Galilean and special relativity HL Paper 1&2

A.5 Galilean and Special Relativity

IB-style HL Paper 1 and Paper 2 question bank

This original practice bank contains 15 multiple-choice questions, 15 extended-response questions and a lab-style data investigation. Each answer expands below its question. These are practice questions, not official IB examination questions.

Instructions and reference frames

A.5 is an HL-only topic. Show working, units and clear explanations.
Use $c=3.00\times10^8\,\mathrm{m\,s^{-1}}$ and give answers to 2–3 significant figures. Retain unrounded values during calculations.
Unless stated otherwise, S′ moves at signed velocity +v along the x-axis of S. The origins coincide at $t=t^{\prime}=0$ .
Use signed velocities in transformation equations. Proper time $\Delta t_0$ is measured between events occurring at the same position in the clock’s rest frame.
Proper length $L_0$ is measured in the object’s rest frame. A moving length requires simultaneous endpoint measurements in the frame measuring that length.
Space–time diagrams use x horizontally and ct vertically, with equal length scales for those two axes. Diagram angles are measured from the stated axis.

IB formula reference

$x^{\prime}=x-vt,\qquad t^{\prime}=t$

$u^{\prime}=u-v\quad\text{(Galilean)}$

$\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$

$x^{\prime}=\gamma(x-vt)$

$t^{\prime}=\gamma\left(t-\frac{vx}{c^2}\right)$

$u^{\prime}=\frac{u-v}{1-uv/c^2}$

$(\Delta s)^2=(c\Delta t)^2-(\Delta x)^2$

$\Delta t=\gamma\Delta t_0$

$L=\frac{L_0}{\gamma}$

$\tan\theta=\frac{v}{c}\quad\text{(angle from the }ct\text{ axis)}$

Inverse velocity transformation: solving the IB velocity equation for u gives the useful equivalent form below.

$u=\frac{u^{\prime}+v}{1+u^{\prime}v/c^2}$

Classification of intervals: $(\Delta s)^2>0$ is timelike, $(\Delta s)^2=0$ is lightlike, and $(\Delta s)^2<0$ is spacelike. For a timelike interval, $\Delta t_0=\sqrt{(\Delta t)^2-(\Delta x/c)^2}$ .

Common misconceptions

“All time intervals measured on a spacecraft are proper times.” The two events must occur at the same position in that frame.
“Length contraction affects every dimension.” It affects the dimension parallel to relative motion.
“Simultaneous events are simultaneous everywhere.” Events separated in space may be simultaneous in one frame and occur at different times in another.
“A moving clock appears slow only because light takes time to arrive.” Time dilation concerns frame measurements after accounting for signal travel times.
“Every muon decays after its mean lifetime.” Individual decays are random; the lifetime is a statistical mean.
“Relativity starts at a fixed speed such as 0.1c.” The required precision determines whether relativistic corrections can be neglected.

Part 1 — Multiple-choice questions

Multiple choice 1 — An inertial reference frame [1 mark]

Which reference frame is inertial, to a suitable approximation?

A train moving at constant velocity along a straight track
A car accelerating along a straight road
A rotating carousel
A train rounding a curve at constant speed

Show worked answer

Correct option: A.

An inertial frame is non-accelerating. Constant speed around a curve still involves changing velocity and therefore acceleration.

Multiple choice 2 — Galilean position transformation [1 mark]

S′ moves at +8.0 m/s relative to S. An event occurs at x = 50 m and t = 4.0 s in S. What is its Galilean x′ coordinate?

82 m
50 m
32 m
18 m

Show worked answer

Correct option: D.

$x^{\prime}=x-vt=50-(8.0)(4.0)=18\,\mathrm{m}$

Multiple choice 3 — Einstein’s postulates [1 mark]

Which statement correctly summarises special relativity?

The laws of physics apply only in stationary frames.
The laws of physics are the same in all inertial frames, and vacuum light speed is the same for all inertial observers.
Time is absolute, but light speed depends on the source.
Light travels faster when its source approaches an observer.

Show worked answer

Correct option: B.

No inertial frame is privileged. Vacuum light speed c does not depend on the motion of the source or inertial observer.

Multiple choice 4 — Lorentz factor [1 mark]

What is the Lorentz factor at speed 0.60c?

0.80
1.67
1.25
0.60

Show worked answer

Correct option: C.

$\gamma=\frac{1}{\sqrt{1-0.60^2}}=1.25$

Multiple choice 5 — Proper time [1 mark]

Which measurement is a proper time interval?

The interval between simultaneous flashes at opposite ends of a moving train
Any interval recorded by an observer on Earth
Any interval measured by two clocks at different positions
The interval between two ticks of one clock in that clock’s rest frame

Show worked answer

Correct option: D.

