked solution

The Scene at this step

End of the bar

12 cmhidden

Before reading / Read

Read

Accepted within

± 0

Computed from the Scene at this step’s setting

Ruler: significant figures in the length as written to the marks (12.0 has 3)/nodes/board/outputs/figures

Wrong answers it recognises

• 2 ± 0

  Misconception: Does not count a zero at the end after a decimal point: 12.0 as two significant figures.

  Feedback from the catalogue

  A zero at the end, after a decimal point, counts: someone took the trouble to write it because it was measured. $12.0$⁠ says the millimetres were checked and came out zero; $12$⁠ says the measurement went only to the centimetre. As numbers they are equal; as measurements they are not.

Hint

Worked solution

The Scene at this step

End of the bar

12 cmhidden

Before reading / Read

Read

Accepted within

± 0

A fixed value in the question

Wrong answers it recognises

• 4 ± 0

  Misconception: Counts the zeros at the start of a number as significant: 0.120 as four figures.

  Feedback from the catalogue

  Zeros at the start of a number only show where the decimal point is. 0.120 m and 12.0 cm are the same measurement, and changing the unit did not add a measured digit. So three significant figures, not four.

• 2 ± 0

  Misconception: Does not count a zero at the end after a decimal point: 12.0 as two significant figures.

  Feedback from the catalogue

  A zero at the end, after a decimal point, counts: someone took the trouble to write it because it was measured. $12.0$⁠ says the millimetres were checked and came out zero; $12$⁠ says the measurement went only to the centimetre. As numbers they are equal; as measurements they are not.

Hint

Worked solution

The Scene at this step

End of the bar

12 cmhidden

Before reading / Read

Read

Choices

Shown in a random order.

• Correct $1.20 \times 10^3$⁠ metres

• Wrong $1200$⁠ metres

  Feedback

  $1200$⁠ does not say whether the zeros were measured. A reader could think it was measured to the nearest hundred, or to the nearest metre.

• Wrong $1.2 \times 10^3$⁠ metres

  Misconception: Does not count a zero at the end after a decimal point: 12.0 as two significant figures.

  Feedback from the catalogue

  A zero at the end, after a decimal point, counts: someone took the trouble to write it because it was measured. $12.0$⁠ says the millimetres were checked and came out zero; $12$⁠ says the measurement went only to the centimetre. As numbers they are equal; as measurements they are not.

• Wrong $1.200 \times 10^3$⁠ metres

  Misconception: Thinks more digits make a result more precise, even digits the instrument did not measure.

  Feedback from the catalogue

  A digit the instrument did not read is not precision, it is a guess. A millimetre ruler cannot tell 3.42 cm from 3.43 cm, so a digit beyond the tenths claims something nobody measured. Write what was measured, and no more.

Hint

Worked solution

The Scene at this step

End of the bar

12 cmdefaulthidden

Before reading / Read

Readhidden

Other settings

Mode/nodes/board/props/mode = "expression"

Other settings

Expression/nodes/board/props/expression = "a × b"

Other settings

A/nodes/board/props/a = 7.3

Other settings

B/nodes/board/props/b = 4.6

Accepted within

± 0.001

A fixed value in the question

Wrong answers it recognises

• 33.58 ± 0.001

  Misconception: Writes every digit the calculator shows, though the measurements had fewer.

  Feedback from the catalogue

  The calculator does not know how many digits were measured; it only calculates. A result cannot be more precise than the measurements it came from, so round: in a product by the number of significant figures, in a sum by the place of the last digit.

• 33.6 ± 0.001

  Misconception: Rounds a product to the decimal places of the measurements instead of their significant figures.

  Feedback from the catalogue

  In multiplying and dividing, count significant figures, not decimal places. $7.3$⁠ and $4.6$⁠ have two figures each, so the area does too: $34$⁠. The decimal-places rule belongs to adding and subtracting.

