Define proper length.

A charged pion decays spontaneously in a time of 26 ns as measured in the frame of reference in which it is stationary. The pion moves with a velocity of 0.96c relative to the Earth. Calculate the pion’s lifetime as measured by an observer on the Earth.

In the pion reference frame, the Earth moves a distance X before the pion decays. In the Earth reference frame, the pion moves a distance Y before the pion decays. Demonstrate, with calculations, how length contraction applies to this situation.

the length of an object in its rest frame

$\frac{1}{\sqrt{\left(1-{0.96}^2\right)}}$ OR $\gamma =3.6$
ECF for wrong $\gamma$

93 «ns»
Award [2] for a bald correct answer.

«X is» 7.5 «m» in frame on pion

«Y is» 26.8 «m» in frame on Earth 

identifies proper length as the Earth measurement
OR
identifies Earth distance according to pion as contracted length
OR
a statement explaining that one of the length is shorter than the other one

Define frame of reference.

In the reference frame of the laboratory the force on X is magnetic.

(i) State the nature of the force acting on X in this reference frame where X is at rest.

(ii) Explain how this force arises.

a coordinate system
OR
a system of clocks and measures providing time and position relative to an observer

OWTTE

i

electric
OR
electrostatic

ii

«as the positive ions are moving with respect to the charge,» there is a length contraction

therefore the charge density on ions is larger than on electrons

so net positive charge on wire attracts X

For candidates who clearly interpret the question to mean that X is now at rest in the Earth frame accept this alternative MS for bii
the magnetic force on a charge exists only if the charge is moving
an electric force on X , if stationary, only exists if it is in an electric field
no electric field exists in the Earth frame due to the wire
and look back at b i, and award mark for there is no electric or magnetic force on X

Calculate the time taken for the journey in the reference frame of twin A as measured on Earth.

Determine the time taken for the journey in the reference frame of twin B.

Draw, for the reference frame of twin A, a spacetime diagram that represents the worldlines for both twins.

Suggest how the twin paradox arises and how it is resolved.

«0.6 ct = 6 ly» so t = 10 «years»

 Accept: If the 6 ly are considered to be measured from B, then the answer is 12.5 years.

ALTERNATIVE 1

10² − 6² = t² − 0²

so t is 8 «years»
Accept: If the 6 ly are considered to be measured from B, then the answer is 10 years.

ALTERNATIVE 2

gamma is $\frac{5}{4}$

10 × $\frac{4}{5}$ = 8 «years»

Allow ECF from a
Allow ECF for incorrect γ in mp1

three world lines as shown

Award mark only if axes OR world lines are labelled.

according to both twins, it is the other one who is moving fast therefore clock should run slow

Allow explanation in terms of spacetime diagram.

«it is not considered a paradox as» twin B is not always in the same inertial frame of reference

OR

twin B is actually accelerating «and decelerating»

State which of the two time intervals is a proper time.

Calculate the speed v of the train for the ratio $\frac{\mathrm{\Delta }{t}_{\text{P}}}{\mathrm{\Delta }{t}_{\text{Q}}}=0.30$.

Later the train is travelling at a speed of 0.60c. Observer P measures the length of the train to be 125 m. The train enters a tunnel of length 100 m according to observer Q.

Show that the length of the train according to observer Q is 100 m.

Draw the time $c{t}^{\prime }$ and space $x^{\prime }$ axes for observer P’s reference frame on the spacetime diagram.

Deduce, using the spacetime diagram, which light was turned on first according to observer P.

Apply a Lorentz transformation to show that the time difference between events B and F according to observer P is 2.5 × 10^–7 s.

Demonstrate that the spacetime interval between events B and F is invariant.

A second train is moving at a velocity of –0.70c with respect to the ground.

Calculate the speed of the second train relative to observer P.

Δt_P / observer sitting in the train

[1 mark]

$\gamma =\frac{\mathrm{\Delta }{t}_{\text{Q}}}{\mathrm{\Delta }{t}_{\text{P}}}=$ «$=\frac{1}{0.30}$» = 3.3

to give v = 0.95c

[2 marks]

$\gamma$ = 1.25

«length of train according Q» = 125/1.25

«giving 100m»

[2 marks]

axes drawn with correct gradients of $\frac{5}{3}$ for $c{t}^{\prime }$ and 0.6 for $x^{\prime }$

Award [1] for one gradient correct and another approximately correct.

[1 mark]

lines parallel to the $x^{\prime }$ axis and passing through B and F

intersections on the $c{t}^{\prime }$ axis at $\text{B'}$ and $\text{F'}$ shown

light at the front of the train must have been turned on first

[3 marks]

$\mathrm{\Delta }{t}^{\prime }=1.25\times \frac{0.6\times 100}{3\times {10}^8}$

«2.5 × 10⁻⁷»

Allow ECF for gamma from (c).

[1 mark]

according to P: ${\left(3\times {10}^8\times 2.5\times {10}^{-7}\right)}^2-{125}^2=$ «−» 10000

according to Q: ${\left(3\times {10}^8\times 0\right)}^2-{100}^2=$ «−» 10000

[2 marks]

$u^{\prime }=\frac{-0.7-0.6}{1+0.7\times 0.6}$ c

= «−» 0.92c

[2 marks]

State one prediction of Maxwell’s theory of electromagnetism that is consistent with special relativity.

