RuukkilabraUpper secondary physics
FY1 · at the end of the module

Apply independently

In these tasks nobody tells you which formula to use. Decide yourself which phenomena and models belong to the situation, solve the task on paper and justify your solution, as in the matriculation examination.

Two tasks bring together the skills of measuring, graphs and investigating from this module. In the third you assess your own investigation report.

Solve the task yourself first, on paper or on a computer. You may use a calculator, a data booklet and graphing software. Write out your reasoning, formulas, substitutions and answers with units.

A rolling ball on video 12 points

Made-up example data · The data was made up as an example of this kind of measurement. It is not the result of a real measurement.

A student lets a ball roll down a short ramp onto the floor and films the motion from the side with a phone. A video analysis app gives the position $s$ of the centre of the ball, measured from the starting point, every $0.20\,\text{s}$. The lower end of the ramp is about $16\,\text{cm}$ from the starting point. In the video, one pixel corresponds to $2.5\,\text{mm}$ on the floor.

$t\ (\text{s})$ $s\ (\text{mm})$
$0.0$ $0.0$
$0.2$ $38.9$
$0.4$ $157.9$
$0.6$ $320.9$
$0.8$ $477.4$
$1.0$ $640.2$
$1.2$ $799.2$
$1.4$ $957.3$
$1.6$ $1120.0$
$1.8$ $1277.2$
$2.0$ $1439.6$
  1. a) Draw the $(t, s)$ graph of the data and fit a model to the points it suits. 3 p
  2. b) Use the graph to determine the speed of the ball on the floor. Justify which measurement points you used. 3 p
  3. c) The app gives the positions to a precision of $0.1\,\text{mm}$. Estimate how precisely the position of the ball is really known. Give the speed from part b to a justified precision. 3 p
  4. d) The student's partner extends the fitted line and concludes that after $10\,\text{s}$ the ball is about $7.8\,\text{m}$ from the starting point. Evaluate this conclusion. Also explain why the fitted line does not pass through the origin. 3 p
Show the model answer and marking Open only when your own solution is ready.

Model answer

a)

Find

the $(t, s)$ graph of the data

Graph

The horizontal axis shows time $t\ (\text{s})$ and the vertical axis shows position $s$. It is a good idea to convert the positions to metres, for example $1439.6\,\text{mm} = 1.4396\,\text{m}$, so that the slope comes out directly in $\text{m/s}$. Each row is one separate point.

$0.4$
$0.8$
$1.2$
$1.6$
$2.0$
$0.4$
$0.8$
$1.2$
$1.6$
$0$
$t\ (\text{s})$
$s\ (\text{m})$
$\text{ramp}$
$v = \dfrac{1.12\,\text{m}}{1.4\,\text{s}} = 0.80\,\tfrac{\text{m}}{\text{s}}$

Model answer

The points from $0.4\,\text{s}$ onwards lie on a line, so a line is fitted to them: the ball rolls steadily on the floor. At $0.4\,\text{s}$ the ball is at the lower end of the ramp (about $16\,\text{cm}$) and has already reached its speed on the floor. The first two points are from the ramp, where the speed is still increasing: they stay above the line and are not part of the fit. The points are not joined into a broken line, because every measurement has a small error.

b)

Given

the graph from part a and the points taken on the floor, $t = 0.4\text{–}2.0\,\text{s}$

Find

the speed $v$ of the ball on the floor

Principle

On the floor the points lie on a line, so the ball moves uniformly: its position is $s = s_0 + vt$, and the speed is the slope of the fitted line. The points from the ramp are left out, because there the speed is still increasing.

Reading the graph

Two points far apart are read from the line, $(0.6\,\text{s}$; $0.32\,\text{m})$ and $(2.0\,\text{s}$; $1.44\,\text{m})$. They are points of the line, not measurement points.

