RuukkilabraUpper secondary physics
FY4 · 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.

Three problems bring together the motion, forces, energy and momentum of the module. In each one, you decide which laws and models fit the situation and justify your choice.

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.

Comparing ski waxes 19 points

Simulated example data · The data are calculated from models and are not the result of a real measurement. In the first table the acceleration is constant; in the second the acceleration decreases as the speed grows because of air resistance. A little random variation is added to both, as in a measurement.

A coach at a ski club compares two glide waxes, A and B. The same skier, whose mass with equipment is $72\,\text{kg}$, skis down the same even slope with each wax. The slope has an angle of inclination of $4.0^\circ$. The skier stands upright and does not push with poles. A camera at the side of the slope films the runs, and the skier's speed is determined from the video every second. The timing starts when the skier passes the start mark.

$t\ (\text{s})$ $v_\text{A}\ (\text{m/s})$ $v_\text{B}\ (\text{m/s})$
$0$ $2.10$ $1.90$
$1$ $2.52$ $2.12$
$2$ $2.86$ $2.42$
$3$ $3.31$ $2.62$
$4$ $3.63$ $2.90$
$5$ $4.06$ $3.09$
$6$ $4.40$ $3.40$
$7$ $4.86$ $3.59$
$8$ $5.23$ $3.86$

To estimate the effect of air resistance, the skier made six more runs with wax A down the same slope, starting from different points, so that the skier passed a measuring section in the middle of the slope at different speeds. For each run the speed $v$ in the measuring section and the acceleration $a$ were determined over so short a distance that the speed changed only a little.

$v\ (\text{m/s})$ $a\ (\text{m/s}^2)$
$2.0$ $0.43$
$3.0$ $0.41$
$4.5$ $0.36$
$6.0$ $0.30$
$7.5$ $0.20$
$9.0$ $0.10$
  1. a) Draw the $(t, v)$ graph of both runs in the same coordinate system and fit a model to each set of points. 3 p
  2. b) Draw the free-body diagram of the skier during the run. Air resistance does not need to be taken into account. 3 p
  3. c) Determine the coefficient of kinetic friction for each wax. 5 p
  4. d) The coach concludes that wax A glides better. Evaluate how reliable the conclusion is. Take into account the uncertainty of the results and the fact that air resistance was left out. 3 p
  5. e) Draw the $(v, a)$ graph of the additional measurement. Determine from the graph the acceleration the skier would have without air resistance, and use it to estimate how much air resistance changed the coefficient of friction $\mu_\text{A}$ of part c. 5 p
Show the model answer and marking Open only when your own solution is ready.

Model answer

a)

Find

the $(t, v)$ graph of both runs and the model fitted to them

Graph

The horizontal axis shows time $t\ (\text{s})$ and the vertical axis shows speed $v\ (\text{m/s})$. The points of each wax are marked with their own markers, for example circles and squares.

$2$
$4$
$6$
$8$
$1$
$2$
$3$
$4$
$5$
$6$
$0$
$t\ (\text{s})$
$v\ (\text{m/s})$
$\text{wax A: } 0.39\,\tfrac{\text{m}}{\text{s}^2}$
$\text{wax B: } 0.245\,\tfrac{\text{m}}{\text{s}^2}$

Model answer

The points of both runs are almost on a straight line, so the speed grows steadily and the acceleration is almost constant. A separate straight line is fitted to each set of points, passing between the points, and the points are not joined into a broken line. The small scatter of the points on both sides of the line is the random variation of the measurement. The line of wax A is steeper, so the skier accelerates faster with it.

b)

Find

the free-body diagram of the skier during the run, when air resistance is not taken into account

Principle

The object is chosen to be the skier with equipment, and we look for what interacts with it: the Earth (weight) and the snow of the slope (normal force and friction). Air is left out according to the problem. The skier does not push with poles, so there is no forward force: the motion down the slope does not need a force of its own.

$\text{situation}$
$\overline{v}$
$\text{forces}$
$\overline{G}$
$\overline{N}$
$\overline{F}_{\mu}$
$\overline{G}_{x}$
$\overline{G}_{y}$
$\overline{a}$
$x$
$y$

Model answer

The skier is drawn as a rectangle parallel to the slope.
- The weight $\overline{G}$ starts from the centre and points straight down, not perpendicular to the slope.
- The normal force $\overline{N}$ starts from the lower surface and points perpendicular to the slope, away from it. It is equal to the component of the weight perpendicular to the slope, so it is slightly smaller than the weight.
- The kinetic friction $\overline{F}_\mu$ starts from the upper corner of the lower surface and points up the slope, because the skier slides down the slope.

