74 free, high-quality past-paper-style questions on Galilean and special relativity (HL) for IB Physics HL, each with a full mark scheme. Covers: The Lorentz factor; Postulates of special relativity; Proper time and proper length; and more. For when you have finished the past papers, or want to practise one topic at a time.Read the A.5 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.316 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1A-1A-27The Lorentz factorPaper 1AEasy1 mark
A particle moves with a speed v for which the Lorentz factor is γ.
Which expression gives v/c?
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Step 1γ = 1/√(1 − v²/c²), so 1/γ² = 1 − v²/c².
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2v²/c² = 1 − 1/γ², so v/c = √(1 − 1/γ²).
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Step 3Check: γ = 2 gives v = √0.75 c = 0.87c, less than c as it must be.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis rearranges γ as if it were √(1 + v²/c²); for γ = 2 it gives 1.7c, faster than light.
BThis is v²/c²: the square root has not been taken.
CThis forgets to square γ when taking the reciprocal of both sides.
DCorrect: v/c = √(1 − 1/γ²), which is always less than 1.
2A-1A-52Postulates of special relativityPaper 1AEasy1 mark
A sealed laboratory without windows is on a train that moves along a straight track at constant velocity relative to the ground.
Which experiment, carried out entirely inside the laboratory, could be used to determine the speed of the train relative to the ground?
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Step 1The laboratory is an inertial frame of reference.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2First postulate: the laws of physics are the same in all inertial frames, so no mechanical experiment inside the laboratory can reveal uniform motion.
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Step 3Second postulate: the speed of light is c in every inertial frame, so the light experiment also gives the same result whatever the speed of the train.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThe speed of light in a vacuum is c for every inertial observer; the measurement gives c whatever the speed of the train.
BThe pendulum obeys the same laws in every inertial frame, so its period is the same as on the ground. Only acceleration of the train would change it.
CA dropped ball falls with acceleration g relative to the laboratory, exactly as on the ground; uniform motion has no effect.
DCorrect: by the postulates of special relativity, no experiment inside an inertial frame can detect its uniform motion.
3IBP2H-008-00-01Postulates of special relativityPaper 1AEasy1 mark
A communications satellite moves directly away from a ground station at a speed of 0.50c relative to the station. The satellite fires a laser pulse back towards the station. What is the speed of the pulse measured by an observer at the ground station?
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The second postulate of special relativity states that the speed of light in a vacuum is the same, c, for all inertial observers, whatever the motion of the source. The station therefore measures c. 0.50c comes from Galilean subtraction of velocities, 1.5c from Galilean addition, and 0.87c wrongly applies a Lorentz factor to the speed of light.
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Answer C
Answer: C
Common error: Do not add or subtract the source speed for light. The motion of the source never changes the measured speed of light in a vacuum.
4IBP2H-008-00-02Proper time and proper lengthPaper 1AEasy1 mark
An astronaut on a fast-moving spacecraft wears a heart monitor that records the time between two consecutive heartbeats. Several observers measure this time interval. Which observer measures the proper time between the two heartbeats?
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Proper time is the time interval between two events measured in the frame in which the events occur at the same position. Both heartbeats happen at the astronaut's chest, so she, and the monitor at rest with her, measure the proper time. No frame is absolutely at rest, so the Earth frame is not special. All other observers measure a longer, dilated interval. The duration of each event does not matter.
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Answer B
Answer: B
Exam tip: To find proper time, ask which frame sees both events at the same location. That observer measures the shortest time interval.
5IBP2H-008-01-01Galilean and Lorentz transformationsPaper 1AEasy1 mark
A delivery drone flies along the aisle of a moving cargo train. Observers on the train and on the ground both measure the drone's motion. According to the Galilean transformation, which quantity has the same value for both observers?
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In Galilean relativity t′ = t, so time intervals are absolute and every inertial observer agrees on them. Velocities transform as u′ = u − v and positions as x′ = x − vt, so both depend on the frame. The speed of light would also change under a Galilean transformation; it is invariant only under the Lorentz transformation.
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Answer B
Answer: B
Common error: Choosing the speed of light confuses the two transformations. Light speed is invariant in special relativity, but the Galilean transformation predicts it would change from frame to frame.
6IBP2H-008-01-07Postulates of special relativityPaper 1AEasy1 mark
A spacecraft approaches Earth at 0.50c. It sends a laser pulse towards Earth. What speed does an observer on Earth measure for the laser pulse?
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The second postulate of special relativity states that the speed of light in a vacuum is the same, c, for all inertial observers, whatever the motion of the source. Option 1.5c is the Galilean result. Option 0.50c confuses the speed of the source with the speed of the light. Option c/γ wrongly applies time dilation to the speed of light.
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Answer B
Answer: B
Common error: Adding the source's speed to c is Galilean thinking. The motion of the source changes the frequency of the light (Doppler shift), not its speed.
7IBP3H-A5-20r43-01Proper time and proper lengthPaper 1AEasy1 mark
A long railway carriage moves at constant high speed along a straight track past a platform. A lamp fixed at the midpoint of the carriage emits a brief flash of light. Which statement about the light reaching the two ends of the carriage is correct?
Carriage moving past the platform; the lamp is at the midpoint (not to scale)
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In the carriage frame the lamp is at the midpoint and the light travels at c in both directions, so both ends are reached together. In the platform frame the light also travels at c in both directions, but the rear end moves towards the point where the light was emitted while the front end moves away from it, so the rear end is reached first. Simultaneity of events at different places is not absolute, so the statement that both frames agree is wrong. The front-end-first statement reverses the effect of the carriage's motion. The last statement swaps the roles of the frames: the lamp is at rest in the carriage frame, where the symmetry gives simultaneous arrival.
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Answer B
Answer: B
Common error: Events that are simultaneous in one frame but at different positions are not simultaneous in another frame; do not treat the time order of such events as absolute.