Hint

Worked solution

The Scene at this step

End of the bar

12 cmdefaulthidden

Before reading / Read

Readhidden

Other settings

Mode/nodes/board/props/mode = "expression"

Other settings

Expression/nodes/board/props/expression = "a + b"

Other settings

A/nodes/board/props/a = 12.3

Other settings

B/nodes/board/props/b = 0.27

Accepted within

± 0.001

A fixed value in the question

Wrong answers it recognises

• 12.57 ± 0.001

  Misconception: Writes every digit the calculator shows, though the measurements had fewer.

  Feedback from the catalogue

  The calculator does not know how many digits were measured; it only calculates. A result cannot be more precise than the measurements it came from, so round: in a product by the number of significant figures, in a sum by the place of the last digit.

• 13 ± 0.001

  Misconception: Rounds a sum to the fewest significant figures instead of the fewest decimal places.

  Feedback from the catalogue

  In a sum what matters is which place was measured, not how many digits. The stack was measured to tenths of a centimetre, so the sum is known to the tenths too: $12.6$⁠. Rounding to $13$⁠ throws away tenths that were measured.

Hint

Worked solution

The Scene at this step

Length of u

5 m

Direction of u

30 °

Length of v

0 mhidden

Given / Added

Added

Other settings

Vector2 angle/nodes/board/props/vector2Angle = 90

Choices

Shown in a random order. More than one choice is correct: every one must be selected.

• Correct Displacement: where you moved from the starting point

• Correct Velocity: how fast and which way

• Correct Force

• Wrong Mass

  Feedback

  Mass has no direction: a kilogram is a kilogram whichever way it faces.

• Wrong Time

  Feedback

  Time is a scalar: ten seconds are ten seconds, with no direction.

• Wrong Speed: how fast, without which way

  Misconception: Treats speed and velocity as the same quantity, ignoring direction.

  Feedback from the catalogue

  Speed says only how fast: a scalar. Velocity says how fast and which way: a vector. Two cyclists at 3 m/s, one going north and one going south, have the same speed and different velocities.

Hint

Worked solution

The Scene at this step

Length of u

5 m

Direction of u

30 °

Length of v

0 mhidden

Given / Added

Addedhidden

Other settings

Vector2 angle/nodes/board/props/vector2Angle = 90

Read from the Scene at

Vector: the length of u/nodes/board/outputs/vectorLength

Read from the Scene at

Vector: the direction of u, degrees from 0 up to 360/nodes/board/outputs/vectorAngle

One setting that reaches it

Length of u

4 m

Direction of u

90 °

Hint

Worked solution

The Scene at this step

Length of u

3 mfixed

Direction of u

90 °fixed

Length of v

3 mfixed

Given / Added

Givenfixed

Other settings

Vector2 angle/nodes/board/props/vector2Angle = 270

Choices

Shown in a random order.

• Correct Their speeds are equal, and their velocities are different

• Wrong Their speeds and their velocities are both equal

  Misconception: Treats speed and velocity as the same quantity, ignoring direction.

  Feedback from the catalogue

  Speed says only how fast: a scalar. Velocity says how fast and which way: a vector. Two cyclists at 3 m/s, one going north and one going south, have the same speed and different velocities.

• Wrong Their speeds are different

  Feedback

  Both arrows are the same length, $3$⁠: both cover 3 m every second.

Hint

Worked solution

The Scene at this step

Length of u

5 mfixed

Direction of u

0 °fixed

Length of v

5 mfixed

Given / Added

Given

Other settings

Vector2 angle/nodes/board/props/vector2Angle = 180

Other settings

Unknown/nodes/board/props/unknown = "sumLength"

Accepted within

± 0.01 m

Computed from the Scene at this step’s setting

Vector: the length of u + v, by Pythagoras from its components/nodes/board/outputs/sumLength

Any unit of the same quantity is accepted; it is converted before checking.

Wrong answers it recognises

• 10 m ± 0.01 m

  Misconception: Gives the distance travelled as the displacement.