A current is established in a long straight wire that is at rest in a laboratory.

A proton is at rest relative to the laboratory and the wire.

Observer X is at rest in the laboratory. Observer Y moves to the right with constant speed relative to the laboratory. Compare and contrast how observer X and observer Y account for any non-gravitational forces on the proton.

the speed of light is a universal constant/invariant
OR
c does not depend on velocity of source/observer

electric and magnetic fields/forces unified/frame of reference dependant

[1 mark]

observer X will measure zero «magnetic or electric» force

observer Y must measure both electric and magnetic forces

which must be equal and opposite so that observer Y also measures zero force

Allow [2 max] for a comment that both X and Y measure zero resultant force even if no valid explanation is given.

[3 marks]

Calculate the length of the rocket according to X.

A space shuttle is released from the rocket. The shuttle moves with speed 0.20c to the right according to X. Calculate the velocity of the shuttle relative to the rocket.

the time interval between the lamps turning on.

which lamp turns on first.

On the diagram label the coordinates x and ct.

State and explain whether the ct coordinate in (c)(i) is less than, equal to or greater than 1.0 m.

Calculate the value of c ²t ² – x ².

the gamma factor is $\frac{5}{3}$ or 1.67

L = $\frac{450}{\frac{5}{3}}$ = 270 «m»

Allow ECF from MP1 to MP2.

[2 marks]

u' = «$\frac{u-v}{1-\frac{uv}{c^2}}=$» $\frac{0.20c-0.80c}{1-0.20\times 0.80}$
OR
0.2c = $=\frac{0.80c+u^{\prime }}{1+0.80u^{\prime }}$

u' =  «–»0.71c

Check signs and values carefully.

[2 marks]

Δt' = «$\gamma \left(\mathrm{\Delta }t-\frac{v\mathrm{\Delta }x}{c^2}\right)=$» $\frac{5}{3}\times \left(0-\frac{\left(0.80c\times 9000\right)}{c^2}\right)$

Δt' = «–»4.0 x 10^–5 «s»

Allow ECF for use of wrong $\gamma$ from (a)(i).

[2 marks]

lamp 2 turns on first

Ignore any explanation

[1 mark]

x coordinate as shown

ct coordinate as shown

Labels must be clear and unambiguous.

Construction lines are optional.

[2 marks]

«in any other frame» ct is greater

the interval ct' = 1.0 «m» is proper time
OR
ct is a dilated time
OR
ct = $\gamma$ct' «= $\gamma$»

MP1 is a statement

MP2 is an explanation

[2 marks]

use of c ²t ² – x ² =  c ²t' ² – x' ²

c ²t ² – x ² = 1² – 0² = 1 «m²»

for MP1 equation must be used.

Award [2] for correct answer that first finds x (1.33 m) and ct (1.66 m)

[2 marks]

Muons are unstable particles with a proper lifetime of 2.2 μs. Muons are produced 2.0 km above ground and move downwards at a speed of 0.98c relative to the ground. For this speed $\gamma$ = 5.0. Discuss, with suitable calculations, how this experiment provides evidence for time dilation.

ALTERNATIVE 1 — for answers in terms of time
overall idea that more muons are detected at the ground than expected «without time dilation»

«Earth frame transit time = $\frac{2000}{0.98c}$» = 6.8 «μs»

«Earth frame dilation of proper half-life = 2.2 μs x 5» = 11 «μs»
OR
«muon’s proper transit time = $\frac{6.8\mu s}{5}$» = 1.4 «μs»

ALTERNATIVE 2 – for answers in terms of distance
overall idea that more muons are detected at the ground than expected «without time dilation»

«distance muons can travel in a proper lifetime = 2.2 μs x 0.98c» = 650 «m»

«Earth frame lifetime distance due to time dilation = 650 m x 5» = 3250 «m»
OR
«muon frame distance travelled = $\frac{2000}{5}$» = 400 «m»

Accept answers from one of the alternatives.

[3 marks]

Calculate the angle between the worldline of S and the worldline of the Earth.

Draw, on the diagram, the x′-axis for the reference frame of S.

An event Z is shown on the diagram. Label the co-ordinates of this event in the reference frame of S.

angle = tan^–1 «$\frac{0.8}{1}$» = 39 «^o» OR 0.67 «rad»

adds x′-axis as shown

Approximate same angle to v = c as for ct′.

Ignore labelling of that axis.

adds two lines parallel to ct′ and x′ as shown indicating coordinates

State what is meant by a reference frame.

State and explain whether the force experienced by P is magnetic, electric or both, in reference frame S.

State and explain whether the force experienced by P is magnetic, electric or both, in the rest frame of P.

a set of coordinate axes and clocks used to measure the position «in space/time of an object at a particular time»
OR
a coordinate system to measure x,y,z, and t / OWTTE

[1 mark]

magnetic only

there is a current but no «net» charge «in the wire»

[2 marks]

electric only

P is stationary so experiences no magnetic force

relativistic contraction will increase the density of protons in the wire

[3 marks]

Maxwell’s equations led to the constancy of the speed of light. Identify what Maxwell’s equations describe.

State a postulate that is the same for both special relativity and Galilean relativity.