The two tick events occur at the same spatial position in the clock’s rest frame.

Multiple choice 6 — Time dilation [1 mark]

A moving clock records 3.0 s between two ticks. Its speed relative to a laboratory is 0.80c. What interval does the laboratory measure between the tick events?

5.0 s
1.8 s
3.0 s
2.4 s

Show worked answer

Correct option: A.

$\gamma=\frac53,\qquad\Delta t=\gamma\Delta t_0=\frac53(3.0)=5.0\,\mathrm{s}$

Multiple choice 7 — Length contraction [1 mark]

A rod has proper length 100 m and moves parallel to its length at 0.60c. What length is measured in the laboratory?

125 m
60 m
80 m
100 m

Show worked answer

Correct option: C.

$L=\frac{100}{1.25}=80\,\mathrm{m}$

Multiple choice 8 — A perpendicular dimension [1 mark]

A spacecraft moves along its length. What happens to its width, measured perpendicular to the motion in another inertial frame?

It contracts by a factor γ.
It is unchanged.
It increases by a factor γ.
It becomes zero near light speed.

Show worked answer

Correct option: B.

Lorentz length contraction applies along the direction of relative motion; perpendicular dimensions are unchanged.

Multiple choice 9 — Forward velocity addition [1 mark]

A spacecraft moves at +0.80c relative to Earth. A probe moves at +0.60c relative to the spacecraft. What is the probe’s Earth-frame velocity?

+0.946c
+1.40c
+0.200c
+0.480c

Show worked answer

Correct option: A.

$\frac{u}{c}=\frac{0.60+0.80}{1+(0.60)(0.80)}=0.946$

Multiple choice 10 — Transforming a light ray [1 mark]

In S, a vacuum light ray moves along +x with velocity c. S′ moves at +0.50c. What light velocity does S′ measure?

0.50c
1.50c
c
0

Show worked answer

Correct option: C.

$u^{\prime}=\frac{c-0.50c}{1-0.50}=c$

Multiple choice 11 — Relativity of simultaneity [1 mark]

Two events are simultaneous in S, with event B at a larger x-coordinate than event A. S′ moves at positive v. Which statement is correct?

B occurs after A in S′.
They remain simultaneous in S′.
Their time order cannot be determined.
B occurs before A in S′.

Show worked answer

Correct option: D.

$\Delta t^{\prime}=\gamma\left(0-\frac{v\Delta x}{c^2}\right)<0$

Here the differences are event B minus event A.

Multiple choice 12 — Classifying an interval [1 mark]

Two events have $(c\Delta t)^2>(\Delta x)^2$ . Their separation is:

Spacelike
Timelike
Lightlike
Impossible in special relativity

Show worked answer

Correct option: B.

The squared interval is positive. A subluminal object can travel from the earlier event to the later event, and a frame exists in which both occur at the same position.

Multiple choice 13 — A space–time worldline [1 mark]

A constant-velocity worldline has gradient 2.5 on a graph of ct vertically against x horizontally. What is the speed?

2.5c
0.625c
0.40c
0.80c

Show worked answer

Correct option: C.

$\frac{\Delta(ct)}{\Delta x}=\frac{c}{v}=2.5\quad\Rightarrow\quad v=0.40c$

Multiple choice 14 — Proper time from an interval [1 mark]

Two events are separated by 5.0 μs and 900 m in S. What is the proper time interval between them?

4.0 μs
5.83 μs
2.0 μs
6.67 μs

Show worked answer

Correct option: A.

$\frac{\Delta x}{c}=3.0\,\mu\mathrm{s}$

$\Delta t_0=\sqrt{5.0^2-3.0^2}=4.0\,\mu\mathrm{s}$

Multiple choice 15 — Muon-frame explanation [1 mark]

Why can fast atmospheric muons reach Earth’s surface, as described in the muons’ rest frame?

Their intrinsic mean lifetime increases in their own frame.
The atmosphere moves past them faster than c.
The Earth-frame distance is longer in their frame.
The distance through the atmosphere is contracted, while their proper mean lifetime is unchanged.

Show worked answer

Correct option: D.

The same events can be described by Earth-frame time dilation or muon-frame length contraction. The muons’ rest-frame mean lifetime remains the proper lifetime.

Part 2 — Extended-response questions

Extended response 1 — Galilean transformations at low speed [8 marks]

A train moves at +20.0 m/s relative to the ground. The train frame is S′ and the ground frame is S. An event occurs at x = 150 m and t = 5.00 s in S. A ball is thrown forward at +8.00 m/s relative to the train.