Formula

$v$$= \dfrac{\Delta s}{\Delta t}$
$= \dfrac{s_2 - s_1}{t_2 - t_1}$

Conversions

The positions have already been read in metres: $320\,\text{mm} = 0.32\,\text{m}$ and $1440\,\text{mm} = 1.44\,\text{m}$.

Substitution

$v$$= \dfrac{1.44\,\text{m} - 0.32\,\text{m}}{2.0\,\text{s} - 0.6\,\text{s}}$
$= \dfrac{1.12\,\text{m}}{1.4\,\text{s}}$
$= 0.80\,\text{m/s}$

Answer

$v \approx 0.80\,\text{m/s}$

The precision is justified in part c.

Reasonableness

On the floor the ball travels about $16\,\text{cm}$ during every $0.20\,\text{s}$, that is, $0.16\,\text{m} : 0.20\,\text{s} = 0.80\,\text{m/s}$. The speed is of the order of a slow walk, which fits a ball rolling down a low ramp.

c)

Find

the real precision of the position and the justified precision of the speed

Principle

The digits shown on the screen do not tell the precision of a measurement. In a video the uncertainty of the position can be estimated from the size of a pixel: the position of the edge of the ball can be off by a pixel or two.

Model answer

One pixel corresponds to $2.5\,\text{mm}$, so the position of the ball is known to a precision of about $\Delta s \approx 3\,\text{mm}$. The app calculates the centre of the ball from the pixels of its edge and can give the position in parts of a pixel, but the edge of a moving ball is slightly blurred. The $0.1\,\text{mm}$ digit shown by the app comes from calculation, when pixels are converted to millimetres, and it says nothing about the precision of the measurement: even the millimetre digit is uncertain. The same can be seen in the scatter of the points: on the floor the points deviate from the fitted line by a few millimetres. The speed was determined from a distance of $1.12\,\text{m}$, whose uncertainty is a few millimetres, that is, less than one per cent, so the speed is given as $v = 0.80\,\text{m/s}$. A least-squares fit gives $0.799\,\text{m/s}$: the third significant figure can still be justified, but the fourth cannot.

d)

Find

an evaluation of the extrapolation and an explanation for the constant term of the line

Principle

The model has been tested only over the interval that was measured. The constant term of the fitted line is the point where the line crosses the $s$-axis.

Model answer

The conclusion is not reliable. The line has been tested only over the interval $0.4\text{–}2.0\,\text{s}$, that is, over $1.6\,\text{s}$, and the conclusion extends it more than six times further. In reality rolling resistance gradually slows the ball, and over a distance of $7.8\,\text{m}$ the ball would probably hit a wall or a piece of furniture. So the model of uniform motion can be used for a longer time only after it has been tested with a longer measurement.
The constant term of the fitted line is about $s_0 = -0.16\,\text{m}$. It is not the position of the ball at the moment $0\,\text{s}$, because the ball started from $s = 0$. On the ramp the ball first moved more slowly than on the floor. If it had moved all the time at its floor speed of $0.80\,\text{m/s}$, it would have got further. So the line describing the motion on the floor has shifted downwards and crosses the $s$-axis below the origin.

Common mistakes

  • The points are joined into a broken line, or one line is fitted to all the points including the ramp points. Then the slope is too small.
  • The slope is calculated from two adjacent measurement points, so the error of a single point has a large effect on the result.
  • The speed is given with all the digits of the app, for example $0.7993\,\text{m/s}$.
  • The constant term of the line is interpreted as the starting position of the ball, although the ball started from the origin.