Identify the nature of the attractive force recorded by an observer stationary with respect to the wires.

A second observer moves at the drift velocity of the electron current in the wires. Discuss how this observer accounts for the force between the wires.

mention of electric AND magnetic fields ✓
OR
mention of electromagnetic radiation/wave/fields ✓

the laws of physics are the same in all «inertial» frames of reference/for all «inertial» observers ✓

OWTTE

magnetic ✓

«In observer frame» protons «in the two wires» move in same/parallel direction ✓

these moving protons produce magnetic attraction ✓

there is also a smaller electrostatic repulsion due to wires appearing positive due to length contraction «of proton spacing» ✓

OWTTE

Easy introduction fairly well answered by most candidates.

With a few exceptions referring to Newton's first, this was very well answered.

Most scored by recognizing the force as magnetic.

Many failed to recognize that the magnetic force would still be present due to the current produced by the relative motion of the protons in both wires, and only focused on the repulsive electrostatic due to length contraction.

The Lorentz transformations assume that the speed of light is constant. Outline what the Galilean transformations assume.

Deduce the length of the probe as measured by an observer in the spaceship.

Explain which of the lengths is the proper length.

Calculate the speed of the probe in terms of $c$, relative to Earth.

constancy of time
OR
speed of light > c is possible ✓

OWTTE

$\gamma =1.15$ ✓

length = $6.9 «\mathrm{m}»$ ✓

Allow length in the range $\mathit{6}\mathit{.}\mathit{7}$ to $\mathit{7}\mathit{.}\mathit{0} m$.

Allow ECF from wrong $\gamma$

Award [2] marks for a bald correct answer in the range indicated above.

$8.0 \mathrm{m}$ / measurement made on the probe ✓

the measurement made by an observer at rest in the frame of the probe ✓

$u=\frac{0.5c+0.8c}{1+{\displaystyle \frac{0.5c\times 0.8c}{c^2}}}$ ✓

$u=0.93\text{c}$ ✓

Allow all negative signs for velocities

Award [2] marks for a bald correct answer

Although the expected answers was the constancy of time, the markscheme allowed references to the speed of light not being constant, as this was a common answer, deriving from the stem used in the question.

Very well answered.

"In the same frame" does not highlight the need to be "at rest" in that frame, and was the most frequent wrong answer, although a vast majority scored full marks here.

Very well answered.

Define frame of reference.

Calculate, according to the observer on Earth, the time taken for A and B to meet.

Identify the terms in the formula.

u′ = $\frac{u-v}{1-\frac{uv}{c^2}}$

Determine, according to an observer in A, the velocity of B.

Determine, according to an observer in A, the time taken for B to meet A.

Deduce, without further calculation, how the time taken for A to meet B, according to an observer in B, compares with the time taken for the same event according to an observer in A.

a co-ordinate system in which measurements «of distance and time» can be made

Ignore any mention to inertial reference frame.

closing speed = c

2 «s»

u and v are velocities with respect to the same frame of reference/Earth AND u′ the relative velocity

Accept 0.4c and 0.6c for u and v

$\frac{-0.4-0.6}{1+0.24}$

«–» 0.81c

$\gamma$ = 1.25

so the time is t = 1.6 «s»

gamma is smaller for B

so time is greater than for A

Calculate, according to Galilean relativity, the time taken for a muon to travel to the ground.

Deduce why only a small fraction of the total number of muons created is expected to be detected at ground level according to Galilean relativity.

Calculate, according to the theory of special relativity, the time taken for a muon to reach the ground in the reference frame of the muon.

Discuss how your result in (b)(i) and the outcome of the muon decay experiment support the theory of special relativity.

«$\frac{{10}^4}{0.995\times 3\times {10}^8}=$» 34 «μs»

Do not accept 10⁴/c = 33 μs.

[1 mark]

time is much longer than 10 times the average life time «so only a small proportion would not decay»

[1 mark]

$\gamma =10$

$\mathrm{\Delta }{t}_0=$ «$\frac{\mathrm{\Delta }t}{\gamma }=\frac{34}{10}=$» 3.4 «μs»

[2 marks]

the value found in (b)(i) is of similar magnitude to average life time

significant number of muons are observed on the ground

«therefore this supports the special theory»

[2 marks]

Draw on the spacetime diagram the worldline of B according to the Earth observer and label it B.

Deduce, showing your working on the spacetime diagram, the value of ct according to the Earth observer at which the rocket B emitted its flash of light.

Explain whether or not the arrival times of the two flashes in the Earth frame are simultaneous events in the frame of rocket A.

Calculate the velocity of rocket B relative to rocket A.

straight line with negative gradient with vertical intercept at ct = 1.2 «km»

through (–0.6, 2.2) ie gradient = –1.67

Tolerance: Allow gradient from interval –2.0 to –1.4, (at ct = 2.2, x from interval 0.5 to 0.7).

If line has positive gradient from interval 1.4 to 2.0 and intercepts at ct = 1.2 km then allow [1 max].

[2 marks]

line for the flash of light from A correctly drawn

line for the flash of light of B correctly drawn

correct reading taken for time of intersection of flash of light and path of B, ct = 2.4 «km»

Accept values in the range: 2.2 to 2.6.