Determine the Galilean coordinates x′ and t′ of the event. [3]
Calculate the ball’s velocity relative to the ground. [2]
State why Galilean transformations are an excellent approximation for the ball. [1]
Explain why applying the same velocity addition to a light beam would conflict with special relativity. [2]

Show worked answer

(a) [3]

$x^{\prime}=150-(20.0)(5.00)=50.0\,\mathrm{m}$

$t^{\prime}=t=5.00\,\mathrm{s}$

(b) [2] Rearrange $u^{\prime}=u-v$ .

$u=u^{\prime}+v=8.00+20.0=+28.0\,\mathrm{m\,s^{-1}}$

(c) [1] All speeds are extremely small compared with c, so the relativistic corrections are negligible at the stated precision.

(d) [2] Classical addition would predict c + 20.0 m/s for a forward train-frame light beam. Special relativity requires every inertial observer to measure the same vacuum light speed c.

Answer: Galilean event coordinates: x′ = 50.0 m, t′ = 5.00 s; ball velocity = +28.0 m/s.

Extended response 2 — Classical and relativistic velocity addition [5 marks]

A spacecraft travels at +0.20c relative to Earth. A probe is fired forward at +0.30c relative to the spacecraft.

Calculate the Earth-frame velocity predicted by Galilean addition. [1]
Calculate the correct relativistic Earth-frame velocity. [2]
Calculate the percentage by which the Galilean result exceeds the relativistic result, and state the underlying reason. [2]

Show worked answer

(a) [1]

$u_{\mathrm{Galilean}}=0.20c+0.30c=0.50c$

(b) [2] Use the inverse relativistic velocity transformation.

$u=\frac{0.30c+0.20c}{1+(0.30)(0.20)}=0.471698c$

(c) [2] Use the relativistic result as the reference value.

$\text{Percentage excess}=\frac{0.50-0.471698}{0.471698}\times100=6.0\%$

Galilean addition assumes absolute time. The Lorentz transformation accounts for relativity of time and simultaneity and preserves invariant vacuum light speed.

Answer: Classical prediction = +0.50c; relativistic velocity ≈ +0.472c; classical excess = 6.0%.

Extended response 3 — Postulates and the Lorentz factor [6 marks]

A spacecraft travels at 0.85c relative to Earth.

State the two postulates of special relativity. [2]
Calculate the Lorentz factor. [2]
Determine whether Galilean relativity is adequate if time and length predictions must be accurate to 1%. Explain. [2]

Show worked answer

(a) [2] The laws of physics are the same in all inertial reference frames. Vacuum light speed c is the same for all inertial observers, independent of source motion.

(b) [2]

$\gamma=\frac{1}{\sqrt{1-0.85^2}}=1.89832\approx1.90$

(c) [2] No. γ differs greatly from 1, giving substantial time dilation and length contraction. These changes are much larger than 1%. For example, a moving length is only $1/\gamma=0.527$ times its proper length.

Answer: Lorentz factor ≈ 1.90; relativistic transformations are required for 1% accuracy.

Extended response 4 — Time dilation and proper time [7 marks]

Two ticks occur at the same onboard clock on a spacecraft travelling at 0.90c relative to Earth. The astronaut measures 4.0 s between the ticks.

Identify the proper time interval and justify your choice. [2]
Calculate the interval between the tick events in the Earth frame. [3]
Explain why inertial observers can each describe the other’s clocks as running slowly without a contradiction. [2]

Show worked answer

(a) [2] The ticks occur at the same position in the clock’s rest frame, so $\Delta t_0=4.0\,\mathrm{s}$ .

(b) [3]

$\gamma=\frac{1}{\sqrt{1-0.90^2}}=2.294157$

$\Delta t=\gamma\Delta t_0=(2.294157)(4.0)=9.18\,\mathrm{s}$

(c) [2] Each comparison uses that observer’s simultaneity convention and synchronised clocks. The reciprocal statements compare different sets of events and distant-clock readings; simultaneity is not shared between the frames. They do not assign contradictory intervals to one common pair of local tick events.

Answer: Proper interval = 4.0 s; Earth-frame interval = 9.18 s.

Extended response 5 — Length contraction and measurement [7 marks]

A spacecraft has proper length 120 m and proper width 20.0 m. It moves at 0.75c relative to Earth, parallel to its length.

Define proper length. [1]
Calculate the Earth-frame length of the spacecraft. [3]
Explain how its endpoints must be measured in the Earth frame. [2]
State its Earth-frame width. [1]

Show worked answer

(a) [1] Proper length is the length measured in the object’s rest frame.

(b) [3]

$\gamma=\frac{1}{\sqrt{1-0.75^2}}=1.511858$

$L=\frac{120}{1.511858}=79.4\,\mathrm{m}$

(c) [2] Record the positions of both endpoints at the same Earth-frame time, using synchronised Earth-frame instruments. An ordinary photograph does not automatically give that simultaneous length because light travel times differ.