Marking and self-assessment

Compare your solution with the model answer and, for each item, choose the points you think your solution earns. Part marks are possible, and any other physically correct method is accepted too.

a)3 p

The axes show the correct quantities and units, and the scales are evenly spaced. 1 p
Each measurement is marked as a separate point in the right place. 1 p
The model is fitted to the points it suits, and the points have not been joined into a broken line. The line is fitted to the points taken on the floor; the points at the start of the ramp do not affect it. 1 p

b)3 p

Justifies why the speed is determined from the points taken on the floor ($t \geq 0.4\,\text{s}$ or $t \geq 0.6\,\text{s}$): on the ramp the speed is still increasing. 1 p
The slope or other quantity is read visibly from the graph, using two points on the line that are far apart. 1 p
The answer has the correct unit and a precision that matches the given data. $v \approx 0.80\,\text{m/s}$ 1 p

c)3 p

Estimates the uncertainty of the position as about one pixel, that is, a few millimetres, and bases the estimate on the pixel size of the video. 1 p
Explains that the decimals on the screen come from calculation and do not show the precision of the measurement. 1 p
Gives the speed with two or three significant figures and justifies the number by the uncertainty of the position and the scatter of the points. 1 p

d)3 p

States that the model has been tested only over the measurement interval $0.4\text{–}2.0\,\text{s}$ and that extending the line to $10\,\text{s}$ is uncertain. 1 p
Names a physical reason why the motion does not continue the same way, for example rolling resistance slows the ball or the floor ends. 1 p
Explains the negative constant term: on the ramp the ball moved more slowly than its speed on the floor, so the line of the floor motion crosses the $s$-axis below the origin. 1 p

Self-assessed points: 0 / 12 (12 items not assessed)

The points are your own estimate, not a checked grade. The marking is Ruukkilabra's own, written according to the features of a good matriculation examination answer, and it is not official marking by the Matriculation Examination Board (YTL).

A race between two balls 15 points

Made-up example data · The lab report and the measurements were made up as an example.

A group investigated whether the mass of a ball affects how long the ball takes to roll down a ramp. Here is the group's lab report.

Research question. Do heavy balls roll faster?

Hypothesis. A heavy ball is faster, because gravity pulls on it more.

Equipment. A ramp $1.2\,\text{m}$ long, a pile of books, a steel ball ($110\,\text{g}$), a solid plastic ball of the same size ($16\,\text{g}$) and a phone stopwatch app that shows thousandths of a second.

Method. We raised the upper end of the ramp on a pile of books. We released the steel ball from the upper end of the ramp, and Eetu started the timing when the ball set off. The plastic ball hardly started to move, so we gave it a small push from roughly the same spot. In between, one book was taken from the pile, because it was needed in another lesson. The timing was stopped when the ball hit the wall at the end of the ramp.

Results. Steel ball $2.437\,\text{s}$, plastic ball $2.109\,\text{s}$.

Conclusion. The plastic ball was $0.328\,\text{s}$ faster, so the hypothesis is wrong: a light ball rolls faster. Light objects always move faster than heavy ones.

After the teacher's feedback, the group corrected their method and measured again:

Measurement steel ball $t\ (\text{s})$ plastic ball $t\ (\text{s})$
$1$ $2.31$ $2.40$
$2$ $2.45$ $2.33$
$3$ $2.38$ $2.47$
$4$ $2.29$ $2.36$
$5$ $2.42$ $2.44$
  1. a) Find four significant weaknesses in the first lab report. For each one, explain why it weakens the investigation and how to fix it. 8 p
  2. b) From the new measurements, calculate the mean rolling time of each ball and estimate the uncertainty of the results. Can you say from the measurements that the rolling times of the balls differ? 4 p
  3. c) Write a new conclusion for the group. 3 p
Show the model answer and marking Open only when your own solution is ready.

Model answer

a)

Find

four significant weaknesses, why each weakens the investigation and how it is fixed

Principle

In a fair test only one quantity is varied at a time, the measurements are repeated, the uncertainty is estimated and the conclusion is drawn only from what the measurements show.