[3 marks]

the two events take place in the same point in space at the same time

so all observers will observe the two events to be simultaneous / so zero difference

Award the second MP only if the first MP is awarded.

[2 marks]

$u^{\prime }=\frac{-0.6-0.8}{1-(-0.6)\times 0.8}$

= «–»0.95 «c»

[2 marks]

Explain what is meant by the statement that the spacetime interval is an invariant quantity.

calculate the spacetime interval.

determine the time between them according to observer B.

Outline why the observed times are different for A and B.

quantity that is the same/constant in all inertial frames

[1 mark]

spacetime interval = 27² – 15² = 504 «m²»

[1 mark]

ALTERNATIVE 1

Evidence of x′ = 0

t′ «$\frac{\sqrt{504}}{c}$» = 7.5 × 10^–8 «s»

ALTERNATIVE 2

γ = 1.2

t′ «= $\frac{9\times {10}^{-8}}{1.2}$» = 7.5 × 10^–8 «s»

[2 marks]

observer B measures the proper time and this is the shortest time measured

OR

time dilation occurs «for B's journey» according to A

OR

observer B is stationary relative to the particle, observer A is not

[1 mark]

Explain what is meant by proper length.

Draw a spacetime diagram for this situation according to an observer at rest relative to the tunnel.

Calculate the velocity of the train, according to an observer at rest relative to the tunnel, at which the train fits the tunnel.

For an observer on the train, it is the tunnel that is moving and therefore will appear length contracted. This seems to contradict the observation made by the observer at rest to the tunnel, creating a paradox. Explain how this paradox is resolved. You may refer to your spacetime diagram in (b).

the length of an object in its rest frame

OR

the length of an object measured when at rest relative to the observer

world lines for front and back of tunnel parallel to ct axis

world lines for front and back of train

which are parallel to ct′ axis

realizes that gamma = 1.25

0.6c

ALTERNATIVE 1

indicates the two simultaneous events for t frame

marks on the diagram the different times «for both spacetime points» on the ct′ axis «shown as Δt′ on each diagram»

ALTERNATIVE 2: (no diagram reference)

the two events occur at different points in space

statement that the two events are not simultaneous in the t′ frame

State what is meant by a reference frame.

explain why the time coordinate of E in frame S is $t=\frac{L}{c}$.

hence show that the space coordinate of E in frame S is $x=L+\frac{vL}{c}$.

a set of rulers and clocks / set of coordinates to record the position and time of events ✔

ALTERNATIVE 1:

the time in frame S′ is $t^{\prime }=\frac{L}{c}$ ✔

but time is absolute in Galilean relativity so is the same in S ✔

ALTERNATIVE 2:

In frame S, light rays travel at c + v ✔

so $t=\frac{L}{\left(c+v\right)-v}=\frac{L}{c}$ ✔

In Alternative 1, they must refer to S'

x = x' + vt and x' = L ✔

«substitution to get answer»

Calculate the speed of the probe relative to the ground.

Determine the time it takes the probe to reach the front of the rocket according to an observer at rest in the rocket.

Determine the time it takes the probe to reach the front of the rocket according to an observer at rest on the ground.

$\frac{0.82c+0.40c}{1+\frac{0.82c\times 0.40c}{c^2}}$ ✔

0.92c ✔

$\mathrm{\Delta }{t}^{\prime }=\frac{120}{0.40c}$ ✔

$\mathrm{\Delta }{t}^{\prime }=1.0\times {10}^{-6}$ «s» ✔

$\gamma =$ «$\frac{1}{\sqrt{1-{0.82}^2}}=$» 1.747 ✔

Δt = «$\gamma \left(\mathrm{\Delta }{t}^{\prime }+\frac{v\mathrm{\Delta }{x}^{\prime }}{c^2}\right)$» = $1.747\times \left(1.0\times {10}^{-6}+\frac{0.82c\times 120}{c^2}\right)$

OR

Δt = $\frac{120}{1.747\times \left(0.92-0.82\right)c}$ ✔

2.3 × 10⁻⁶ «s» ✔

x′ = 1.5 m.

ct′ = –1.1 m.

Label, on the diagram, the space coordinate of event E in the S′ frame. Label this event with the letter P.

Label, on the diagram, the event that has coordinates x′ = 1.0 m and ct′ = 0. Label this event with the letter Q.

Using the spacetime diagram, outline without calculation, why observers in frame S′ measure the length of the
rod to be less than 1.0 m.

Using the spacetime diagram, estimate, in m, the length of this rod in the S′ frame.

$\gamma =$ «$\frac{1}{\sqrt{1-{0.745}^2}}=$» 1.499 ✔

x′ = «$\gamma \left(x-vt\right)=$» 1.499 × (1.0 − 0) ✔

«x′ = 1.5 m»

t′ = «$\gamma \left(t-\frac{vx}{c^2}\right)$ =» $1.499\times \left(0-\frac{0.745c\times 1}{c^2}\right)$ «= $-\frac{1.11}{c}$»

«ct′ = –1.1 m»

OR

using spacetime interval 0 − 1² = (ct′)² − 1.5² ⇒ «ct′ = –1.1» ✔

line through event E parallel to ct′ axis meeting x' axis and labelled P ✔

point on x' axis about $\frac{2}{3}$ of the way to P labelled Q ✔

ends of rod must be recorded at the same time in frame S′ ✔

any vertical line from E crossing x’, no label required ✔

right-hand end of rod intersects at R «whose co-ordinate is less than 1.0 m» ✔

0.7 m ✔

State whether the field around the wire according to observer X is electric, magnetic or a combination of both.