The acceleration $\overline{a}$ points down the slope, so the sum of the forces also points down the slope: the component of the weight along the slope is larger than the friction. The acceleration arrow is drawn next to the object in its own colour, because it is not a force. The dashed arrows $G_x$ and $G_y$ in the model figure are the components of the weight along the slope and perpendicular to it, not new forces: together they replace the weight and help with the equations of motion in part c. They do not need to be drawn in your own free-body diagram. In the model figure the slope is drawn steeper and the friction larger than in the problem so that the arrows can be told apart: on a $4.0^\circ$ slope the normal force is almost equal to the weight and the friction is very short.

c)

Given

the angle of inclination of the slope $\alpha = 4.0^\circ$, $g = 9.81\,\text{m/s}^2$ and the graph of part a; mass $m = 72\,\text{kg}$

Find

the coefficients of kinetic friction $\mu_\text{A}$ and $\mu_\text{B}$

Principle

The graphs are straight lines, so the acceleration is constant and it is the slope of the line. By Newton's second law the acceleration is caused by the sum of the forces, so the friction force is deduced from the acceleration and the coefficient of friction from that. On a slope it is best to choose the axes along the slope, because the motion is along the slope.

Free-body diagram

The diagram of part b: the $x$-axis down the slope and the $y$-axis perpendicular to the slope, away from it. The components of the weight are $G_x = mg\sin\alpha$ down the slope and $G_y = mg\cos\alpha$ towards the slope. The normal force is in the direction of the $y$-axis and the friction opposite to the $x$-axis.

$x$
$y$
$\overline{G}$
$\overline{F}_{\mu}$
$\overline{N}$
$\overline{G}_{x}$
$\overline{G}_{y}$
$\overline{a}$

Reading the graph

The line of wax A passes through the points $(0\,\text{s}$; $2.10\,\text{m/s})$ and $(8\,\text{s}$; $5.22\,\text{m/s})$, so

$a_\text{A}$$= \dfrac{3.12\,\text{m/s}}{8\,\text{s}}$
$= 0.39\,\text{m/s}^2$

The line of wax B passes through the points $(0\,\text{s}$; $1.90\,\text{m/s})$ and $(8\,\text{s}$; $3.86\,\text{m/s})$, so

$a_\text{B}$$= \dfrac{1.96\,\text{m/s}}{8\,\text{s}}$
$= 0.245\,\text{m/s}^2$

Formula

$y$: $N - mg\cos\alpha = 0$, so $N = mg\cos\alpha$. $x$: $mg\sin\alpha - \mu N = ma$, that is, $mg\sin\alpha - \mu mg\cos\alpha = ma$. The mass cancels, and

$\mu$$= \dfrac{g\sin\alpha - a}{g\cos\alpha}$

Conversions

Not needed, the accelerations are in the unit $\text{m/s}^2$.

Substitution

$g\sin\alpha$$= 9.81\,\text{m/s}^2 \cdot \sin 4.0^\circ$
$= 0.684\,\text{m/s}^2$
$g\cos\alpha$$= 9.81\,\text{m/s}^2 \cdot \cos 4.0^\circ$
$= 9.786\,\text{m/s}^2$
$\mu_\text{A}$$= \dfrac{(0.684 - 0.39)\,\text{m/s}^2}{9.786\,\text{m/s}^2}$
$= 0.030$
$\mu_\text{B}$$= \dfrac{(0.684 - 0.245)\,\text{m/s}^2}{9.786\,\text{m/s}^2}$
$= 0.045$

Answer

$\mu_\text{A} \approx 0.030$

And $\mu_\text{B} \approx 0.045$. The coefficient of friction has no unit. Two significant figures, because the slopes are read from the graph to about two figures.

Reasonableness

The coefficient of friction between a waxed ski and snow is a few hundredths, much smaller than, for example, between a shoe and a floor. Both accelerations are smaller than $g\sin\alpha = 0.684\,\text{m/s}^2$, which the skier would get without friction. The mass $72\,\text{kg}$ was not needed, because the weight, the normal force and the friction are all proportional to the mass.

d)

Find

how reliable the conclusion "wax A glides better" is

Principle

The conclusion is reliable if the difference between the waxes is clearly larger than the uncertainty of the results and if the sources of error cannot turn the difference the other way round.

Model answer

Uncertainty. The points are close to the lines, so the slope can be read to an accuracy of about $\pm 0.02\,\text{m/s}^2$. In the coefficient of friction this means $0.02\,\text{m/s}^2 : 9.786\,\text{m/s}^2 \approx 0.002$. An uncertainty in the angle of the slope, for example $\pm 0.1^\circ$, changes both coefficients by about $0.002$ in the same direction, so it does not change the difference. The difference between the waxes, $0.045 - 0.030 = 0.015$, is many times larger.

Air resistance. Air resistance points up the slope like friction, so it reduces the acceleration. When it is left out, all of the slowing effect is counted as friction, and both coefficients of friction come out larger than the true ones. Air resistance grows with the speed, and in run A the skier moves faster, so the coefficient of wax A is overestimated more. The true advantage of wax A is therefore at least as large as the calculated one.

Conclusion. With this data wax A glides better. However, each wax has been tried only once, and the snow or the wind may change between the runs. A more certain result is obtained by repeating the runs alternately with each wax.

e)

Find

the $(v, a)$ graph of the additional measurement, the acceleration without air resistance, and the effect of air resistance on the coefficient of friction $\mu_\text{A}$ of part c

Graph

The speed $v\ (\text{m/s})$ is on the horizontal axis and the acceleration $a\ (\text{m/s}^2)$ on the vertical axis. Each run is marked as a point of its own, and a curved line is fitted to the points and extended as a dashed line to the vertical axis.