(d) [1] The width is perpendicular to relative motion and remains 20.0 m.

Answer: Earth-frame length = 79.4 m; Earth-frame width = 20.0 m.

Extended response 6 — Muon mean lifetime and travel distance [8 marks]

A sample of muons travels at 0.98c relative to Earth. Its proper mean lifetime is 2.2 μs. Treat the muons as moving at constant speed.

Calculate the Lorentz factor. [2]
Calculate the Earth-frame mean lifetime. [2]
Calculate the mean distance travelled in that time and compare with the prediction obtained without time dilation. [2]
Explain why the calculated distance is not a fixed distance at which every muon decays. [2]

Show worked answer

(a) [2]

$\gamma=\frac{1}{\sqrt{1-0.98^2}}=5.025189$

(b) [2]

$\Delta t=(5.025189)(2.2\times10^{-6})=1.11\times10^{-5}\,\mathrm{s}$

(c) [2] Retain the unrounded lifetime when calculating distance.

$d=v\Delta t=(0.98)(3.00\times10^8)(5.025189)(2.2\times10^{-6})$

$d=3.25\times10^3\,\mathrm{m}$

$d_{\mathrm{without\ dilation}}=(0.98)(3.00\times10^8)(2.2\times10^{-6})=647\,\mathrm{m}$

(d) [2] Decay is random for individual muons. Some decay earlier and some later; the calculated distance is a statistical mean for the population, not a universal decay position.

Answer: Lorentz factor ≈ 5.03; Earth mean lifetime ≈ 11.1 μs; mean distance ≈ 3.25 km, compared with 0.647 km without time dilation.

Extended response 7 — Interstellar travel in two frames [8 marks]

A spacecraft moves at 0.92c from Earth toward a star. Earth and the star are at rest in the same inertial frame, separated by $4.0\times10^{16}\,\mathrm{m}$ . Ignore the brief acceleration and deceleration stages.

Calculate the distance between Earth and the star in the spacecraft frame. [3]
Calculate the Earth-frame travel time. [2]
Calculate the time elapsed on the spacecraft clock. [2]
Explain why the spacecraft clock records a proper time for the departure and arrival events. [1]

Show worked answer

(a) [3] The Earth–star rest-frame distance is the proper length.

$\gamma=\frac{1}{\sqrt{1-0.92^2}}=2.551552$

$L=\frac{4.0\times10^{16}}{2.551552}=1.57\times10^{16}\,\mathrm{m}$

(b) [2]

$\Delta t_{\mathrm{Earth}}=\frac{4.0\times10^{16}}{0.92(3.00\times10^8)}=1.45\times10^8\,\mathrm{s}$

(c) [2]

$\Delta t_0=\frac{\Delta t_{\mathrm{Earth}}}{\gamma}=5.68\times10^7\,\mathrm{s}$

Equivalently, use the contracted distance divided by 0.92c in the spacecraft frame.

(d) [1] Departure and arrival occur at the same position of the spacecraft’s clock in its inertial cruising frame.

Answer: Contracted distance ≈ 1.57 × 10¹⁶ m; Earth time ≈ 1.45 × 10⁸ s; onboard time ≈ 5.68 × 10⁷ s.

Extended response 8 — Lorentz transformation of an event [7 marks]

S′ moves at +0.60c relative to S. The origins coincide at zero time. Event A is their common origin event. Event B has coordinates x = 900 m and t = 5.00 μs in S.

Calculate the Lorentz factor. [1]
Calculate the x′ coordinate of B. [2]
Calculate the t′ coordinate of B. [2]
Calculate the squared space–time interval from A to B in both frames and verify invariance. [2]

Show worked answer

(a) [1] $\gamma=1.25$ .

(b) [2]

$x^{\prime}=1.25\left[900-(0.60)(3.00\times10^8)(5.00\times10^{-6})\right]=0\,\mathrm{m}$

(c) [2]

$t^{\prime}=1.25\left(5.00\times10^{-6}-\frac{(0.60)(900)}{3.00\times10^8}\right)$

$t^{\prime}=4.00\times10^{-6}\,\mathrm{s}=4.00\,\mu\mathrm{s}$

(d) [2]

$(\Delta s)^2_S=1500^2-900^2=1.44\times10^6\,\mathrm{m^2}$

$(\Delta s)^2_{S^{\prime}}=1200^2-0^2=1.44\times10^6\,\mathrm{m^2}$

Answer: Event B: x′ = 0 m, t′ = 4.00 μs; both frames give a squared interval of 1.44 × 10⁶ m².