8IBP3H-A5-20r43-02Spacetime diagramsPaper 1AEasy1 mark
A spacetime diagram for frame S has ct on the vertical axis and x on the horizontal axis, with the same scale on both axes. A particle moves along the x-axis at a constant speed of 0.25c. What is the gradient, Δ(ct)/Δx, of its worldline?
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In a time Δt the particle moves Δx = 0.25cΔt, so Δ(ct)/Δx = cΔt/(0.25cΔt) = 4. The value 0.25 is the gradient the line would have if x were plotted against ct. A gradient of 1 is the worldline of light, which no massive particle can match. The value 0.75 comes from computing 1 minus the speed fraction, which has no meaning here.
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Answer C
Answer: C
Exam tip: On a diagram with ct vertical, the gradient of a worldline is 1/β, so slower objects have steeper lines and light has gradient 1.
9A-1A-09Muon decay experimentsPaper 1AMedium1 mark
Muons produced in the upper atmosphere travel towards the ground at a speed close to c. Far more of them reach the ground than would be expected from their mean lifetime at rest and the thickness of the atmosphere measured by an observer on the ground.
Which row gives a correct explanation in the frame of the ground and in the rest frame of the muons?
Frame of the groundRest frame of the muons
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Step 1Frame of the ground: the muons are moving clocks, so their mean lifetime is dilated to γτ0; the atmosphere is at rest, so its thickness is the proper length.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Rest frame of the muons: the muons are at rest, so their mean lifetime is the proper time τ0; the atmosphere moves past them, so its thickness is contracted by the factor γ.
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Step 3Both descriptions give the same number of mean lifetimes for the journey, so both observers predict the same number of muons reaching the ground.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: on the ground the moving muons’ lifetime is dilated; in the muon frame the moving atmosphere is contracted.
BThis reverses the two frames: a body’s own length or lifetime is never changed in its own rest frame.
CIn their own rest frame the muons are at rest, so their lifetime there is the proper lifetime and cannot be dilated.
DThe atmosphere is at rest relative to the ground, so in the ground frame its thickness is the proper length and is not contracted.
A train moves at constant speed v in the positive x-direction relative to a platform (frame S). The train frame S′ has its origin at the front of the train, and the origins of S and S′ coincide at t = 0. At t = 0 a passenger is at x′ = L and walks towards the rear of the train, in the negative x′-direction, at a constant speed w relative to the train. Both speeds are much less than c.
What is the position x of the passenger in frame S at time t?
Not to scale; O′ is the front of the train
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Step 1In the train frame the passenger starts at x′ = L and moves in the −x′ direction: x′ = L − wt.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Galilean transformation: x′ = x − vt, so x = x′ + vt.
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Step 3x = L − wt + vt = L + (v − w)t. Check: if w = v the passenger stays at x = L relative to the platform.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: x = x′ + vt with x′ = L − wt.
BThis uses x = x′ − vt (the transformation applied the wrong way round) and also reverses the walking direction: L + wt − vt.
CThis ignores the direction of walking and adds the two speeds, as if the passenger walked towards the front of the train.
DThis uses x = x′ − vt (the transformation applied the wrong way round): the platform observer would then see the passenger move backwards faster than the train moves forwards.
11A-1A-53Proper time and proper lengthPaper 1AMedium1 mark
A train moves at constant velocity past a long platform. Event 1 is the front of the train passing the start of the platform; event 2 is the front of the train passing the end of the platform. Observer F sits at the front of the train. Observer S stands on the platform at its start.
Which row identifies the observer who measures the proper time between events 1 and 2 and the observer who measures the proper length of the platform?
Proper time between events 1 and 2Proper length of the platform
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Step 1The proper time between two events is measured by a single clock that is present at both events. Both events happen at the front of the train, where F is; S is present only at event 1.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The proper length of an object is measured in the frame in which the object is at rest. The platform is at rest relative to S.
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Step 3So F measures the proper time and S the proper length; F measures a contracted platform, and the platform frame measures a dilated time between the events.
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThe proper time is right, but the platform moves relative to F, so F measures a contracted length, not the proper length.
BCorrect: F is present at both events; S is at rest relative to the platform.
CBoth are reversed: S is not present at event 2, and F is moving relative to the platform.
DThe proper length is right, but S is not present at event 2, so the platform frame needs a second clock and measures a dilated time.
13A-1A-55Relativistic velocity additionPaper 1AMedium1 mark
A spacecraft moves away from the Earth at 0.50c. It fires a projectile straight ahead. An observer on the Earth measures the speed of the projectile to be 0.80c.
What is the speed of the projectile relative to the spacecraft?
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Step 1In the Earth frame u = 0.80c; the spacecraft frame moves at v = 0.50c.
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All 3 steps must be completed — there is no mark for a part-answer.
The space-time diagram shows the axes of an inertial frame S and of a frame S′ moving at 0.50c relative to S. Five events A, B, C, D and E are marked.
Which two events are simultaneous in frame S′?
Axes of S (black) and of S′ (gold), which moves at 0.50c relative to S. Grid squares are 1 unit × 1 unit in S (drawn to scale).Show mark scheme
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Step 1Events simultaneous in S′ lie on a line parallel to the x′ axis. For v = 0.50c the x′ axis has gradient Δ(ct)/Δx = 0.50.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2From A (1, 1) to B (3, 2): Δ(ct)/Δx = 1/2 = 0.50, so AB is parallel to the x′ axis and A and B are simultaneous in S′.
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Step 3A and D are simultaneous in S (same ct); A and C occur at the same place in S; A and E lie on a line parallel to the ct′ axis, so they occur at the same place in S′.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: AB is parallel to the x′ axis, a line of constant t′.
BA and C have the same x: they occur at the same place in S, not at the same time in S′.