$2$
$4$
$6$
$8$
$10$
$0.1$
$0.2$
$0.3$
$0.4$
$0.5$
$0$
$v\ (\text{m/s})$
$a\ (\text{m/s}^2)$
$a_0 \approx 0.45\,\tfrac{\text{m}}{\text{s}^2}$

Model answer

The acceleration decreases more and more steeply as the speed increases, so a straight line does not fit the points; a curve does. The friction force does not depend on the speed, but air resistance grows with the speed, so the decrease in the acceleration is caused by air resistance. At a very low speed air resistance is almost zero. When the curve is extended to the speed $v = 0$, it gives the acceleration without air resistance: $a_0 \approx 0.45\,\text{m/s}^2$.

Substituting $a_0$ into the equation of part c: $\mu_\text{A} = \dfrac{g\sin\alpha - a_0}{g\cos\alpha} = \dfrac{(0.684 - 0.45)\,\text{m/s}^2}{9.786\,\text{m/s}^2} \approx 0.024$.

The value $0.030$ of part c is therefore about $0.006$ too large, an error of about a fifth of the value of part c, because there the slowing effect of air resistance was counted as friction. This is a systematic error: it always changes the result in the same direction, and repeating the runs does not remove it. The error is about three times the uncertainty of reading the slope, $0.002$, but smaller than the difference between the waxes, $0.015$. In the slower run B air resistance has a smaller effect, so wax A glides better also after the correction.

Also accepted

  • In part c the slope can also be obtained from a straight-line fit by a calculator or spreadsheet ($0.390$ and $0.245\,\text{m/s}^2$) or from two other points of the line far from each other. Slopes of $0.37\text{–}0.41\,\text{m/s}^2$ and $0.23\text{–}0.26\,\text{m/s}^2$ are accepted.
  • In part b the air resistance $\overline{F}_\text{v}$ may be drawn up the slope when it is named, but the problem does not require it.
  • In part e, $a_0 = 0.43\text{–}0.47\,\text{m/s}^2$ is accepted as the point where the curve meets the vertical axis, and the $\mu_\text{A} \approx 0.022\text{–}0.026$ calculated from it.

Common mistakes

  • The coefficient of friction is calculated as $\mu = a : g$ as on a horizontal surface, which forgets the inclination of the slope. Then wax A would seem to glide worse, because its acceleration is larger.
  • The normal force is taken to be $N = mg$, although on a slope $N = mg\cos\alpha$. At a small angle the error is small, but the justification is wrong.
  • A forward force is drawn in the free-body diagram because the skier moves down. The motion along the slope is caused by the component of the weight.
  • The components of the weight are drawn as forces of their own in addition to the weight, so the weight is counted twice.
  • The slope is read from two adjacent measurement points, so the random error of a single point has a large effect on the result.
  • In part e a straight line is fitted to the points, or the curve is stopped at the lowest measured speed, so the acceleration without air resistance is not found.

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. The points of the runs can be told apart, for example by different markers. 1 p
The model is fitted to the points it suits, and the points have not been joined into a broken line. A separate straight line has been fitted to each set of points. 1 p

b)3 p

Every force acting on the object is included, and there are no invented forces. Weight, the normal force from the slope and kinetic friction; no forward force. 1 p
Every force points in the physically correct direction. Weight straight down, the normal force perpendicular to the slope and away from it, and friction up the slope. 1 p
Each force arrow starts at a sensible point on the object, and each force has its own unambiguous name. The skier is a rectangle parallel to the slope; the weight starts from the centre, the normal force and friction from the contact surface, and each force has its own symbol. 1 p

c)5 p

The slope or other quantity is read visibly from the graph, using two points on the line that are far apart. The accelerations as the slopes of the lines: $a_\text{A} \approx 0.39\,\text{m/s}^2$ and $a_\text{B} \approx 0.245\,\text{m/s}^2$. 1 p
The equations of motion come from the free-body diagram: the axes and the positive direction are chosen, and the signs are right. $x$-axis down the slope: $mg\sin\alpha - \mu N = ma$, and $y$: $N - mg\cos\alpha = 0$. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $\mu = \dfrac{g\sin\alpha - a}{g\cos\alpha}$; the mass cancels. 1 p
The numerical values are substituted with their units, and the calculation is correct. The coefficients for both waxes are calculated in the same way. 1 p
The answer has the correct unit and a precision that matches the given data. $\mu_\text{A} \approx 0.030$ and $\mu_\text{B} \approx 0.045$, without a unit. 1 p

d)3 p

The uncertainty of the measurement has been estimated. For example, an uncertainty of $\pm 0.02\,\text{m/s}^2$ in reading the slope changes the coefficient of friction by only about $\pm 0.002$, which is much less than the difference between the waxes, $0.015$. 1 p
The main sources of error are named, and it is explained how they affect the result. Air resistance reduces the acceleration, so the calculated coefficient of friction is larger than the true one; in the faster run A the effect is larger. 1 p
The conclusion answers the question and is based on the data. The difference is larger than the uncertainty, so wax A glides better; certainty is obtained by repeating the runs. 1 p

e)5 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. A curved line that levels off at low speeds is fitted to the points; a straight line does not fit them. 1 p
The slope or other quantity is read visibly from the graph, using two points on the line that are far apart. The curve is extended to the speed $v = 0$, and the point where it meets the vertical axis is read: $a_0 \approx 0.45\,\text{m/s}^2$. 1 p
The main sources of error are named, and it is explained how they affect the result. $\mu_\text{A} \approx 0.024$. Air resistance is a systematic error: it made the value of part c larger by about $0.006$, which is more than the reading uncertainty $0.002$ but less than the difference between the waxes $0.015$, and repeating the runs does not remove it. 1 p

Self-assessed points: 0 / 19 (19 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).