  Feedback from the catalogue

  Distance is how far you went along the path, and displacement is the arrow from where you started to where you ended. Someone who walked 5 m and came back covered a distance of 10 m, but stands exactly where they started: the displacement is zero.

Hint

Worked solution

The Scene at this step

Length of u

6 mfixed

Direction of u

0 °fixed

Length of v

1 m

Given / Added

Addedfixed

Other settings

Vector2 angle/nodes/board/props/vector2Angle = 180

Read from the Scene at

Vector: the horizontal component of u + v/nodes/board/outputs/sumX

Read from the Scene at

Vector: the vertical component of u + v/nodes/board/outputs/sumY

One setting that reaches it

Length of v

4 m

Hint

Worked solution

The Scene at this step

Length of u

3 mfixed

Direction of u

0 °fixed

Length of v

4 mfixed

Given / Added

Givenfixed

Other settings

Vector2 angle/nodes/board/props/vector2Angle = 90

Other settings

Unknown/nodes/board/props/unknown = "sumLength"

Choices

• Correct Less than 7 m

• Wrong Exactly 7 m

  Misconception: Adds the lengths of two moves in different directions to get the displacement: 3 m + 4 m = 7 m.

  Feedback from the catalogue

  The lengths of vectors add like that only when they point the same way. When they point in different directions, the displacement is a straight line that cuts the corner, and it is shorter than the sum of the lengths. 3 m east and then 4 m north is a displacement of 5 m.

• Wrong More than 7 m

  Feedback

  A straight line is the shortest way between two points, so the displacement cannot be longer than the distance.

Once the prediction is locked in, the Scene changes

Given / Added

Addedfixed

The Scene then shows: 5 m · Vector: the length of u + v, by Pythagoras from its components/nodes/board/outputs/sumLength

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Worked solution

The Scene at this step

Hypothesis

Heavier swings fasterfixed

Trials per pendulum

5

Run the test

Prediction

Choices

• Correct All five bobs will swing at about the same beat, one swing in about 2 s

• Wrong The heavy bobs will swing faster, like the ghosts

  Misconception: Thinks a heavier pendulum bob swings faster, so the mass sets the period.

  Feedback from the catalogue

  A pendulum’s period does not depend on its mass. Gravity pulls four times as hard on a bob four times as heavy, but its inertia is four times larger too, so it accelerates exactly like a light one. The period depends on the length of the string: $T = 2\pi\sqrt{L/g}$⁠.

• Wrong The light bobs will swing faster

  Feedback

  That would also refute the hypothesis, but the mass does not change the period either way: gravity pulls harder on a heavy bob, and its inertia is larger by the same factor.

Once the prediction is locked in, the Scene changes

Run the test

Test

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Worked solution

The Scene at this step

Hypothesis

Heavier swings fasterfixed

Trials per pendulum

5

Run the test

Verdict

Choices

Shown in a random order.

• Correct It is falsified: one clear miss is enough

• Wrong It is partly right, because one pendulum agrees with the prediction

  Misconception: Thinks agreeing cases count against a clear miss, so a hypothesis that fits some measurements is partly right.

  Feedback from the catalogue

  A general hypothesis claims something about every case. A case that fits it does not cancel one that contradicts it: one clear miss is enough to refute it, and any number of agreements does not change that.

• Wrong We cannot tell yet: it needs more trials

  Feedback

  More trials only shrink the error bars. The misses here are many times larger than the bars, so they will not go away: they will only stand out more.

Hint

Worked solution

The Scene at this step

Hypothesis

T = 2π√(L/g)fixed

Trials per pendulum

20

Run the test

Verdict

Choices

Shown in a random order.

• Correct The hypothesis passed this test. It is not proven, and a sharper test could still refute it

• Wrong The hypothesis is proven: five different pendulums agree with it

  Misconception: Thinks a hypothesis whose predictions agree with the measurements has been proven true.