Extended response 9 — Simultaneity and the length-measurement trap [7 marks]

In S, event A occurs at x = 0 and event B at x = 900 m. Both occur at t = 0. S′ moves at +0.60c. Define all differences as B minus A.

Calculate the transformed event separations Δx′ and Δt′. [4]
State which event occurs first in S′. [1]
Explain why Δx′ is not automatically the S′ length of a rod whose endpoints were recorded in these two events. [2]

Show worked answer

(a) [4] Use $\Delta t=0$ , $\Delta x=900\,\mathrm{m}$ and $\gamma=1.25$ .

$\Delta x^{\prime}=\gamma(\Delta x-v\Delta t)=(1.25)(900)=1125\,\mathrm{m}$

$\Delta t^{\prime}=\gamma\left(\Delta t-\frac{v\Delta x}{c^2}\right)$

$\Delta t^{\prime}=-(1.25)\frac{(0.60)(900)}{3.00\times10^8}=-2.25\,\mu\mathrm{s}$

(b) [1] B occurs before A because $t_B^{\prime}-t_A^{\prime}<0$ .

(c) [2] These transformed endpoint events are not simultaneous in S′. A length in S′ requires endpoint positions at the same S′ time, so a different pair of measurement events is needed.

Answer: Δx′ = 1125 m; Δt′ = −2.25 μs; B occurs first. The transformed separation is not a simultaneous S′ rod-length measurement.

Extended response 10 — A timelike interval and its rest frame [8 marks]

Two events have separation Δx = +600 m and Δt = +4.00 μs in S.

Calculate the squared interval and classify it. [2]
Calculate the proper time interval. [2]
Find the velocity of a frame in which the events occur at the same position. [2]
Explain why all inertial frames agree on the time order of these events. [2]

Show worked answer

(a) [2]

$(\Delta s)^2=1200^2-600^2=1.08\times10^6\,\mathrm{m^2}$

The positive squared interval is timelike.

(b) [2]

$\Delta t_0=\sqrt{(4.00)^2-(2.00)^2}=3.46\,\mu\mathrm{s}$

(c) [2] Set $\Delta x^{\prime}=\gamma(\Delta x-v\Delta t)=0$ .

$v=\frac{\Delta x}{\Delta t}=\frac{600}{4.00\times10^{-6}}=1.50\times10^8\,\mathrm{m\,s^{-1}}=+0.500c$

(d) [2] For a timelike separation, $c\Delta t>|\Delta x|$ . Since every inertial-frame relative speed has magnitude below c, the Lorentz time transformation cannot make this positive interval negative. A causal influence could connect the events, although the interval alone does not prove that one caused the other.

Answer: Squared interval = 1.08 × 10⁶ m², timelike; proper time = 3.46 μs; same-position frame velocity = +0.500c.

Extended response 11 — A spacelike interval and reversed order [8 marks]

Event B occurs 1200 m farther along +x than event A and 2.00 μs later in S.

Calculate the squared interval and classify it. [2]
Find the frame velocity for which the events are simultaneous. [2]
Calculate Δt′ for a frame moving at +0.80c. [2]
Explain why the reversed order in that frame does not violate causality. [2]

Show worked answer

(a) [2]

$(\Delta s)^2=600^2-1200^2=-1.08\times10^6\,\mathrm{m^2}$

The negative squared interval is spacelike.

(b) [2] Set $\Delta t^{\prime}=0$ .

$v=\frac{c^2\Delta t}{\Delta x}=\frac{(3.00\times10^8)^2(2.00\times10^{-6})}{1200}=+0.500c$

(c) [2] At 0.80c, $\gamma=5/3$ .

$\Delta t^{\prime}=\frac53\left(2.00-0.80\times4.00\right)\,\mu\mathrm{s}=-2.00\,\mu\mathrm{s}$

(d) [2] Light or any slower signal cannot connect these two spacelike-separated events. Their relative time order can therefore change without reversing the order of a possible cause and effect.

Answer: Squared interval = −1.08 × 10⁶ m², spacelike; simultaneous frame velocity = +0.500c; at +0.80c, B occurs 2.00 μs before A.

Extended response 12 — Approaching spacecraft and relative speed [8 marks]

In the Earth frame, spacecraft A moves at +0.70c and spacecraft B at −0.60c along the same line.

Calculate the rate at which their separation decreases in the Earth frame and explain its meaning. [2]
Calculate B’s velocity in A’s rest frame. [3]
A emits a forward vacuum light pulse along +x. Calculate the pulse’s Earth-frame velocity and explain why it is not 1.70c. [3]

Show worked answer

(a) [2] The Earth-frame separation decreases at $0.70c+0.60c=1.30c$ . This is a rate involving two positions measured in one frame, not the velocity of one spacecraft measured in the other’s rest frame.