Zip line 14 points

Made-up example data · The zip line and the measurement result are made-up examples.

On the zip line of an activity centre, the rider hangs in a harness under a trolley with wheels and glides along a steel cable over a hillside. The starting platform is $15.0\,\text{m}$ higher than the beginning of the brake, and the cable from the starting platform to the beginning of the brake is $180\,\text{m}$ long. The rider starts from rest. From the beginning of the brake onwards the cable is almost horizontal, and a spring brake stops the rider there.

The instructor tells the riders: "You arrive at the brake at over $60\,\text{km/h}$." With a radar meter, the speed of a $62\,\text{kg}$ rider at the beginning of the brake was measured as $11.2\,\text{m/s}$.

  1. a) What kind of calculation can the instructor's claim be based on? Calculate the speed at the beginning of the brake with this model. 3 p
  2. b) The measurement does not match the instructor's claim. Based on the measurement, determine how large a force on average opposes the rider's motion on the cable. 4 p
  3. c) The instructor estimates the speed of a $95\,\text{kg}$ rider at the beginning of the brake, assuming that the same force opposes the motion as in part b. Calculate the estimate. 3 p
  4. d) Another instructor claims: "A heavy rider does not arrive any faster, because friction also grows with the mass." Evaluate the model of each instructor. According to whose estimate should the brake be designed? 4 p
Show the model answer and marking Open only when your own solution is ready.

Model answer

a)

Given

the height difference from the starting platform to the beginning of the brake $h = 15.0\,\text{m}$, the rider starts from rest, $g = 9.81\,\text{m/s}^2$

Find

the speed $v$ at the beginning of the brake with the instructor's model

Principle

The instructor's figure is obtained when friction and air resistance are not taken into account. Then only the weight and the force of the cable through the trolley act on the rider. The force of the cable is perpendicular to the cable, that is, to the motion, so it does no work. Only the weight does work, so the mechanical energy is conserved: potential energy changes into kinetic energy. The shape and length of the cable then do not affect the final speed, only the height difference.

Formula

$mgh = \tfrac{1}{2}mv^2$, so $v = \sqrt{2gh}$

Conversions

Not needed in the calculation. For comparison $1\,\text{m/s} = 3.6\,\text{km/h}$.

Substitution

$v$$= \sqrt{2 \cdot 9.81\,\text{m/s}^2 \cdot 15.0\,\text{m}}$
$= \sqrt{294.3}\,\text{m/s}$
$= 17.2\,\text{m/s}$

$17.2 \cdot 3.6\,\text{km/h} = 61.8\,\text{km/h}$

Answer

$v \approx 17.2\,\text{m/s} \approx 62\,\text{km/h}$

Three significant figures like the height $15.0\,\text{m}$. The instructor's claim is thus based on a model in which mechanical energy is conserved.

Reasonableness

The same speed would be reached if the rider fell freely $15\,\text{m}$. It is the greatest possible speed, because in reality part of the energy changes into heat.

b)

Given

$m = 62\,\text{kg}$, $h = 15.0\,\text{m}$, the length of the cable from the starting platform to the beginning of the brake $s = 180\,\text{m}$ and the measured speed $v = 11.2\,\text{m/s}$

Find

the force $F_\text{res}$ opposing the motion on average

Principle

The measured $11.2\,\text{m/s} \approx 40\,\text{km/h}$ is much smaller than the $17.2\,\text{m/s}$ of part a, so mechanical energy is not conserved. The friction of the trolley wheels and air resistance oppose the motion for the whole distance and do negative work. By the work–energy principle $E_{\text{k},1} + E_{\text{p},1} + W_\text{other} = E_{\text{k},2} + E_{\text{p},2}$. The zero level is chosen to be the beginning of the brake, where the speed was measured. When the combined effect of the resisting forces is described by an average force $F_\text{res}$ pointing against the direction of motion, its work is $W_\text{other} = -F_\text{res}\,s$.

Formula

$mgh - F_\text{res}\,s = \tfrac{1}{2}mv^2$, so

$F_\text{res}$$= \dfrac{mgh - \tfrac{1}{2}mv^2}{s}$

Conversions

Not needed, all quantities are in base units.

Substitution

$E_\text{p}$$= 62\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 15.0\,\text{m}$
$= 9123\,\text{J}$
$E_\text{k}$$= \tfrac{1}{2} \cdot 62\,\text{kg} \cdot (11.2\,\text{m/s})^2$
$= 3889\,\text{J}$
$F_\text{res}$$= \dfrac{9123\,\text{J} - 3889\,\text{J}}{180\,\text{m}}$
$= \dfrac{5234\,\text{J}}{180\,\text{m}}$
$= 29.1\,\text{N}$

Answer

$F_\text{res} \approx 29\,\text{N}$

Two significant figures like the mass $62\,\text{kg}$.