Discuss the change in d according to observer Y.

Deduce whether the overall field around the wire is electric, magnetic or a combination of both according to observer Y.

magnetic field

[1 mark]

«according to Y» the positive charges are moving «to the right»

d decreases

For MP1, movement of positive charges must be mentioned explicitly.

[2 marks]

positive charges are moving, so there is a magnetic field

the density of positive charges is higher than that of negative charges, so there is an electric field

The reason must be given for each point to be awarded.

[2 marks]

Calculate, for the reference frame of rocket A, the speed of rocket B according to the Galilean transformation.

Calculate, for the reference frame of rocket A, the speed of rocket B according to the Lorentz transformation.

Outline, with reference to special relativity, which of your calculations in (a) is more likely to be valid.

1.25c

[1 mark]

ALTERNATIVE 1

$u^{\prime }=\frac{(0.50+0.75)}{1+0.5\times 0.75}c$

0.91c

ALTERNATIVE 2

$u^{\prime }=\frac{-0.50-0.75}{1-(-0.5\times 0.75)}c$

–0.91c

[2 marks]

nothing can travel faster than the speed of light (therefore (a)(ii) is the valid answer)

OWTTE

[1 mark]

Calculate the velocity of the spaceship relative to the Earth.

The spaceship passes the space station 90 minutes later as measured by the spaceship clock. Determine, for the reference frame of the Earth, the distance between the Earth and the space station.

As the spaceship passes the space station, the space station sends a radio signal back to the Earth. The reception of this signal at the Earth is event A. Determine the time on the Earth clock when event A occurs.

Construct event A and event B on the spacetime diagram.

Estimate, using the spacetime diagram, the time at which event B occurs for the spaceship.

0.60c

OR

1.8 × 10⁸ «m s^–1»

[1 mark]

ALTERNATIVE 1

time interval in the Earth frame = 90 × γ = 112.5 minutes

«in Earth frame it takes 112.5 minutes for ship to reach station»

so distance = 112.5 × 60 × 0.60c

1.2 × 10m¹² «m»

ALTERNATIVE 2

Distance travelled according in the spaceship frame = 90 × 60 × 0.6c

= 9.72 × 10¹¹ «m»

Distance in the Earth frame «= 9.72 × 10¹¹ × 1.25» = 1.2 × 10¹² «m»

[3 marks]

signal will take «112.5 × 0.60 =» 67.5 «minutes» to reach Earth «as it travels at c»

OR

signal will take «$\frac{1.2\times {10}^{12}}{3\times {10}^8}$ =» 4000 «s»

total time «= 67.5 + 112.5» = 180 minutes or 3.00 h or 3:00am

[2 marks]

line from event E to A, upward and to left with A on ct axis (approx correct)

line from event A to B, upward and to right with B on ct' axis (approx correct)

both lines drawn with ruler at 45 (judge by eye)

eg:

[3 marks]

ALTERNATIVE 1

«In spaceship frame»

Finds the ratio $\frac{OB}{OE}$ (or by similar triangles on x or ct axes), value is approximately 4

hence time elapsed ≈ 4 × 90 mins ≈ 6h «so clock time is ≈ 6:00»

Alternative 1:

Allow similar triangles using x-axis or ct-axis, such as $\frac{distance2}{distance1}$ from diagrams below

ALTERNATIVE 2

«In Earth frame»

Finds the ratio

$\frac{ct\text{ coordinate of B}}{ct\text{ coordinate of A}}$, value is approximately 2.5

hence time elapsed ≈ $\frac{2.5\times 3\text{h}}{1.25}$ ≈ 6h

«so clocktime is ≈ 6:00»

ALTERNATIVE 2:

[2 marks]

Define an inertial reference frame.

As the spaceship passes the Earth it emits a flash of light that travels in the same direction as the spaceship with speed c as measured by an observer on the spaceship. Calculate, according to the Galilean transformation, the speed of the light in the Earth’s reference frame.

Use your answer to (a)(ii) to describe the paradigm shift that Einstein’s theory of special relativity produced.

a coordinate system which is not accelerating/has constant velocity/Newtons 1st law applies ✔

OWTTE

Both “inertial” and “reference frame” need to be defined

1.5c ✔

c is the same in all frames

OR

c is maximum velocity possible ✔

velocity addition frame dependent ✔

length/time/mass/fields relative measurements ✔

Newtonian/Galilean mechanics valid only at low speed ✔

In defining an inertial frame of reference far too many candidates started with the words ‘ a frame of reference that...... ’ instead of ‘a coordinate system that.....’

Almost no incorrect answers were seen.

Most candidates correctly stated that in special relativity the velocity of light, c, is the maximum possible velocity or is invariant. Only a few added that Galilean relativity only applies at speeds much less than the speed of light.

Show, using the spacetime diagram, that the speed of the spaceship relative to the Earth is 0.80c.

Label, with the letter E, the event of the spaceship going past the planet.