CA and D have the same ct: they are simultaneous in S, but not in S′ (relativity of simultaneity).
DAE is parallel to the ct′ axis, so A and E occur at the same place in S′; this confuses the two S′ axes.
A spacecraft of proper length 200 m moves at constant speed along its length through an open-ended straight hangar, of proper length 160 m, on a space station. In the rest frame of the hangar, there is an instant at which the spacecraft fits exactly inside the hangar.
What is the length of the hangar in the rest frame of the spacecraft?
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Step 1In the hangar frame the spacecraft is contracted from 200 m to 160 m, so γ = 200/160 = 1.25 (v = 0.60c).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2In the spacecraft frame the hangar is the moving object, so it is contracted by the same factor.
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Step 3Length of the hangar = 160/1.25 = 128 m. (In this frame the spacecraft does not fit: the two observers disagree about whether the ends of the spacecraft are inside the hangar at the same time.)
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis takes the ratio of the lengths, 160/200 = 0.80, as v/c instead of 1/γ: γ = 1.67 and 160/1.67 = 96 m.
BCorrect: γ = 1.25 and the moving hangar is contracted to 160/1.25 = 128 m.
CThis assumes that only the spacecraft is contracted. Length contraction is symmetric: each observer measures the other's object to be shorter.
DThis multiplies the proper length of the hangar by γ (160 × 1.25), so that the hangar again "fits" the spacecraft. Moving objects are shorter, never longer, than their proper length.
17IBP2H-008-00-03Spacetime intervalPaper 1AMedium1 mark
A deep-space probe moves at constant velocity. It leaves a relay station (event 1) and later passes a navigation beacon (event 2). In the frame of the relay station, the events are separated by 5.0 light-years in distance and 13 years in time. What time interval does the probe's own clock record between the two events?
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The probe is present at both events, so its clock records the proper time. This is found from the invariant spacetime interval: (cΔτ)2 = (cΔt)2 − Δx2 = 132 − 5.02 = 144, so Δτ = 12 years. 8.0 years subtracts the numbers directly. 14 years adds the squares. 13 years ignores time dilation.
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Answer B
Answer: B
Exam tip: Working in years and light-years sets c = 1. The interval then becomes simply Δt² − Δx².
18IBP2H-008-00-04The Lorentz factorPaper 1AMedium1 mark
A shuttle has a proper length of 120 m. An observer on a space station measures its length as 72 m as it flies past parallel to its length. What is the speed of the shuttle relative to the station?
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From length contraction, L = L0/γ, so 1/γ = 72/120 = 0.60. Then √(1 − v2/c2) = 0.60, which gives v2/c2 = 0.64 and v = 0.80c. 0.60c takes the length ratio as the speed. 0.40c uses 1 − 0.60. 0.64c forgets the square root.
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Answer D
Answer: D
Common error: The ratio L/L₀ equals 1/γ, not v/c. Square it, subtract from 1, then take the square root.
19IBP2H-008-00-05Relativistic velocity additionPaper 1AMedium1 mark
Two probes are launched from Earth in opposite directions. Each moves at 0.50c relative to Earth. What is the speed of one probe measured in the rest frame of the other?
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Use the relativistic velocity transformation u′ = (u − v)/(1 − uv/c2). Take u = −0.50c and v = +0.50c. Then u′ = −1.0c/(1 + 0.25) = −0.80c, so the speed is 0.80c, which is less than c. 1.0c is the Galilean result. 1.3c uses the wrong sign in the denominator. 0.50c ignores the motion of the second probe.
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Answer B
Answer: B
Common error: Keep track of signs. If the result exceeds c, the sign of the uv/c² term is wrong.
20IBP2H-008-00-07Galilean and Lorentz transformationsPaper 1AMedium1 mark
Two events are observed from several inertial frames moving relative to one another. Which quantity has the same value in every one of these frames?
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The spacetime interval (cΔt)2 − Δx2 is invariant under the Lorentz transformation. Time intervals and distances change from frame to frame because of time dilation and length contraction. Simultaneity of separated events is also relative. In Galilean relativity time intervals are invariant, but they are not in special relativity.
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Answer C
Answer: C
Common error: Galilean intuition says time is absolute. In special relativity only the combination (cΔt)² − Δx² is shared by all inertial observers.
21IBP2H-008-01-02Spacetime diagramsPaper 1AMedium1 mark
A spacetime diagram for frame S has axes ct and x drawn to the same scale. The ct′ and x′ axes of frame S′ are added to the diagram. The angle between the ct′ axis and the x′ axis is 30°. What is the speed of S′ relative to S?
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The ct′ axis is tilted by θ away from the ct axis. The x′ axis is tilted by the same angle θ away from the x axis. The angle between them is therefore 90° − 2θ = 30°, so θ = 30°. The worldline of S′ has tan θ = v/c, so v = c tan 30° = 0.58c. Option 0.27c comes from wrongly taking θ = 15°. Options 0.50c and 0.87c come from using sin θ or cos θ instead of tan θ.
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Answer C
Answer: C
Exam tip: Both primed axes close in towards the 45° light line by the same angle θ, with tan θ = v/c. Find θ first, then the speed.
22IBP2H-008-01-03Relativistic velocity additionPaper 1AMedium1 mark
A spacecraft moves away from Earth at 0.50c. It launches a probe backwards, towards Earth, at 0.80c relative to the spacecraft. What is the velocity of the probe relative to Earth?
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Take the direction away from Earth as positive. Then u′ = −0.80c and v = +0.50c. u = (u′ + v)/(1 + u′v/c2) = (−0.80 + 0.50)c/(1 − 0.40) = −0.30c/0.60 = −0.50c. So the probe moves at 0.50c towards Earth. Option 0.30c is the Galilean result. Option 0.21c comes from writing the denominator as 1 + 0.40, which ignores the signs. Option 0.50c away from Earth has the direction wrong.