Reasonableness

The force is about $5\,\%$ of the rider's weight $608\,\text{N}$, which fits the friction of the wheels and air resistance. However, the distance is long, so more than half of the energy changes into heat: $5234\,\text{J} : 9123\,\text{J} \approx 0.57$.

c)

Given

$m = 95\,\text{kg}$, $h = 15.0\,\text{m}$, $s = 180\,\text{m}$ and the work of the resisting force from part b, $F_\text{res}\,s = 5234\,\text{J}$ ($F_\text{res} \approx 29\,\text{N}$)

Find

the speed $v$ of the heavier rider at the beginning of the brake with the instructor's assumption

Principle

The same energy principle as in part b, but now the unknown is the speed. According to the assumption the work of the resisting force is the same as for the light rider, but the potential energy is larger in proportion to the mass.

Formula

$mgh - F_\text{res}\,s = \tfrac{1}{2}mv^2$, so

$v$$= \sqrt{\dfrac{2\,(mgh - F_\text{res}\,s)}{m}}$

Conversions

Not needed.

Substitution

$mgh$$= 95\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 15.0\,\text{m}$
$= 13\,979\,\text{J}$

and from part b $F_\text{res}\,s = 5234\,\text{J}$, so

$E_\text{k}$$= 13\,979\,\text{J} - 5234\,\text{J}$
$= 8745\,\text{J}$
$v$$= \sqrt{\dfrac{2 \cdot 8745\,\text{J}}{95\,\text{kg}}}$
$= \sqrt{184.1}\,\text{m/s}$
$= 13.6\,\text{m/s}$

Answer

$v \approx 14\,\text{m/s}$

(more precisely $13.6\,\text{m/s} \approx 49\,\text{km/h}$). Two significant figures, because the mass $95\,\text{kg}$ and the resisting force $29\,\text{N}$ are known to two figures.

Reasonableness

The speed is between the $11.2\,\text{m/s}$ of the light rider and the $17.2\,\text{m/s}$ of the frictionless model, as it should be: the same resisting work takes a smaller share of the energy of the heavy rider.

d)

Find

the estimate of each instructor's model and according to which estimate the brake is designed

Principle

The models differ in how the resisting force depends on the mass of the rider. From the energy equation we can see what each assumption means for the speed.

Model answer

The other instructor's model. If the whole resisting force were friction proportional to the normal force and thus to the mass, $F_\text{res} = k\,mg$, the energy equation would be $mgh - k\,mg\,s = \tfrac{1}{2}mv^2$. The mass cancels, and every rider would get the same speed, $11.2\,\text{m/s}$.

The resisting forces separated. The friction of the trolley wheels grows with the mass, because the normal force between the cable and the wheels grows with the rider's weight. Air resistance, on the other hand, depends on the speed and on the size and posture of the rider, not on the mass. For a heavy rider air resistance is therefore a smaller part of the weight, and they arrive faster than a light rider.

The estimates. The estimate $14\,\text{m/s}$ of part c is too large, because in it friction also stays the same, although in reality it grows with the mass. The other instructor's estimate $11.2\,\text{m/s}$ is too small, because air resistance does not grow with the mass. The true speed is between these.

Design. The brake is designed for the larger estimate, $14\,\text{m/s}$: if the true speed is smaller, the brake is certainly enough, but a brake designed for too small a speed could fail. The best information is obtained by measuring the speeds of heavy riders with the radar as well.

Common mistakes

  • In part b the whole difference between the potential energy and the kinetic energy is divided by the height $15.0\,\text{m}$ and not by the distance travelled by the resisting force, $180\,\text{m}$.
  • In part a the acceleration and distance along the cable are calculated using the angle, although conservation of energy needs only the height difference.
  • In part c the mass is assumed to cancel as in part a, which gives the same $11.2\,\text{m/s}$ without justifying why the resisting force would grow with the mass.
  • In part d it is only stated that "the heavy one goes faster" or "the mass does not matter", without comparing either claim with the resisting forces.

Also accepted

  • In parts b and c the rounded resisting force $F_\text{res} = 29\,\text{N}$ may be used: then $F_\text{res}\,s = 29\,\text{N} \cdot 180\,\text{m} = 5220\,\text{J}$, and the speed of part c is again $13.6\,\text{m/s} \approx 14\,\text{m/s}$. In part c, $13.6\,\text{m/s}$ and $14\,\text{m/s}$ are accepted.
  • In part d the brake may be designed for the upper limit of the frictionless model, $17.2\,\text{m/s}$, or for the estimate of part c with a safety margin for heavier riders. Either is a correct and safe conclusion when justified, and earns the point for the conclusion.