  Feedback from the catalogue

  Agreeing with measurements does not prove a hypothesis: there are always cases nobody measured, or a precision nobody reached. A hypothesis that passed a test has survived it, and a sharper test, with more trials or other conditions, can still refute it.

• Wrong Nothing: a test that agrees with the prediction teaches nothing

  Feedback

  The test could have failed, and it did not. That teaches a lot: the formula took a real risk and survived it, so there is more reason to trust it.

Worked solution

The Scene at this step

Hypothesis

Width does not matterfixed

Trials per pendulum

1

Run the test

Verdict

Choices

• Correct It will be falsified: the wide swings are measurably slower

• Wrong It will survive again: it has already passed the test

  Misconception: Thinks a hypothesis whose predictions agree with the measurements has been proven true.

  Feedback from the catalogue

  Agreeing with measurements does not prove a hypothesis: there are always cases nobody measured, or a precision nobody reached. A hypothesis that passed a test has survived it, and a sharper test, with more trials or other conditions, can still refute it.

• Wrong It will be falsified: the wide swings are faster

  Feedback

  A wide swing travels a longer path, and spends much of it far from the bottom, where the bob is slow. So it takes longer, not less.

Once the prediction is locked in, the Scene changes

Trials per pendulum

20

The Scene then shows: 2 · Pendulums whose mean period is more than two standard errors from the prediction/nodes/bench/outputs/misses

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Worked solution

The Scene at this step

Hypothesis

Width does not matterfixed

Trials per pendulum

1

Run the test

Verdictfixed

Accepted within

± 0.4

Computed from the Scene at this step’s setting

The fewest trials per pendulum at which the test falsifies the hypothesis; NaN if it survives even 20/nodes/bench/outputs/trialsToFalsify

Hint

Worked solution

The Scene at this step

Hypothesis

T = 2π√(L/g)

Trials per pendulum

5

Run the test

Verdict

Choices

Shown in a random order.

• Correct Repeat the measurement, ideally other people with other equipment, before giving up the formula

• Wrong Abandon the formula: one miss is enough

  Misconception: Thinks any single result just outside the error bars refutes a well-tested idea at once, without repeating it.

  Feedback from the catalogue

  Even a true value lands outside two standard errors about once in twenty measurements, by chance. So one surprising result is repeated, ideally by other people with other equipment. A clear miss is one that comes back, and only that refutes.

• Wrong Ignore the miss: the formula is in every textbook

  Feedback

  If no result could ever shake an idea, it is no longer a scientific hypothesis. You may suspect the measurement, but you have to check it, not throw it away.

Hint

Worked solution

The Scene at this step

Drops

25

Timing jitter

4 msfixed

Magnet delay

1.5 msfixed

Work out the true value

Spread

Choices

• Correct It will stay about the same

• Wrong It will halve, because there are four times as many readings

  Misconception: Thinks more readings make the spread of single readings narrower.

  Feedback from the catalogue

  The spread $s$⁠ describes how far one reading lies from the mean. It depends on the instrument and the method, not on the number of readings: more drops fill out the bell curve but do not narrow it. What shrinks with more readings is the standard error of the mean, $\dfrac{s}{\sqrt{n}}$⁠.

• Wrong It will grow, because there are more far-off readings

  Feedback

  More drops add far-off readings, but also many more close to the middle. The shape of the bell stays the same; only its height grows.

Once the prediction is locked in, the Scene changes

Drops

100

The Scene then shows: 0.1792 m/s² · S, the spread of the readings (dividing by n − 1); NaN for one drop/nodes/tally/outputs/spread

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Worked solution

The Scene at this step

Drops

25fixed

Timing jitter

4 msfixed

Magnet delay

1.5 msfixed

Work out the true value

Spreadfixed

Accepted within

± 0.003 m/s²

Computed from the Scene at this step’s setting

S/√n, the standard error of the mean; NaN for one drop/nodes/tally/outputs/standardError

Any unit of the same quantity is accepted; it is converted before checking.