(b) [3] In the IB formula, $u=-0.60c$ and $v=+0.70c$ .

$u^{\prime}=\frac{-0.60c-0.70c}{1-(-0.60)(0.70)}=-0.915c$

The negative sign means B moves in the −x direction in A’s frame.

(c) [3] In A’s frame the light velocity is $u^{\prime}=c$ .

$u=\frac{c+0.70c}{1+0.70}=c$

Relativistic velocity addition preserves invariant vacuum light speed. Adding 0.70c classically would be inappropriate.

Answer: Earth closing rate = 1.30c; B’s velocity in A’s frame ≈ −0.915c; Earth-frame light velocity = +c.

Extended response 13 — Worldlines and tilted coordinate axes [9 marks]

On an equally scaled x–ct diagram, a particle’s straight worldline passes through the origin and the event x = 900 m, ct = 1500 m. Take S′ as the particle’s rest frame.

Calculate the particle’s velocity in S. [2]
Calculate the angle of its worldline from the ct-axis. [2]
State the worldlines of light pulses emitted from the origin in both x-directions and their angles from the ct-axis. [2]
Describe the directions of the ct′ and x′ axes on this diagram, and explain why ordinary Euclidean ruler distances cannot directly calibrate the primed axes. [3]

Show worked answer

(a) [2]

$\frac{v}{c}=\frac{x}{ct}=\frac{900}{1500}=0.600$

$v=+1.80\times10^8\,\mathrm{m\,s^{-1}}$

(b) [2]

$\tan\theta=\frac{v}{c}=0.600\quad\Rightarrow\quad\theta=31.0^\circ$

(c) [2] The light worldlines are $x=ct$ and $x=-ct$ . Each makes an angle of 45° from the positive ct-axis, on opposite sides.

(d) [3] The ct′ axis is the particle’s worldline, tilted 31.0° toward +x from ct. The x′ axis consists of events with $t^{\prime}=0$ , so $ct=(v/c)x=0.600x$ ; it tilts 31.0° above the x-axis. Primed-axis scales are set by the invariant space–time interval, not by the ordinary Euclidean length of a slanted line segment. The primed axes are not a simple Euclidean rotation of S’s axes.

Answer: Particle velocity = +0.600c; worldline tilt = 31.0° from ct; light rays lie at ±45° from ct.

Extended response 14 — Challenge: Light reaches the ends of a train [10 marks]

A train of proper length 300 m moves at +0.60c relative to Earth. At the shared origin event, a lamp at the train’s midpoint emits two light pulses toward the front and rear. Let x = 0 and t = 0 at emission. Both pulses travel in vacuum.

Calculate the train-frame time for either pulse to reach an end. [2]
Calculate the Earth-frame train length and the endpoint positions at emission. [2]
Calculate the Earth-frame arrival times at the front and rear, allowing the endpoints to move. [4]
Use the Lorentz time transformation to verify that the two arrival events are simultaneous in the train frame. [2]

Show worked answer

(a) [2] Each end is 150 m from the midpoint in the train frame.

$t^{\prime}_{\mathrm{arrival}}=\frac{150}{c}=0.500\,\mu\mathrm{s}$

(b) [2] $\gamma=1.25$ .

$L=\frac{300}{1.25}=240\,\mathrm{m}$

At Earth time zero, the front is at +120 m and the rear at −120 m.

(c) [4] The front pulse catches an endpoint moving in the same direction; the rear endpoint moves toward the backward pulse.

$t_f=\frac{120}{c-0.60c}=1.00\,\mu\mathrm{s}$

$t_r=\frac{120}{c+0.60c}=0.250\,\mu\mathrm{s}$

At these arrival events, $x_f=ct_f=300\,\mathrm{m}$ and $x_r=-ct_r=-75.0\,\mathrm{m}$ . The rear arrival is earlier in Earth’s frame.

(d) [2] Transform the time of each arrival, retaining the signed positions.

$t_f^{\prime}=1.25\left(1.00-0.60\times1.00\right)\,\mu\mathrm{s}=0.500\,\mu\mathrm{s}$

$t_r^{\prime}=1.25\left(0.250-0.60\times(-0.250)\right)\,\mu\mathrm{s}=0.500\,\mu\mathrm{s}$

Answer: Train arrivals are simultaneous at 0.500 μs. Earth arrivals occur at 1.00 μs at the front and 0.250 μs at the rear.

Extended response 15 — Challenge: When is the classical approximation adequate? [6 marks]

For a particular calculation, a student decides to neglect time dilation only when $\gamma-1\leq0.010$ . This sets a 1% threshold relative to proper time.