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

Chooses conservation of mechanical energy and justifies it: when friction and air resistance are left out, only the weight does work, because the force of the cable is perpendicular to the motion. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $mgh = \tfrac{1}{2}mv^2$, so $v = \sqrt{2gh}$ 1 p
The answer has the correct unit and a precision that matches the given data. $v \approx 17.2\,\text{m/s} \approx 62\,\text{km/h}$ 1 p

b)4 p

Explains the difference: friction and air resistance do negative work, so the mechanical energy decreases, and therefore uses the work–energy principle. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $mgh - F_\text{res}\,s = \tfrac{1}{2}mv^2$ solved for $F_\text{res}$ 1 p
The numerical values are substituted with their units, and the calculation is correct. $E_\text{p} \approx 9.12\,\text{kJ}$ and $E_\text{k} \approx 3.89\,\text{kJ}$ 1 p
The answer has the correct unit and a precision that matches the given data. $F_\text{res} \approx 29\,\text{N}$ 1 p

c)3 p

The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. The same energy equation with the new mass, solved for $v$ 1 p
The numerical values are substituted with their units, and the calculation is correct. $E_\text{k} \approx 13.98\,\text{kJ} - 5.23\,\text{kJ} \approx 8.74\,\text{kJ}$ 1 p
The answer has the correct unit and a precision that matches the given data. $v \approx 14\,\text{m/s}$ with two significant figures like the mass and the resisting force; $13.6\,\text{m/s}$ is also accepted. 1 p

d)4 p

Shows that if the whole resisting force were proportional to the mass, the mass would cancel from the energy equation and the speed would be the same $11.2\,\text{m/s}$ as for the light rider. 1 p
Separates the resisting forces: the friction of the trolley wheels grows with the normal force and thus with the mass, but air resistance depends on the speed and the size of the rider, not on the mass. 1 p
The assumption is named, and its effect on the result is justified. The true speed is between the estimates: the estimate of part c is too large, because friction grows with the mass, and the other instructor's estimate is too small, because air resistance does not grow with the mass. 1 p
The conclusion answers the question and is based on the data. The brake is designed for the larger speed, because for safety it is better to prepare for a speed that is too large than one that is too small. Designing it for the upper limit of the frictionless model, $17.2\,\text{m/s}$, is also correct. 1 p

Self-assessed points: 0 / 14 (14 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).

Rear-end collision at traffic lights 16 points

Made-up example data · The accident and the figures in the report are made-up examples.

Car A is standing at a red light, and its driver is pressing the brake. Car B coming from behind runs into the back of it. The cars lock together and slide together straight ahead until they stop. The wheels of both cars are locked for the whole slide.

From the police report:

  • the mass of car A with its driver is $1250\,\text{kg}$ and the mass of car B with its driver is $1700\,\text{kg}$
  • the cars' common sliding distance from the point of collision to the stop is $4.2\,\text{m}$ on horizontal, dry asphalt
  • the coefficient of friction between the locked tyres and the road is $0.75$
  • the speed limit is $40\,\text{km/h}$; the driver of car B says they were driving at less than $40\,\text{km/h}$.
  1. a) How large was the cars' common speed right after the collision? 4 p
  2. b) Estimate the speed of car B just before the collision. Is the driver's account true? 4 p
  3. c) The mass of the driver of car A is $68\,\text{kg}$. The collision lasted about $0.12\,\text{s}$, and during that time the seat back accelerated the driver along with the car to the speed of part a. How large an average force did the seat back exert on the driver? Compare it with the driver's weight. 3 p
  4. d) Explain why a head restraint adjusted close to the back of the head reduces neck injuries in a rear-end collision, and why a slightly flexible seat back is better than a completely rigid one. 3 p
  5. e) Evaluate the reliability of the result of part b. Name two assumptions and tell in which direction each one changes the estimate of the speed of car B. 2 p
Show the model answer and marking Open only when your own solution is ready.

Model answer

a)

Given

$m_\text{A} = 1250\,\text{kg}$, $m_\text{B} = 1700\,\text{kg}$, sliding distance $s = 4.2\,\text{m}$, coefficient of friction $\mu = 0.75$ and $g = 9.81\,\text{m/s}^2$

Find

the cars' common speed $v'$ right after the collision

Principle

After the collision the cars move as one object with the mass $m = m_\text{A} + m_\text{B}$. The road is horizontal, so there is no acceleration in the vertical direction and the normal force is equal to the weight, $N = mg$. The wheels are locked, so the kinetic friction is $F_\mu = \mu N = \mu mg$, and it is the only horizontal force. Friction does negative work until the kinetic energy has been used up (the work–energy principle). Newton's second law gives the same result: the deceleration is $\mu g$.

Formula

$\tfrac{1}{2}mv'^2 - \mu mg\,s = 0$, so $v' = \sqrt{2\mu g s}$

Conversions

Not needed. The masses cancel, so they are not needed in this part.

Substitution

$v'$$= \sqrt{2 \cdot 0.75 \cdot 9.81\,\text{m/s}^2 \cdot 4.2\,\text{m}}$
$= \sqrt{61.8}\,\text{m/s}$
$= 7.86\,\text{m/s}$

Answer

$v' \approx 7.9\,\text{m/s}$

Two significant figures like the distance $4.2\,\text{m}$ and the coefficient of friction $0.75$.