Determine, according to an observer on the spaceship as the spaceship passes the planet, the time shown by the clock on the spaceship.

Determine, according to an observer on the spaceship as the spaceship passes the planet, the time shown by the clock on the planet.

On passing the planet a probe containing the spaceship’s clock and an astronaut is sent back to Earth at a speed of 0.80c relative to Earth. Suggest, for this situation, how the twin paradox arises and how it is resolved.

Evidence of finding 1/gradient such as:

use of any correct coordinate pair to find v - eg $\frac{4}{5}$ or $\frac{6}{7.5}$.

OR

measures tan of angle between ct and ct’ as about 39° AND tan 39 ≈ 0.8 ✔

Answer 0.8c given, so check coordinate values carefully.

E labelled at $x$ = 4, ct = 5 ✔

Check that E is placed on the worldline of S.

OR

Allow solutions involving the use of Lorentz equations.

t = 5 years OR ct = 5 ly ✔

On return to Earth the astronaut will have aged less than Earthlings «by 4 years»

OR

time passed on Earth is greater than time passed for the astronaut «by 4 years» ✔

astronaut accelerated/changed frames but Earth did not

OR

for astronaut the Earth clock jumps forward at turn-around ✔

OWTTE

Treat as neutral any mention of both the Earth and astronaut seeing each other’s clock as running slow.

Most candidates could show that the velocity of the spacecraft was 0.8c.

Event E was usually correctly labelled on the space-time diagram.

A very simple time dilation question which most candidates got wrong at SL but the question was better answered at HL.

Many candidates tried to use time dilation again without realising that the clock on P must also read 5 years at event E because that is the time on the Earth clock in P’s frame for the event.

The twin paradox is now well understood and there were some good quality answers. Some candidates even knew that the Earth clock jumps forward when the Astronaut turns around.

Define proper length.

In the reference frame of the train a ball travels with speed 0.50c from the back to the front of the train, as the train passes the platform. Calculate the time taken for the ball to reach the front of the train in the reference frame of the train.

In the reference frame of the train a ball travels with speed 0.50c from the back to the front of the train, as the train passes the platform. Calculate the time taken for the ball to reach the front of the train in the reference frame of the platform.

the length measured «in a reference frame» where the object is at rest ✔

ALTERNATIVE 1:

ALTERNATIVE 2:

v of ball is 0.846c for platform ✔

length of train is 68m for platform ✔

ALTERNATIVE 3:

Proper length is quite well understood. A common mistake is to mention that it is the length measured by a reference frame at rest.

Because there were three frames of reference in this question many candidates struggled to find the simple value for the time of the ball’s travel down the train in the train’s frame of reference.

Almost no candidates could use a Lorentz transformation to find the time of the ball’s travel in the frame of reference of the platform. Most just applied some form of t=γt’. Elapsed time and instantaneous time in different frames were easily confused. Candidates rarely mention which reference frame is used when making calculations, however this is crucial in relativity.

State the speed of the flash of light according to an observer on the ground using Galilean relativity.

State the speed of the flash of light according to an observer on the ground using Maxwell’s theory of electromagnetism.

State the speed of the flash of light according to an observer on the ground using Einstein’s theory of relativity.

c–v ✔

c ✔

c ✔

The speed of a flash of light from different viewpoints. Most of the prepared candidates well stated the speed of the flash using Galilean relativity, Maxwell’s theory and Einstein’s theory.

The speed of a flash of light from different viewpoints. Most of the prepared candidates well stated the speed of the flash using Galilean relativity, Maxwell’s theory and Einstein’s theory.

The speed of a flash of light from different viewpoints. Most of the prepared candidates well stated the speed of the flash using Galilean relativity, Maxwell’s theory and Einstein’s theory.

Plot, on the axes, the point corresponding to event 2.

Suggest whether the rocket launched by the spacecraft might be the cause of the explosion of the asteroid.

Show that the value of the invariant spacetime interval between events 1 and 2 is 9600 ly².

An observer in the spacecraft measures that events 1 and 2 are a distance of 120 ly apart. Determine, according to the spacecraft observer, the time between events 1 and 2.

Using the spacetime diagram, determine which event occurred first for the spacecraft observer, event 1 or event 2.

Determine, using the diagram, the speed of the spacecraft relative to the galaxy.

point as shown ✔

ALTERNATIVE 1
the rocket would have to travel faster than the speed of light ✔

so impossible ✔

ALTERNATIVE 2
drawing of future lightcone at origin ✔

and seeing that the asteroid explodes outside the lightcone so impossible ✔

ALTERNATIVE 3
the event was observed at +20 years, but its distance (stationary) is 100 ly ✔

so the asteroid event happened 80 years before t = 0 for the galactic observer ✔

100² − 20² = 9600 «ly²» ✔

Also accept 98 (the square root of 9600).

Allow negative value.

9600 = 120² − c²t² ✔

ct = «−» 69.3 «ly» / t = «−» 69.3 «y» ✔

Allow approach with Lorentz transformation.

line from event 2 parallel to $x$’ axis intersects ct’ axis at a negative value ✔

event 2 occurred first ✔

use of tan θ = $\frac{v}{c}$ with the angle between the time axes ✔

to get (0.70 ± 0.02)c ✔

Almost all candidates were able to plot the event on the diagram.