Calculate the largest speed allowed by this criterion. [3]
Calculate the fractional time-dilation correction at 0.10c. [2]
Explain why the statement “relativistic effects become significant at exactly 0.10c” is misleading. [1]

Show worked answer

(a) [3] At the boundary $\gamma=1.010$ . Rearrange the Lorentz factor.

$\frac{v}{c}=\sqrt{1-\frac{1}{\gamma^2}}=\sqrt{1-\frac{1}{1.010^2}}=0.14037$

$v_{\mathrm{max}}=0.140c=4.21\times10^7\,\mathrm{m\,s^{-1}}$

(b) [2]

$\gamma=\frac{1}{\sqrt{1-0.10^2}}=1.0050378$

$(\gamma-1)\times100=0.504\%$

(c) [1] There is no sharp physical threshold. The size of the correction must be compared with the required accuracy; very precise experiments can detect relativity at much lower speeds.

Answer: For this criterion, maximum speed ≈ 0.140c; time-dilation correction at 0.10c ≈ 0.504%.

Part 3 — Lab-style data investigation

Testing time dilation using muon mean lifetimes [20 marks]

A research team measures mean lifetimes of muon samples selected into narrow speed ranges. Moving samples travel through a straight decay channel; a separate stationary sample provides the rest-frame reference. The table contains simulated practice data, not results from an actual experiment.

The independently measured proper mean lifetime is $\Delta t_0=(2.20\pm0.02)\,\mu\mathrm{s}$ . The quoted uncertainties in the table are absolute uncertainties in the fitted mean lifetime. Treat speed uncertainty as negligible for the numerical calculations below.

v/c	Measured mean lifetime / μs	γ	Predicted lifetime / μs
0.00	2.21 ± 0.04	Calculate	Calculate
0.60	2.74 ± 0.05	Calculate	Calculate
0.80	3.68 ± 0.07	Calculate	Calculate
0.90	5.06 ± 0.10	Calculate	Calculate
0.95	7.02 ± 0.14	Calculate	Calculate

Identify the independent and dependent variables and state two controls or experimental conditions that must remain comparable. [2]
Complete the γ and predicted-lifetime columns. Show a calculation for the fastest sample. [3]
Describe a suitable linear graph, including error bars, and estimate its gradient and intercept. [3]
Estimate an uncertainty range for the gradient using the extreme allowed endpoint values. Compare it with the independently measured proper lifetime. [3]
For the fastest sample, compare the measurement with the classical prediction and the relativistic prediction, including uncertainties. [3]
Identify one possible systematic error and suggest two specific improvements. [3]
Explain why individual decay times must be treated statistically and what conclusions simulated data can and cannot establish. [3]

Show worked answer

(a) [2] Independent variable: selected muon speed v/c. Dependent variable: fitted laboratory mean lifetime. Keep the particle species and decay-time analysis method fixed; maintain comparable detector calibration, selection criteria and background treatment.

(b) [3] For the fastest sample:

$\gamma=\frac{1}{\sqrt{1-0.95^2}}=3.202563$

$\Delta t_{\mathrm{predicted}}=\gamma\Delta t_0=(3.202563)(2.20)=7.05\,\mu\mathrm{s}$

v/c	Measured lifetime / μs	γ	Predicted lifetime / μs
0.00	2.21 ± 0.04	1.000	2.20
0.60	2.74 ± 0.05	1.250	2.75
0.80	3.68 ± 0.07	1.667	3.67
0.90	5.06 ± 0.10	2.294	5.05
0.95	7.02 ± 0.14	3.203	7.05

(c) [3] Plot measured laboratory mean lifetime vertically against γ horizontally, with the stated lifetime error bars. The relativistic model predicts a straight line through the origin with gradient equal to proper mean lifetime. The origin is an extrapolation, since physical γ values are at least 1. Use a best-fit line; endpoint estimates are:

$b\approx\frac{7.02-2.21}{3.202563-1.000}=2.18\,\mu\mathrm{s}$

$a\approx2.21-(2.18382)(1.000)=0.026\,\mu\mathrm{s}$

The small estimated intercept is consistent with the predicted zero intercept within the data uncertainties. A complete assessment uses the fit and its uncertainty, not just a single extrapolated value.

(d) [3] A conservative endpoint estimate uses the largest and smallest possible slopes from the two endpoint error bars.

$b_{\mathrm{min}}=\frac{(7.02-0.14)-(2.21+0.04)}{3.202563-1.000}=2.10\,\mu\mathrm{s}$

$b_{\mathrm{max}}=\frac{(7.02+0.14)-(2.21-0.04)}{3.202563-1.000}=2.27\,\mu\mathrm{s}$

Thus an approximate gradient estimate is $b=(2.18\pm0.08)\,\mu\mathrm{s}$ . Its range overlaps $(2.20\pm0.02)\,\mu\mathrm{s}$ , supporting consistency with the proper mean lifetime. This endpoint method is an estimate, not a full regression uncertainty calculation.