Reasonableness

The deceleration

$\mu g$$= 0.75 \cdot 9.81\,\text{m/s}^2$
$\approx 7.4\,\text{m/s}^2$

is typical for braking on dry asphalt. The speed $7.9\,\text{m/s} \approx 28\,\text{km/h}$ is smaller than the speed of car B before the collision, as it should be, because B had to accelerate car A as well.

b)

Given

$m_\text{A} = 1250\,\text{kg}$ at rest, $m_\text{B} = 1700\,\text{kg}$ and $v' = 7.86\,\text{m/s}$ from part a

Find

the speed $v_\text{B}$ of car B just before the collision, and whether the account is true

Principle

In the collision the cars act on each other with large forces which, by Newton's third law, are equally large and opposite. These internal forces do not change the cars' total momentum. The external force, the friction of the road, is much smaller than the collision forces, and the collision lasts only a fraction of a second, so the impulse of friction during the collision is small. Therefore momentum is almost conserved. The positive direction is chosen to be the direction of motion of car B.

$\text{before}$
$\text{B}$
$\text{A}$
$\overline{v}_\text{B}$
$\overline{p}_\text{B}$
$v_\text{A} = 0$
$\text{after}$
$\text{B}$
$\text{A}$
$\overline{v}{}'$
$\overline{p}{}'$
$+$

Formula

$m_\text{B}v_\text{B} + m_\text{A} \cdot 0 = (m_\text{A} + m_\text{B})\,v'$, so

$v_\text{B}$$= \dfrac{(m_\text{A} + m_\text{B})\,v'}{m_\text{B}}$

Conversions

For comparison $1\,\text{m/s} = 3.6\,\text{km/h}$.

Substitution

$v_\text{B}$$= \dfrac{2950\,\text{kg} \cdot 7.86\,\text{m/s}}{1700\,\text{kg}}$
$= 13.6\,\text{m/s}$

$13.6 \cdot 3.6\,\text{km/h} = 49\,\text{km/h}$

Answer

$v_\text{B} \approx 14\,\text{m/s}$

(more precisely $13.6\,\text{m/s} \approx 49\,\text{km/h}$). Two significant figures, because the sliding distance $4.2\,\text{m}$ and the coefficient of friction $0.75$ are known to two figures. The account is not true: the speed of car B at the moment of collision was about $9\,\text{km/h}$ over the limit. The wheels were locked right after the collision, so the driver had probably braked already before the collision, and before braking the speed was even greater.

Reasonableness

The common speed is smaller than $v_\text{B}$, because the same momentum is shared among a larger mass. The kinetic energy decreases in the collision from about $158\,\text{kJ}$ to about $91\,\text{kJ}$: about $40\,\%$ of it is used in crumpling the bodywork, which fits a perfectly inelastic collision.

c)

Given

the mass of the driver $m = 68\,\text{kg}$, the change in velocity $\Delta v = v' - 0 = 7.86\,\text{m/s}$ and the duration of the collision $\Delta t = 0.12\,\text{s}$

Find

the average force $F_\text{avg}$ of the seat back on the driver and its comparison with the weight

Principle

The driver is at first at rest and after the collision moves with the car at the speed $v'$. In the horizontal direction the driver's momentum is changed by the force of the seat back, so by the impulse principle its impulse is equal to the change in the driver's momentum. The force is not constant, so the average force is calculated.

Formula

$F_\text{avg}\,\Delta t = m\,\Delta v$, so

$F_\text{avg}$$= \dfrac{m\,\Delta v}{\Delta t}$

and the comparison $\dfrac{F_\text{avg}}{mg}$

Conversions

Not needed.

Substitution

$F_\text{avg}$$= \dfrac{68\,\text{kg} \cdot 7.86\,\text{m/s}}{0.12\,\text{s}}$
$= 4455\,\text{N}$
$\dfrac{F_\text{avg}}{mg}$$= \dfrac{4455\,\text{N}}{68\,\text{kg} \cdot 9.81\,\text{m/s}^2}$
$= \dfrac{4455\,\text{N}}{667\,\text{N}}$
$= 6.7$

Answer

$F_\text{avg} \approx 4.5\,\text{kN}$

That is, about $6.7$ times the driver's weight. Two significant figures like the duration of the collision $0.12\,\text{s}$.

Reasonableness

The driver's average acceleration is $7.86\,\text{m/s} : 0.12\,\text{s} \approx 66\,\text{m/s}^2$, almost $7g$. It is large but typical for a rear-end collision, and the force is spread over the whole area of the seat back, so the body can withstand it. The peak force is larger than the average.

d)

Find

why a head restraint and a flexible seat back reduce injuries

Principle

The velocity of an object changes only when a force acts on it (Newton's first law). The impulse needed for the change in velocity is $m\,\Delta v$, and the average force is $F_\text{avg} = m\,\Delta v : \Delta t$.

Model answer

In a rear-end collision the seat back pushes the driver's torso forward. The head is not attached to the seat back, so it tends to keep its old velocity, that is, to stay in place. The torso moves forward from under the head, and the neck bends backwards and stretches until the muscles and ligaments of the neck pull the head along. This sudden bending causes a neck injury, so-called whiplash.

When the head restraint is close to the back of the head, it pushes the head forward almost at the same time as the seat back pushes the torso. The head gets the impulse it needs directly from the head restraint, so the neck does not bend and does not have to transmit a large force to the head. A head restraint that is far away does not help, because the neck has time to bend before the head hits the restraint.