Most of the candidates identified, that the spacecraft was launched after the asteroid explosion and better candidates were also able to explain their reasoning with a drawing of light cones on the spacetime graph being the most popular response type.

Most of the candidates well used the formula for invariant spacetime from the data booklet, but only a few strong candidates were able to determine the time between the events according to the spacecraft observer. This implies a lack of understanding of the concept of invariance for different frames of reference. 

Most of the candidates well used the formula for invariant spacetime from the data booklet, but only a few strong candidates were able to determine the time between the events according to the spacecraft observer. This implies a lack of understanding of the concept of invariance for different frames of reference. 

Most of the candidates well used the formula for invariant spacetime from the data booklet, but only a few strong candidates were able to determine the time between the events according to the spacecraft observer. This implies a lack of understanding of the concept of invariance for different frames of reference. Most candidates well determined that event 2 occurred first in d ii), with a lesser number showing this correctly via the spacetime diagram.

Many candidates determined the speed using the spacetime diagram. However, some experienced difficulties reading accurately from the graph, and though their approach was correct, they failed to gain a result within the accepted range of values.

One of the two postulates of special relativity states that the speed of light in a vacuum is the same for all observers in inertial reference frames. State the other postulate of special relativity.

State the nature of the force on the particle P in the reference frame of the laboratory.

Deduce, using your answer to part (a), the nature of the force that acts on the particle P in the rest frame of P.

Explain how the force in part (b)(ii) arises.

The velocity of P is 0.30c relative to the laboratory. A second particle Q moves at a velocity of 0.80c relative to the laboratory.

Calculate the speed of Q relative to P.

laws of physics are the same for all observers
OR
laws of physics are the same in all «inertial» frames ✔

NOTE: OWTTE

magnetic ✔

«from 3a»
force must still be repulsive ✔

for P there is no magnetic force AND force is electric/electrostatic
OR
since P is at rest the force is electric/electrostatic ✔

protons and electrons in the wire move with different velocities «relative to P»
OR
speed of electrons is greater ✔

«for P» the density of protons and electrons in wire will be different «due to length contraction»
OR
«for P» the wire appears to have negative charge «due to length contraction» ✔

«hence electric force arises»

NOTE: Do not award mark for mention of length contraction without details.

$u\text{'}=\frac{0.80+0.30}{1+0.80\times 0.30}c$ ✔

$=0.89c$ ✔

NOTE: Accept 0.89c if all negative values used. Accept –0.89c even though speed is required.

Estimate in the Earth frame the fraction of the original muons that will reach the Earth’s surface before decaying according to Newtonian mechanics.

Estimate in the Earth frame the fraction of the original muons that will reach the Earth’s surface before decaying according to special relativity.

Demonstrate how an observer moving with the same velocity as the muons accounts for the answer to (a)(ii).

time of travel is «$\frac{3230}{0.98\times 3.0\times {10}^8}$» = 1.10 × 10⁻⁵ «s» ✔

which is «$\frac{1.10\times {10}^{-5}}{2.20\times {10}^{-6}}$» = 5.0 half-lives ✔

so fraction arriving as muons is « $\frac{1}{2^5}$ » = $\frac{1}{32}$

OR

3 % ✔

Award [3] for a bald correct answer.

time of travel corresponds to «$\frac{1.10\times {10}^{-5}}{5.0\times 2.20\times {10}^{-6}}$» = 1.0 half-life ✔

so fraction arriving as muons is $\frac{1}{2}$

OR

50 % ✔

Award [2] for a bald correct answer.

observer measures the distance to the surface to be shorter «by a factor of 5.0»/ length contraction occurs ✔

so time of travel again corresponds to «$\frac{\left(\frac{\frac{3230}{5.0}}{0.98\times 3.0\times {10}^8}\right)}{\left(2.20\times {10}^{-6}\right)}$»= 1.0 half-life ✔

Muons. The problem of muons created above the Earth surface moving almost at the speed of light, explicitly mentioned in the Guide, was solved well by prepared candidates. In i) many correctly calculated the time of travel in the earth’s frame though some struggled to recognize and apply the decay half-life aspect of this problem.

Students who correctly answered part i) did so in ii).

Some candidates struggled in b), with only the best candidates identified length contraction. Importantly, the command term here is “demonstrate” which means students must make their response clear by reasoning or evidence. Many who identified length contraction did not provide adequate reasoning to gain full marks.

Calculate in terms of $c$ the velocity of spaceship A relative to observer O.

Draw the $x\text{'}$ axis for the reference frame of spaceship A.

Plot the event E on the spacetime diagram and label it E.

Determine the time, according to spaceship A, when light from event E was observed on spaceship A.

$0.6\mathrm{c}$ ✓

Accept $\mathit{1}\mathit{.}\mathit{8}\mathit{\times }\mathit{10}{\mathit{ }}^{\mathit{8}} m{s}^{\mathit{-}\mathit{1}}$ if unit given.

line through origin and through $(5, 3) \pm$ one small square at this coordinate ✓

Answers shown for 5(a)(ii) and (b)(i) and (b)(ii).