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Answer B
Answer: B
Common error: Keep the signs of both velocities inside the denominator. When the velocities are opposite, uv/c² is negative and the denominator is less than 1.
23IBP2H-008-01-04Muon decay experimentsPaper 1AMedium1 mark
Muons created high in the atmosphere reach detectors at sea level in much greater numbers than non-relativistic physics predicts. Which statement correctly describes this observation from the muon's rest frame?
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In its own rest frame the muon lives for its proper lifetime, so no time dilation applies to it there. Instead, the atmosphere moves past the muon at close to c. Its thickness, which is a proper length in the Earth frame, is contracted by the factor γ. The ground therefore arrives within a few proper lifetimes. Time dilation is the explanation in the Earth frame, not the muon frame. The frames do agree on the relative speed, and no speed exceeds c.
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Answer B
Answer: B
Exam tip: Pair each frame with one effect. In the Earth frame the muon's clock runs slow (time dilation). In the muon frame the atmosphere is contracted (length contraction).
24IBP2H-008-01-06Spacetime intervalPaper 1AMedium1 mark
Two navigation beacons are at rest in the frame of a space station. Each beacon emits one flash. In the station frame the flashes are 5.0 years apart in time and 3.0 light-years apart in space. A courier ship moves at constant velocity and is present at both flashes. What time interval between the flashes does the ship's clock record?
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The spacetime interval is invariant. In the ship's frame both flashes happen at the same place, so the ship's clock measures the proper time. (cΔτ)2 = (cΔt)2 − Δx2 = 5.02 − 3.02 = 16 (light-years)2. Δτ = 4.0 years. This agrees with time dilation: v = 0.6c, γ = 1.25, and 5.0 / 1.25 = 4.0 years. The other options:
2.0 years comes from subtracting the numbers without squaring them.
5.8 years comes from adding the squares instead of subtracting.
5.0 years is the station's time interval, not the proper time.
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Answer B
Answer: B
Exam tip: The observer who is present at both events measures the proper time. Find it directly from the invariant interval, which has a minus sign: (cΔt)² − Δx².
25IBP3H-A5-20r43-03Relativity of simultaneityPaper 1AMedium1 mark
A spacecraft moves directly away from Earth at 0.60c, as measured from Earth. It launches a probe directly away from Earth at 0.60c relative to the spacecraft. What is the speed of the probe relative to Earth?
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Using u = (u′ + v)/(1 + u′v/c2) with u′ = v = 0.60c gives u = 1.20c/(1 + 0.36) = 0.88c. The value 1.20c is the Galilean sum, which exceeds c and so cannot be right. The value 0.75c comes from dividing by 1 + 0.60 instead of 1 + 0.36, forgetting to multiply the speeds together and square c. The value 1.00c wrongly assumes that any combined speed simply saturates at c, but the combined speed of two massive objects is always less than c.
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Answer A
Answer: A
Common error: Do not divide by 1 + v/c; the denominator in the velocity addition formula is 1 + u′v/c², which involves the product of the two speeds.
26IBP3H-A5-20r43-04Lorentz transformation and the invariant intervalPaper 1AMedium1 mark
Frame S′ moves at 0.80c in the +x direction relative to frame S. The origins coincide at t = t′ = 0. In S, a flare occurs at x = 9.0 × 108 m at t = 0. Take c = 3.00 × 108 m s−1. What is the time of the flare in S′?
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With γ = 1/√(1 − 0.802) = 5/3, the Lorentz transformation t′ = γ(t − vx/c2) gives t′ = (5/3)(0 − 0.80 × 9.0 × 108/(3.00 × 108)) = −4.0 s. The positive value comes from using the wrong sign in the bracket. A result of 0 s assumes that because t = 0, t′ must also be zero, which ignores relativity of simultaneity. The value −2.4 s is what you get if the factor γ is left out.
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Answer D
Answer: D
Exam tip: Check the sign convention in the Lorentz transformation: if S′ moves in +x relative to S, then t′ = γ(t − vx/c²) for coordinates measured in S.
27IBP3H-A5-20r43-05Galilean and Lorentz transformationsPaper 1AMedium1 mark
Two starting guns at opposite ends of a straight stadium fire at the same instant in the rest frame of the stadium: gun P at x = 0 and gun Q at x = L, where L > 0. A fast spacecraft flies along the +x direction past the stadium. Which statement describes the firing of the guns in the frame of the spacecraft?
Stadium frame; not to scale
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In the spacecraft frame, t′ = γ(t − vx/c2). Both events have the same t, so the event with the larger x (gun Q) has the smaller t′ and happens first. The suggestion that both fire together would require simultaneity to be absolute. The suggestion that gun P is first reverses the sign of the term vx/c2. The last statement makes the order depend on the position of the spacecraft at one instant, but the order of the two events in the spacecraft frame depends only on the velocity of the spacecraft and the separation of the guns.
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Answer C
Answer: C
Common error: The event further along the direction of relative motion of the other frame happens earlier in that frame; students often get this order reversed.
28IBP3H-A5-20r43-07Relativistic velocity additionPaper 1AMedium1 mark
Two spacecraft approach each other along a straight line. Each moves at 0.80c relative to a space station between them, as measured in the station frame. What is the speed of one spacecraft as measured from the other?
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In the frame of one spacecraft the station moves at 0.80c, and the other spacecraft moves at 0.80c relative to the station. Then u = (0.80 + 0.80)c/(1 + 0.80 × 0.80) = 1.60c/1.64 = 0.98c. The value 1.60c is the Galilean sum, which exceeds c. The value 0.89c comes from dividing by 1 + 0.80 instead of 1 + 0.64. The value 1.00c assumes that the combined speed simply saturates at c, but two massive objects always have a relative speed below c.