The change in the driver's velocity $\Delta v$ is the same whether the seat back is rigid or flexible. A flexible seat back lengthens the time $\Delta t$ in which the velocity changes, so the average force $F_\text{avg} = m\,\Delta v : \Delta t$ is smaller. For example, twice the time halves the average force.

e)

Find

two assumptions of part b and the direction of their effect on the estimate of the speed of car B

Principle

For each assumption we think about whether the true speed would be larger or smaller than the calculated one.

Model answer

Any two, justified:
- Momentum is conserved in the collision. Friction already acts during the collision and reduces the cars' common momentum. Friction is at most $\mu (m_\text{A} + m_\text{B})\,g \approx 21.7\,\text{kN}$, so during $0.12\,\text{s}$ its impulse is at most about $2600\,\text{N}\cdot\text{s}$, about $11\,\%$ of the momentum of car B. The true $v_\text{B}$ may thus be as much as about $15\,\text{m/s}$, so the calculated value is a lower limit.
- The coefficient of friction is $0.75$. If the true coefficient of friction was smaller, the same sliding distance corresponds to a smaller speed, and $v_\text{B}$ is smaller than the calculated one. With a larger coefficient it is larger.
- The wheels were locked for the whole distance. If some wheel turned at times, the friction was smaller, and the true $v_\text{B}$ is smaller than the calculated one.

The conclusion that the limit was exceeded is still strong: the effect of friction during the collision increases the estimate, and the coefficient of friction would have to be clearly smaller than $0.75$ for the speed to be under $40\,\text{km/h}$.

Also accepted

  • Part a can be solved with Newton's second law and the formula for uniformly decelerating motion: $a = \mu g$ and $v'^2 = 2as$.
  • In parts b and c the result of part a may be rounded, for example $7.9\,\text{m/s}$; results of $13.5\text{–}14\,\text{m/s}$ and $4.4\text{–}4.5\,\text{kN}$ are accepted.

Common mistakes

  • In part b conservation of kinetic energy is used, although the cars stay stuck together and kinetic energy is used in crumpling. Then $v_\text{B}$ comes out too small.
  • In part a friction is assumed to depend only on the mass of car B, although both cars slide together.
  • The speed calculated from the sliding distance is interpreted directly as the speed of car B before the collision.
  • In part c the speed is converted incorrectly from $\text{km/h}$, or the force is compared with the mass and not with the weight.
  • In part d it is said that "the head flies backwards". The head does not get a force directed backwards: the torso moves forward from under it.

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)4 p

Justifies the model: while sliding, the cars are slowed only by friction $\mu (m_\text{A} + m_\text{B}) g$, because on a horizontal road the normal force is equal to the weight. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $\tfrac{1}{2}mv'^2 = \mu mgs$, so $v' = \sqrt{2\mu gs}$; the mass cancels. 1 p
The answer has the correct unit and a precision that matches the given data. $v' \approx 7.9\,\text{m/s}$ 1 p
The reasonableness of the result has been assessed. For example, the deceleration $\mu g \approx 7.4\,\text{m/s}^2$ is suitable for braking on dry asphalt, and $7.9\,\text{m/s} \approx 28\,\text{km/h}$. 1 p

b)4 p

Justifies why momentum is almost conserved: the collision is short and the forces between the cars are much larger than friction, so the impulse of friction during the collision is small. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $m_\text{B}v_\text{B} = (m_\text{A} + m_\text{B})v'$ solved for $v_\text{B}$ 1 p
The answer has the correct unit and a precision that matches the given data. $v_\text{B} \approx 14\,\text{m/s} \approx 50\,\text{km/h}$ with two significant figures like the sliding distance and the coefficient of friction; $13.6\,\text{m/s} \approx 49\,\text{km/h}$ is also accepted. 1 p
The conclusion answers the question and is based on the data. The account is not true: the speed at the moment of collision was clearly over $40\,\text{km/h}$, and even greater before braking. 1 p

c)3 p

The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $F_\text{avg}\,\Delta t = m\,\Delta v$, so $F_\text{avg} = \dfrac{m\,\Delta v}{\Delta t}$ 1 p
The answer has the correct unit and a precision that matches the given data. $F_\text{avg} \approx 4.5\,\text{kN}$ 1 p
The reasonableness of the result has been assessed. The force is about $6.7$ times the driver's weight. 1 p

d)3 p

Explains with inertia: the seat back pushes the torso forward, but the head tends to keep its old velocity until the neck pulls it along, so the neck bends backwards. 1 p
Explains that a head restraint close by gives the head the same change in velocity at the same time as the torso, so the neck does not have to transmit a large force. 1 p
Justifies with impulse: the change in velocity is the same, but a flexible seat back lengthens the time $\Delta t$, so the average force $F_\text{avg} = m\,\Delta v : \Delta t$ becomes smaller. 1 p

e)2 p

The assumption is named, and its effect on the result is justified. For example, friction already acts during the collision and reduces the cars' common momentum, so the true $v_\text{B}$ is larger than the calculated one. 1 p
The assumption is named, and its effect on the result is justified. The other assumption, for example the value of the coefficient of friction or the wheels being locked for the whole distance, and the direction of its effect. 1 p

Self-assessed points: 0 / 16 (16 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).