(e) [3] Classical absolute time predicts $2.20\pm0.02\,\mu\mathrm{s}$ , which does not overlap the measured $7.02\pm0.14\,\mu\mathrm{s}$ . Relativity predicts 7.0456 μs. With γ treated as exact:

$\delta(\Delta t_{\mathrm{predicted}})=\gamma\delta(\Delta t_0)=(3.202563)(0.02)=0.064\,\mu\mathrm{s}$

The relativistic prediction is approximately $(7.05\pm0.06)\,\mu\mathrm{s}$ , consistent with the measurement. Under the stated assumptions, the data favour time dilation over absolute-time predictions.

(f) [3] A detector that preferentially misses late or distant decays can bias the fitted lifetime. Calibrate timing and detection efficiency across the relevant decay positions. Collect more events and repeat measurements to reduce statistical uncertainty. Also verify the selected speeds and include their uncertainties when they are not negligible. Accept one concrete systematic error and two relevant improvements.

(g) [3] Muon decay is random for individual particles, so a mean lifetime must be estimated from many decay events. These simulated data demonstrate how to analyse a time-dilation test and are consistent with the relativistic model by construction. They do not provide independent experimental evidence; that requires real measurements and a validated experimental procedure.

Answer: Gradient ≈ 2.18 ± 0.08 μs, consistent with the proper mean lifetime of 2.20 ± 0.02 μs. The simulated fastest-sample result agrees with the relativistic prediction and disagrees with the classical prediction.

Topic A — Final exam readiness checklist

Identify frames and use signed velocities consistently.
State Einstein’s postulates and recognise the low-speed Galilean approximation.
Identify proper time and proper length before substituting values.
Use Lorentz transformations for event coordinates and simultaneity.
Distinguish a closing rate from a velocity measured in another rest frame.
Classify intervals and preserve the time order of causally connectable events.
Interpret x–ct worldlines, light lines and primed coordinate axes.
Explain atmospheric muons using both time dilation and length contraction.
Use uncertainty and genuine experimental data when evaluating a model.

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Galilean and special relativity HL Paper 1&2

Instructions and reference frames

IB formula reference

Common misconceptions

Part 1 — Multiple-choice questions

Multiple choice 1 — An inertial reference frame [1 mark]

Multiple choice 2 — Galilean position transformation [1 mark]

Multiple choice 3 — Einstein’s postulates [1 mark]

Multiple choice 4 — Lorentz factor [1 mark]

Multiple choice 5 — Proper time [1 mark]

Multiple choice 6 — Time dilation [1 mark]

Multiple choice 7 — Length contraction [1 mark]

Multiple choice 8 — A perpendicular dimension [1 mark]

Multiple choice 9 — Forward velocity addition [1 mark]

Multiple choice 10 — Transforming a light ray [1 mark]

Multiple choice 11 — Relativity of simultaneity [1 mark]

Multiple choice 12 — Classifying an interval [1 mark]

Multiple choice 13 — A space–time worldline [1 mark]

Multiple choice 14 — Proper time from an interval [1 mark]

Multiple choice 15 — Muon-frame explanation [1 mark]

Part 2 — Extended-response questions

Extended response 1 — Galilean transformations at low speed [8 marks]

Extended response 2 — Classical and relativistic velocity addition [5 marks]

Extended response 3 — Postulates and the Lorentz factor [6 marks]

Extended response 4 — Time dilation and proper time [7 marks]

Extended response 5 — Length contraction and measurement [7 marks]

Extended response 6 — Muon mean lifetime and travel distance [8 marks]

Extended response 7 — Interstellar travel in two frames [8 marks]

Extended response 8 — Lorentz transformation of an event [7 marks]

Extended response 9 — Simultaneity and the length-measurement trap [7 marks]

Extended response 10 — A timelike interval and its rest frame [8 marks]

Extended response 11 — A spacelike interval and reversed order [8 marks]

Extended response 12 — Approaching spacecraft and relative speed [8 marks]

Extended response 13 — Worldlines and tilted coordinate axes [9 marks]

Extended response 14 — Challenge: Light reaches the ends of a train [10 marks]

Extended response 15 — Challenge: When is the classical approximation adequate? [6 marks]

Part 3 — Lab-style data investigation

Testing time dilation using muon mean lifetimes [20 marks]

Topic A — Final exam readiness checklist

Related

Leave A Reply Cancel reply

Company

Links

Support

Recommend

alphy school publishers

Customise Consent Preferences

Log in with your account

Modal title