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Answer D
Answer: D
Common error: Even for two objects approaching at very high speed, the relative speed is always below c; check that your answer does not exceed 1.00c.
A star is 9.6 ly from the Earth in the Earth frame. A spacecraft leaves the Earth and travels to the star at a constant speed of 0.96c relative to the Earth.
Which statements are correct? I. In the Earth frame the journey takes 10 years. II. The crew measure the journey to take 2.8 years. III. According to the crew, a clock on the Earth advances by 0.78 years during the journey.
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Step 1Earth frame: time = distance/speed = 9.6 ly/0.96c = 10 years, so I is correct.
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All 4 steps must be completed — there is no mark for a part-answer.
Step 2γ = 1/√(1 − 0.96²) = 1/0.28 = 3.57. Leaving the Earth and arriving at the star both happen at the spacecraft, so the crew measure the proper time: 10/3.57 = 2.8 years. II is correct.
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Step 3In the crew frame it is the Earth clock that moves at 0.96c, so the crew judge it to run slow by the same factor: during their 2.8 years it advances 2.8/3.57 = 0.78 years. III is correct.
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Step 4There is no contradiction: the event "Earth clock reads 10 years" is simultaneous with the arrival only in the Earth frame (relativity of simultaneity). All three statements are correct.
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Answer D
Answer: D · 4 stages of work, one mark
Every option, and why
AThis rejects III by assuming the crew must find that the Earth clock has advanced 10 years. Time dilation is reciprocal: for the crew the Earth clock is the moving one, so it advances only 2.8/3.57 = 0.78 years.
BThis rejects II by applying time dilation the wrong way round (10 × 3.57 = 36 years for the crew). The crew are present at both events, so they measure the shortest (proper) time.
CThis rejects I by dilating the Earth time as well (10 × 3.57 = 36 years). In the Earth frame the time is simply the Earth distance divided by the speed.
DCorrect: 10 years in the Earth frame; the crew measure the proper time of 2.8 years and judge the moving Earth clock to advance only 0.78 years.
All 4 steps must be completed — there is no mark for a part-answer.
Step 2A negative value means the interval is space-like: light would need 5.0 years to cross the gap but only 3.0 years elapse, so no signal can connect the events.
—
Step 3For a space-like separation no frame can bring the events to the same place, but a frame moving at v = c²Δt/Δx = c × (3.0/5.0) = 0.60c sees them at the same time.
—
Step 4So the events can be made simultaneous, and only C is consistent with the invariant interval.
✓ 1
Answer C
Answer: C · 4 stages of work, one mark
Every option, and why
ACausation needs Δx ≤ cΔt. Here 5.0 ly > 3.0 ly, so nothing, not even light, could travel from one event to the other.
BA same-place frame exists only for time-like separations, where the "moving clock" travels between the events at less than c.
CCorrect: the interval is space-like, and a boost of 0.60c makes the events simultaneous.
D4.0 ly is √|Δs²|, the invariant distance in the frame where they are simultaneous — the time separation is different in every frame.
31A-1A-22Lorentz transformationsPaper 1AHard1 mark
In frame S, event Y occurs 4.0 μs after event X and 600 m further along the positive x-direction. Frame S′ moves at 0.60c in the positive x-direction relative to S.
What is the time between X and Y measured in S′?
Show mark scheme
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Mark
Notes
Step 1γ = 1/√(1 − 0.60²) = 1.25. The events are at different places in S, so 4.0 μs is not a proper time and Δt = γΔt0 cannot be used.
—
All 3 steps must be completed — there is no mark for a part-answer.
BThis assumes time intervals are the same in all frames (Galilean t′ = t).
CThis applies the time-dilation formula, 1.25 × 4.0 μs, treating 4.0 μs as a proper time. The events are 600 m apart in S, so the vΔx/c² term is needed.
DThis adds the vΔx/c² term instead of subtracting it: 1.25 × (4.0 + 1.2) μs = 6.5 μs.
32A-1A-31Galilean and Lorentz transformationsPaper 1AHard1 mark
Two inertial frames move relative to each other at a constant velocity.
Which quantities have the same value in both frames according to the Galilean transformation, but can have different values according to the Lorentz transformation? I. The time interval between two events II. The length of a rod lying along the direction of the relative motion III. The order in time of two events that occur at the same place in one of the frames
Show mark scheme
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Mark
Notes
Step 1Galilean transformation: t′ = t and x′ = x − vt, so time intervals, lengths and the order of any two events are the same in both frames.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Lorentz transformation: time intervals change (time dilation, relativity of simultaneity) and lengths along the motion change (length contraction), so I and II qualify.
—
Step 3Two events at the same place in one frame have a time-like separation (Δs² = c²Δt² > 0): one could cause the other, and their order is the same in every frame. III does not qualify.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: I and II differ between frames in special relativity; the order of two events at the same place is invariant in both theories.
BThis rejects II by forgetting length contraction, and accepts III by assuming that the order of any two events can be reversed. Only events with a space-like separation can change order.
CThis rejects I, treating time as absolute in special relativity as well; Δt′ = γ(Δt − vΔx/c²) differs from Δt.
DIII is wrong: reversing the order of two events at the same place would allow an effect to precede its cause.
33A-1A-56Relativity of simultaneityPaper 1AHard1 mark
Two beacons P and Q, 1200 m apart, are at rest in frame S with P at x = 0 and Q at x = 1200 m. They flash at the same time in S. A spacecraft moves at 0.80c in the positive x-direction, from P towards Q.
Which row gives the flash that happens first according to an observer on the spacecraft, and the time between the flashes measured by that observer?
Frame S: beacons at rest; not to scale
First flashTime between flashes
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Mark
Notes
Step 1γ = 1/√(1 − 0.80²) = 5/3. With Δt = 0 and Δx = xQ − xP = 1200 m:
—
All 3 steps must be completed — there is no mark for a part-answer.