Model answer

Any four of the following, each justified and fixed:
1. Other conditions changed besides the mass. The plastic ball was pushed to get it moving, it set off only from roughly the same spot and the ramp became lower in between. The difference in time may be due to these and not to the mass. Fix: both balls are released from rest from the same marked spot with the same slope.
2. Each ball was measured only once. A time taken by hand has random error, so a single measurement does not show how much the result varies. Fix: at least five repeats for each ball and their mean.
3. The results are given too precisely. The stopwatch app shows thousandths of a second, but a person's reaction time varies by about $0.1\text{–}0.2\,\text{s}$. Fix: the times are given to the nearest hundredth of a second and the uncertainty is estimated from the variation of the repeats.
4. The uncertainty was not estimated. The difference of $0.33\,\text{s}$ may be entirely measurement error. Fix: the uncertainty is estimated from the repeats and the difference is compared with it.
5. The conclusion generalises too much. Nothing can be concluded about all objects from one ramp and two balls. Fix: the conclusion is limited to the conditions studied.
6. The research question is vague. It does not say what "faster" means or what is measured. Fix: "Does the time a ball takes to roll down a ramp depend on its mass when the balls are of the same size and solid?"

b)

Given

the times of the steel ball $2.31$; $2.45$; $2.38$; $2.29$; $2.42\,\text{s}$ and the times of the plastic ball $2.40$; $2.33$; $2.47$; $2.36$; $2.44\,\text{s}$

Find

the means $\bar{t}$ and estimates of the uncertainty of the results $\Delta t$, and a conclusion about the difference

Principle

The mean of the repeats is the best estimate of the rolling time. Half of the range is a simple estimate of the scatter of the individual measurements, and here it is used as a simple, cautious uncertainty estimate for the comparison. The uncertainty of the mean itself is usually smaller than this. A difference can be said to be shown only if it is clearly larger than these uncertainty estimates.

Formula

$\bar{t}$$= \dfrac{t_1 + t_2 + t_3 + t_4 + t_5}{5}$
$\Delta t$$= \dfrac{t_\text{max} - t_\text{min}}{2}$

Conversions

Not needed, all the times are in seconds.

Substitution

Steel:

$\bar{t}$$= \dfrac{11.85\,\text{s}}{5}$
$= 2.37\,\text{s}$
$\Delta t$$= \dfrac{2.45\,\text{s} - 2.29\,\text{s}}{2}$
$= 0.08\,\text{s}$

Plastic:

$\bar{t}$$= \dfrac{12.00\,\text{s}}{5}$
$= 2.40\,\text{s}$
$\Delta t$$= \dfrac{2.47\,\text{s} - 2.33\,\text{s}}{2}$
$= 0.07\,\text{s}$

The difference of the means is $2.40\,\text{s} - 2.37\,\text{s} = 0.03\,\text{s}$.

Answer

Steel ball $t = (2.37 \pm 0.08)\,\text{s}$ and plastic ball $t = (2.40 \pm 0.07)\,\text{s}$, to the nearest hundredth of a second like the measurements. The difference of $0.03\,\text{s}$ is smaller than either uncertainty, and the intervals overlap. Based on the measurements, a difference between the rolling times cannot be shown with this measurement precision. The conclusion does not change even if the uncertainty of the means is estimated more precisely: for the means of five repeats it is about $0.03\,\text{s}$, and the difference is not clearly larger than that.

Reasonableness

The variation of the repeats, about $0.1\,\text{s}$, is of the order of the reaction time when timing by hand, so the size of the uncertainty is believable.

c)

Find

a conclusion that answers the research question and takes the uncertainty into account

Principle

A conclusion answers the research question, gives the result with its uncertainties and limits the inference to the conditions in which the measurements were made.

Model answer

The measurements did not show that the mass of the ball affects the rolling time: any difference remained smaller than the measurement uncertainty. The rolling time of the steel ball ($110\,\text{g}$) was $(2.37 \pm 0.08)\,\text{s}$ and that of the plastic ball ($16\,\text{g}$) was $(2.40 \pm 0.07)\,\text{s}$. The difference of $0.03\,\text{s}$ is smaller than the uncertainty of the measurements, so the hypothesis that a heavy ball rolls faster received no support. The result applies to solid balls of the same size on this ramp. To detect a smaller difference, more precise timing would be needed, for example light gates.