X value of E at $4$ «$\mathrm{ly}$» ✓

Y value of E at $5$ «$\mathrm{y}$» ✓

light cone from E «crosses $ct$ at $9$ so» intersection on $ct\text{'}=5.6\pm 0.2$ $\mathrm{y}$ «on $ct$ scale» ✓

$\gamma =1.25$ ✓

so $t\text{'}=«\frac{5.6}{1.25}=» 4.5$ «$\mathrm{y}$ after leaving Earth» ✓

MP1 accept use of linear equations to find $t=\mathit{5}\mathit{.}\mathit{625}$

Allow ECF from (b)(i) and (a)

Very well answered.

Most answers successfully drew the correct x' axis.

Event E is the event when the light is emitted (4,5). Some candidates missed that and placed it at (4,9).

To determine these coordinates candidates were expected to construct a light path from (0,9) which intercepts at x=4 and t=5. This is a very common mistake which needs careful explanation by teachers.

Those candidates who placed E at the correct position were then able to calculate the time appropriately.

Calculate, for observer A, the length L_A of the bridge

Calculate, for observer A, the time taken to cross the bridge.

Outline why L_B is the proper length of the bridge.

Draw, on the spacetime diagram, the space axis for the reference frame of observer A. Label this axis $x$'.

Demonstrate using the diagram which lamp, according to observer A, was turned on first.

Demonstrate, using the diagram, which lamp observer A observes to light first.

Determine the time, according to observer A, between X and Y.

$\gamma =1.09$✔

$L_{\mathrm{A}}=«\frac{2.0}{1.09}=»$ 1.8 «km» ✔

ALTERNATIVE 1

time = $\frac{1.8\times {10}^3}{1.2\times {10}^8}$✔

1.5 × 10^–5«s» ✔

ALTERNATIVE 2

$t_{\mathrm{B}}=\frac{2\times {10}^3}{1.2\times {10}^8}=1.66\times {10}^{-5}$ «s» ✔

$t_{\mathrm{A}}=\frac{t_{\mathrm{B}}}{\gamma }=1.5\times {10}^{-5}$ «s» ✔

L_B is the length/measurement «by observer B» made in the reference frame in which the bridge is at rest ✔

NOTE: Idea of rest frame or frame in which bridge is not moving is required.

x′ axis drawn with correct gradient of 0.4 ✔

NOTE: Line must be 1 square below Y, allow ±0.5 square.

Allow line drawn without a ruler.

lines parallel to the x′ axis through X and Y intersecting the worldline ct′ at points shown ✔

so Y/lamp at the end of the bridge turned on first ✔

NOTE: Allow lines drawn without a ruler
Do not allow MP2 without supporting argument or correct diagram.

light worldlines at 45° from X AND Y intersecting the worldline ct′ ✔

so light from lamp X is observed first ✔

NOTE: Allow lines drawn without a ruler.
Do not allow MP2 without supporting argument or correct diagram.

ALTERNATIVE 1

$∆t\text{'}=1.09\times \left(0-\frac{0.4\times 2.0\times {10}^3}{3.0\times {10}^8}\right)$ ✔

= «–»2.9 × 10^–6 «s» ✔

ALTERNATIVE 2

equating spacetime intervals between X and Y

relies on realization that $∆x\text{'}=\gamma \left(∆x-0\right)$  eg: 

${\left(c∆t\text{'}\right)}^2-{\left(1.09\times 2000\right)}^2=0^2-{2000}^2$✔

$∆t\text{'}=«\pm »\frac{\sqrt{{\left(1.09\times 2000\right)}^2-{2000}^2}}{3.0\times {10}^8}=«\pm »2.9\times {10}^{-6}$ «s» ✔

ALTERNATIVE 3

use of diagram from answer to 4(c)(ii) (1 small square = 200 m)

counts 4.5 to 5 small squares (allow 900 – 1000 m) between events for A seen on B’s ct axis ✔

$\frac{950}{\gamma c}=2.9\times {10}^{-6}\pm 0.2\times {10}^{-6}$ «s» ✔

Outline why there is an attractive magnetic force on each proton in the laboratory frame. 

Explain why there is no magnetic force on each proton in its own rest frame.

Explain why there must be a resultant repulsive force on the protons in all reference frames.

moving charges give rise to magnetic fields

OR

magnetic attraction between parallel currents ✔

protons at rest produce no magnetic field

OR

mention of F = Bev where B and/or v =0 ✔

there is a repulsive electric/electrostatic force «in both frames» ✔

the attractive magnetic force «in the lab frame» is smaller than the repulsive electric force ✔

in all frames the net force is repulsive as all must agree that protons move apart

OR

mention of the first postulate of relativity ✔

Candidates usually realised that the magnetic field was due to the motion of the protons and that in the proton rest frame there could be no magnetic field. The answers were too often poorly worded and the candidates appeared to reword the question without providing a clear explanation.

Candidates usually realised that the magnetic field was due to the motion of the protons and that in the proton rest frame there could be no magnetic field. The answers were too often poorly worded and the candidates appeared to reword the question without providing a clear explanation.

A few candidates mentioned that there was an electrostatic repulsive force between the protons in both frames. However very few realised that there had to be an overall repulsive force in both frames because of the relativity postulate.

Outline the conclusion from Maxwell’s work on electromagnetism that led to one of the postulates of special relativity.

light is an EM wave

speed of light is independent of the source/observer