A cargo shuttle of proper length 120 m passes a navigation beacon at a constant speed of 0.80c.
Which row gives the time taken for the shuttle to pass the beacon, as measured by an observer at the beacon and as measured by the crew of the shuttle?
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Beacon frame: the shuttle is contracted to 120/γ = 72 m, so the time is 72/(2.40 × 108) = 0.30 μs. Both events (front passes, rear passes) occur at the beacon, so this is the proper time.
—
Step 3Shuttle frame: the beacon moves 120 m (the proper length) at 0.80c: 120/(2.40 × 108) = 0.50 μs = γ × 0.30 μs, consistent with time dilation.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThe beacon time is right but the crew time has been divided by γ (0.30 × 3/5 = 0.18 μs); the crew, for whom the events are at different places, measures the longer time.
BCorrect: 0.30 μs (proper time, at the beacon) and 0.50 μs = γ × 0.30 μs.
CThe values are swapped: this assumes the crew, rather than the beacon observer, measures the proper time.
DThis uses the proper length in the beacon frame (0.50 μs) and then multiplies by γ (0.83 μs).
35A-1A-73Postulates of special relativityPaper 1AHard1 mark
A rod of proper length L0 moves at constant speed v along its own length relative to a laboratory. A light pulse is emitted from the rear end of the rod towards the front end. The Lorentz factor for speed v is γ.
What is the time taken for the pulse to reach the front end of the rod, measured in the laboratory frame?
Show mark scheme
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Mark
Notes
Step 1In the laboratory the rod is moving, so its length is L0/γ.
—
All 4 steps must be completed — there is no mark for a part-answer.
Step 2By the second postulate the pulse travels at c in the laboratory frame, while the front of the rod moves away from it at v. In this frame the gap closes at c − v.
—
Step 3Time = (L0/γ)/(c − v).
—
Step 4Check with the Lorentz transformation: in the rod frame Δt′ = L0/c and Δx′ = L0, so Δt = γ(L0/c + vL0/c²) = γL0(1 + v/c)/c, which is the same expression.
✓ 1
Answer D
Answer: D · 4 stages of work, one mark
Every option, and why
AThis contracts the rod but ignores the motion of its front end during the flight of the pulse.
BThis applies time dilation to the rod-frame time L0/c. The emission and the arrival happen at different places in the rod frame, so that time is not a proper time.
CThis allows for the front end moving away but forgets that the moving rod is contracted to L0/γ in the laboratory.
DCorrect: contracted length L0/γ, closed at the rate c − v in the laboratory frame.
36IBP2H-008-00-06Spacetime intervalPaper 1AHard1 mark
In frame S, event B occurs 5.0 light-years from event A, in the +x direction, and 3.0 years after A. Which statement is correct?
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Mark
Notes
The spacetime interval is (cΔt)2 − Δx2 = 9 − 25 = −16 ly2, so the interval is spacelike. A signal would need to travel faster than light to link the events, so A cannot cause B and no frame sees them at the same place. Using the Lorentz transformation, Δt′ = γ(Δt − vΔx/c2) = γ(3.0 − 0.60 × 5.0) = 0, so the events are simultaneous in that frame. The value 4 is a proper length (4.0 ly), not a proper time.
✓ 1
Answer C
Answer: C
Exam tip: Check the sign of the interval first. A spacelike interval rules out a causal link and rules out any proper time between the events.
37IBP2H-008-00-08Lorentz transformationsPaper 1AHard1 mark
A spacecraft of proper length L0 moves at 0.60c relative to a planet. A light pulse is sent from the rear of the spacecraft to the front. In the spacecraft frame the pulse takes a time L0/c. What time does the pulse take according to an observer on the planet?
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Notes
At 0.60c, γ = 1.25. On the planet the spacecraft is contracted to 0.80L0. The light travels at c while the front moves away at 0.60c, so it closes the gap at 0.40c. The time is 0.80L0/(0.40c) = 2.0L0/c. The Lorentz transformation agrees: Δt = γ(Δt′ + vΔx′/c2) = 1.25(1 + 0.60)L0/c = 2.0L0/c. Option C applies simple time dilation, which is invalid because the emission and arrival events occur at different places in the spacecraft frame.
✓ 1
Answer D
Answer: D
Common error: Time dilation, Δt = γΔt₀, only applies when the two events occur at the same place in one frame. Otherwise use the full Lorentz transformation.
38IBP2H-008-01-05Relativity of simultaneityPaper 1AHard1 mark
Two landing lights, one at each end of a straight runway of proper length 600 m, flash simultaneously in the runway frame. A spacecraft flies over the runway at 0.80c, moving from the west end towards the east end. In the spacecraft frame, what is the time interval between the flashes, and which flash happens first?
Runway frame; not to scale
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Mark
Notes
Use the Lorentz transformation Δt′ = γ(Δt − vΔx/c2) with Δt = 0, Δx = xeast − xwest = +600 m and γ = 5/3. Δt′ = −(5/3)(0.80)(600)/(3.00×108) = −2.7×10−6 s. The negative sign means the east flash has the earlier t′, so the flash at the front (in the direction of travel) happens first. Option 1.6 μs omits γ. Option 'west first' has the sign wrong. Simultaneity is not absolute in special relativity, so the final option is wrong.
✓ 1
Answer B
Answer: B
Common error: Leaving out γ in the term vΔx/c² gives the wrong magnitude. Use the full transformation Δt′ = γ(Δt − vΔx/c²).
39IBP2H-008-01-08Spacetime intervalPaper 1AHard1 mark
In frame S, two events are separated by a distance Δx and a time Δt, where Δx > cΔt > 0. Which statement is correct?