Wrong answers it recognises

• 0.2007 m/s² ± 0.01 m/s²

  Misconception: Gives the spread of single readings as the uncertainty of their mean.

  Feedback from the catalogue

  The spread $s$⁠ is the uncertainty of one reading. The mean of $n$⁠ readings is surer, because the random errors in it partly cancel: its standard error is $\dfrac{s}{\sqrt{n}}$⁠.

• 0.008 m/s² ± 0.002 m/s²

  Misconception: Divides the spread by n instead of √n for the standard error of the mean.

  Feedback from the catalogue

  The standard error is $\dfrac{s}{\sqrt{n}}$⁠, not $\dfrac{s}{n}$⁠. The random errors only partly cancel, so four times the measurements shrink the error only twofold.

Hint

Worked solution

The Scene at this step

Drops

25

Timing jitter

4 msfixed

Magnet delay

1.5 msfixed

Work out the true value

Error bar

Read from the Scene at

S/√n, the standard error of the mean; NaN for one drop/nodes/tally/outputs/standardError

One setting that reaches it

Drops

59

Hint

Worked solution

The Scene at this step

Drops

25

Timing jitter

4 msfixed

Magnet delay

1.5 msfixed

Work out the true value

True value

Choices

• Correct No: the bar will shrink, but the mean will stay too low

• Wrong Yes, and even better: more drops always bring you closer to the truth

  Misconception: Thinks taking more readings removes a systematic error, as it does a random one.

  Feedback from the catalogue

  A systematic error pushes every reading the same way, so it does not cancel in the mean: the mean of a hundred readings that are all too low is too low by just as much. More readings only shrink the error bar, and the wrong answer looks more certain. Finding a systematic error takes a change of method.

• Wrong Yes: with more drops the bar will widen

  Feedback

  The error bar is $\dfrac{s}{\sqrt{n}}$⁠, so it shrinks with more drops; it does not widen.

Once the prediction is locked in, the Scene changes

Drops

100

The Scene then shows: -3.533 · The error in standard errors; NaN for one drop/nodes/tally/outputs/gap

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Worked solution

The Scene at this step

Drops

100fixed

Timing jitter

4 msfixed

Magnet delay

5 msfixed

Work out the true value

Error barfixed

Accepted within

± 0.01 m/s²

Computed from the Scene at this step’s setting

The systematic error: what the delay alone does to every reading/nodes/tally/outputs/shift

Any unit of the same quantity is accepted; it is converted before checking.

Wrong answers it recognises

• 0.108 m/s² ± 0.01 m/s²

  Misconception: Treats g as inversely proportional to the fall time rather than its square when working out how a timing error shifts g.

  Feedback from the catalogue

  $g = \dfrac{2h}{t^2}$⁠, so $g$⁠ goes as one over the time squared. A small relative error in the time becomes twice that relative error in $g$⁠: a time $1\%$⁠ too long gives a $g$⁠ about $2\%$⁠ too small.

Hint

Worked solution

The Scene at this step

Drops

25

Timing jitter

2 msfixed

Magnet delay

1.5 msfixed

Work out the true value

True value

Choices

Shown in a random order.

• Correct High precision, but not accurate

• Wrong Accurate, but not precise

  Misconception: Takes precision (readings close together, a small error bar) for accuracy (close to the true value).

  Feedback from the catalogue

  Precision is how close the readings are to one another, and accuracy is how close they are to the true value. Random error spoils precision, and systematic error spoils accuracy. A small error bar shows precision, but says nothing about a systematic error.

• Wrong Both accurate and precise: the bar is small

  Misconception: Takes precision (readings close together, a small error bar) for accuracy (close to the true value).

  Feedback from the catalogue

  Precision is how close the readings are to one another, and accuracy is how close they are to the true value. Random error spoils precision, and systematic error spoils accuracy. A small error bar shows precision, but says nothing about a systematic error.

• Wrong Neither

  Feedback

  The readings are very close to one another, so the precision is good. What is missing is accuracy: they have all moved together away from the true value.

Worked solution