Also accepted

  • The uncertainty can also be estimated in another justified way, for example with the standard error of the mean. The conclusion is the same: the difference is smaller than the uncertainty.

Common mistakes

  • Something that does not weaken the investigation is named as a weakness, for example the length of the ramp.
  • A weakness is named, but no fix is given or the fix does not remove the problem.
  • The mean is given with all the digits of the calculator, for example $2.370\,\text{s}$, or the uncertainty is left out.
  • The conclusion "mass does not affect anything" generalises too much, in the same way as the original one.

Marking and self-assessment

Compare your solution with the model answer and, for each item, choose the points you think your solution earns. Part marks are possible, and any other physically correct method is accepted too.

a)8 p

The first weakness is named, and it is justified why it weakens the investigation. 1 p
The fix for the first weakness removes the problem. 1 p
The second weakness is named, and it is justified why it weakens the investigation. 1 p
The fix for the second weakness removes the problem. 1 p
The third weakness is named, and it is justified why it weakens the investigation. 1 p
The fix for the third weakness removes the problem. 1 p
The fourth weakness is named, and it is justified why it weakens the investigation. 1 p
The fix for the fourth weakness removes the problem. 1 p

b)4 p

The means are correct: steel $2.37\,\text{s}$, plastic $2.40\,\text{s}$. 1 p
The uncertainty of the measurement has been estimated. For example half of the range as a simple, cautious estimate: $\Delta t \approx 0.08\,\text{s}$ and $0.07\,\text{s}$. 1 p
Compares the difference of the means, $0.03\,\text{s}$, with the uncertainties and concludes that a difference cannot be shown, because the difference is not clearly larger than the uncertainties. 1 p
The answer has the correct unit and a precision that matches the given data. The means are given to the nearest hundredth of a second, like the measurements. 1 p

c)3 p

The conclusion answers the question and is based on the data. Answers the question of whether mass affects the rolling time. 1 p
States the result with its uncertainties: no difference between the times was observed within the measurement uncertainty. 1 p
Limits the conclusion to the conditions studied (solid balls of the same size, this ramp) and does not generalise it to all objects. 1 p

Self-assessed points: 0 / 15 (15 items not assessed)

The points are your own estimate, not a checked grade. The marking is Ruukkilabra's own, written according to the features of a good matriculation examination answer, and it is not official marking by the Matriculation Examination Board (YTL).

Assess your own investigation report own work

In the chapter Your own experimental investigation: planning, carrying out and reporting you planned and carried out your own experiment and wrote a report about it. Take out your report and assess it with the criteria below. The criteria also show what belongs in a good investigation report, so you can already use them while writing your report.

Open the assessment criteria

For each item, choose the points you think your report earns.

The research question concerns the relationship between two measurable quantities, and the hypothesis is a justified prediction that a measurement can disprove. 2 p
The method shows the varied quantity, the measured quantity and the quantities kept constant, the equipment and the measurement range so precisely that someone else could repeat the experiment. 2 p
The measurements are in a table that shows the quantities, the units and the repeats. 2 p
The axes show the correct quantities and units, and the scales are evenly spaced. 1 p
The model is fitted to the points it suits, and the points have not been joined into a broken line. 1 p
The result (for example a mean or a slope) is calculated correctly, and its uncertainty is estimated. 2 p
The conclusion answers the question and is based on the data. The conclusion compares the result with the hypothesis and does not generalise outside the measurement range. 2 p
The main sources of error are named, and it is explained how they affect the result. The report also explains how the experiment could be improved. 2 p

Self-assessed points: 0 / 14 (8 items not assessed)

The points are your own estimate, not a checked grade. The marking is Ruukkilabra's own, written according to the features of a good matriculation examination answer, and it is not official marking by the Matriculation Examination Board (YTL).