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Notes
Since Δx > cΔt, the interval (cΔt)2 − Δx2 is negative, so the events are spacelike separated. A frame moving at v = c2Δt/Δx, which is less than c, gives Δt′ = γ(Δt − vΔx/c2) = 0, so the events are simultaneous in that frame. No signal can travel between the events, so one cannot cause the other. The order of spacelike events depends on the frame. A frame in which both events happen at the same place would need v = Δx/Δt > c, which is impossible.
✓ 1
Answer C
Answer: C
Exam tip: Classify the interval first. Timelike events can be causally linked and can share a location in some frame. Spacelike events cannot be causally linked and can be made simultaneous in some frame.
40IBP3H-A5-20r43-06Spacetime intervalPaper 1AHard1 mark
In frame S, two flares occur 12 light-minutes apart in space and 13 minutes apart in time. In frame S′, both flares occur at the same position. What is the time interval between the flares in S′?
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Notes
The spacetime interval is invariant: (cΔt)2 − (Δx)2 = 132 − 122 = 25 min2 (using units in which light travels 1 light-minute per minute). In S′ the separation is zero, so (cΔt′)2 = 25 and Δt′ = 5.0 min. The value 1.0 min comes from simply subtracting 12 from 13, and 25 min comes from forgetting to take the square root. The value 12.5 min is the mean of the two given values, which has no physical basis.
✓ 1
Answer A
Answer: A
Exam tip: The quantity (cΔt)² − (Δx)² has the same value in every inertial frame; use it to find the time between two events in a frame where they occur at the same place.
41IBP3H-A5-20r43-08Spacetime intervalPaper 1AHard1 mark
In frame S, event Q occurs 4.0 light-seconds from event P along the x-axis and 3.0 s after it. A frame S′ moves along the x-axis at speed v relative to S, and in S′ the events P and Q are simultaneous. What is v?
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Notes
Setting Δt′ = γ(Δt − vΔx/c2) = 0 gives v = c2Δt/Δx = c(3.0/4.0) = 0.75c. The value 0.43c comes from dividing 3.0 by the sum of 3.0 and 4.0. The value 0.60c comes from dividing 3.0 by the hypotenuse of a 3–4–5 triangle, which mixes up the invariant interval with the condition for simultaneity. The value 1.33c inverts the ratio and is impossible since it exceeds c.
✓ 1
Answer B
Answer: B
Common error: For events separated by more distance than light can cover in the time between them, there is a frame in which they are simultaneous; the frame speed is v = c²Δt/Δx, which is always less than c.
42A-1B-23Postulates of special relativityPaper 1BEasy7 marks
In an accelerator, neutral pions moving at 0.9990c relative to the laboratory decay into gamma-ray photons. Photons emitted in the direction of motion of the pions are detected at a distance d from the decay region. The time t between a reference signal from the accelerator and the detection of the photons is recorded; each time is ±0.5 ns. Each value of t includes the same fixed delay in the electronics.
The graph shows t against d with the line of best fit.
d / m
5.0
10.0
15.0
20.0
25.0
30.0
t / ns
28.4
45.0
62.2
78.1
95.3
111.4
Graph drawn to scale
(a)
(i)
Determine the gradient of the graph. Give its unit.
(2)
(ii)
Hence determine the speed of the photons in the laboratory frame.
(1)
(b)
(i)
Determine the fixed delay, using your gradient and one data point.
(2)
(ii)
The uncertainty in the gradient is ±2 %. Discuss whether the data support the Galilean prediction or special relativity for the speed of the photons.
(2)
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Mark
Notes
Part (a)(i)
Using the full range: (111.4 − 28.4)/(30.0 − 5.0) = 3.32
Allow ECF from (a)(i). Accept 2.9–3.1 × 108 m s−1.
Part (b)(i)
Intercept = t − gradient × d, e.g. 111.4 − 3.32 × 30.0
✓ 1
Allow ECF from (a)(i).
Delay = 12 ns
✓ 1
Accept 11–13 ns. A value read directly from the graph by extrapolation is also accepted.
Part (b)(ii)
Galilean addition predicts c + 0.9990c ≈ 6.0 × 108 m s−1 (gradient ≈ 1.7 ns m−1); special relativity predicts c for all inertial observers
✓ 1
Measured (3.01 ± 0.06) × 108 m s−1 includes c and excludes 2c, so the data support special relativity (the second postulate)
✓ 1
Allow ECF from (a)(ii): the conclusion must use the candidate's value and its uncertainty.
Answers: (a)(i) 3.32 ns m−1 · (a)(ii) 3.0 × 108 m s−1 · (b)(i) 12 ns · (b)(ii) supports special relativity (the remaining parts are explanations — see the table above)
43A-1B-02Galilean velocity additionPaper 1BMedium7 marks
A loudspeaker and a microphone are fixed 1.500 ± 0.002 m apart inside a long straight air duct, with the microphone downstream. Air flows past them at a speed w measured with an anemometer. The loudspeaker emits a short click and a data logger records the time t until the microphone detects it; each reading is ±0.01 ms. When the microphone is pressed against the loudspeaker, the logger reads 0.15 ms.
The speed of the click relative to the duct is u = 1.500 m/(t − 0.15 ms).
w / m s−1
t / ms
u / m s−1
0.0
4.52
343.2
4.0
4.47
347.2
8.0
4.42
351.3
12.0
4.38
16.0
4.33
358.9
20.0
4.28
363.2
Long straight air duct (not to scale)
(a)
(i)
Calculate u for w = 12.0 m s−1.
(2)
(ii)
Determine the absolute uncertainty in your answer to (a)(i).
(2)
(b)
(i)
Sound travels at a fixed speed cs relative to the air. Galilean relativity therefore predicts u = cs + w. Test this prediction using at least three of the data points.
(2)
(ii)
The loudspeaker and the microphone are interchanged so that the click travels against the air flow. Predict the reading of the logger when w = 20.0 m s−1.
(1)
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Notes
Part (a)(i)
Correct for the zero offset: t − 0.15 = 4.38 − 0.15 = 4.23 ms
✓ 1
u = 1.500/(4.23 × 10−3) = 355 m s−1
✓ 1
Accept 354–355 m s−1. 342 m s−1 (offset not subtracted) scores [1 max].
Part (a)(ii)
The corrected time is the difference of two readings: ±0.02 ms, i.e. 0.02/4.23 = 0.47 %; distance 0.002/1.500 = 0.13 %
✓ 1
Award this mark only if the time uncertainty is doubled (two readings).
Total 0.6 %, so Δu = ±2 m s−1
✓ 1
Allow ECF from (a)(i). Absolute uncertainty to 1 s.f.; accept ±2.2. ±1 m s−1 (one reading only) scores [1 max].
Part (b)(i)
Calculates u − w for at least three points, e.g. 343.2, 343.3, 342.6, 343.2 m s−1
✓ 1
Allow ECF from (a)(i) for the 12.0 m s−1 point. A test using two points only scores [1 max] in total.
The values are constant (≈ 343 m s−1) to within the ±2 m s−1 uncertainty, so the data support u = cs + w (with cs ≈ 343 m s−1)
✓ 1
Allow ECF from (a)(ii): the conclusion must refer to the uncertainty.
Part (b)(ii)
u = 343 − 20.0 = 323 m s−1; logger reading = 1.500/323 + 0.15 ms = 4.64 + 0.15 = 4.79 ms
✓ 1
Allow ECF from (b)(i) value of cs. Accept 4.78–4.80 ms. 4.64 ms (offset not added back) scores [0].
Answers: (a)(i) 355 m s−1 · (a)(ii) ±2 m s−1 · (b)(i) cs ≈ 343 m s−1; supported · (b)(ii) 4.79 ms (the remaining parts are explanations — see the table above)
44A-1B-20Time dilation and length contractionPaper 1BMedium7 marks
A beam of unstable particles, all moving at 0.95c relative to the laboratory (γ = 3.20), travels along a straight vacuum pipe. Detectors at distances x from the production point count the number N of particles that reach them in a fixed time. The particles decay exponentially, so in the laboratory frame ln N = ln N0 − x/(γvτ0), where τ0 is the mean lifetime of the particles in their rest frame.
The uncertainty in each value of ln N is negligible. The graph shows ln N against x with the line of best fit.
x / m
N
ln N
2.0
4241
8.35
4.0
3589
8.19
6.0
2902
7.97
8.0
2423
7.79
10.0
2084
7.64
12.0
1804
7.50
Graph drawn to scale
(a)
(i)
Determine the gradient of the graph. Give its unit.
(2)
(ii)
Hence determine τ0.
(2)
(iii)
Predict the number of particles detected by a detector placed at x = 18.0 m.
(2)
(iv)
Every detector records only 80 % of the particles that reach it. State and explain whether this affects your answer to (a)(ii).
(1)
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Notes
Part (a)(i)
Large triangle on the line, e.g. (7.50 − 8.35)/(12.0 − 2.0) = −0.085 or from the best-fit line −0.087
Accept 12–13 ns. Omitting γ (40 ns) scores 0 for this mark.
Part (a)(iii)
Distance over which N halves = ln 2/|gradient| = 0.693/0.087 = 8.0 m
✓ 1
Allow ECF from (a)(i).
18.0 m is two halving distances beyond 2.0 m, so N ≈ 4241/4 ≈ 1.1 × 103
✓ 1
Accept 1000–1100 (or N = N0e−0.087 × 18.0 with N0 from the intercept).
Part (a)(iv)
No: every N is multiplied by 0.80, which subtracts the same amount (ln 0.80) from every ln N; the line moves down but its gradient, from which τ0 is found, is unchanged
✓ 1
Allow ECF from (a)(ii).
Answers: (a)(i) −0.087 m−1 · (a)(ii) 1.3 × 10−8 s · (a)(iii) ≈ 1.1 × 103(the remaining parts are explanations — see the table above)
In an ion-beam experiment, singly charged ions of mass m, initially at rest, are accelerated through a potential difference V and then move at a constant speed v ≪ c. An atomic transition in the ions acts as a clock. The time interval Δt0 between two "ticks" in the ions' rest frame is compared with the interval Δt measured in the laboratory, giving the fractional shift f = (Δt − Δt0)/Δt0.
For v ≪ c, γ ≈ 1 + v²/(2c²). Each value of f has an uncertainty of ±0.3 × 10−7.
V / kV
1.0
2.0
3.0
4.0
5.0
6.0
f / 10−7
1.89
3.05
4.81
6.26
7.80
9.34
Graph drawn to scale, with the line of best fit
(a)
(i)
Show that f = eV/(mc²), where e is the elementary charge.
(2)
(b)
(i)
The lines of maximum and minimum gradient pass through the ends of the error bars of the first and last points. Determine the gradient of the line of best fit and its absolute uncertainty.
(2)
(ii)
Hence determine m, with its absolute uncertainty.
(2)
(iii)
The ions are of a single isotope of lithium. Deduce, using the unified atomic mass unit, whether they are lithium-6 or lithium-7.
(1)
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Notes
Part (a)(i)
Δt = γΔt0, so f = γ − 1 ≈ v²/(2c²)
✓ 1
Energy gained eV = ½mv², so v² = 2eV/m and f = eV/(mc²)
m = e/(c² × gradient) = 1.60 × 10−19/(9.00 × 1016 × 1.51 × 10−10) = 1.18 × 10−26 kg
✓ 1
Allow ECF from (b)(i). Gradient must be per volt.
Same percentage uncertainty (≈ 8 %): Δm = ±0.09 × 10−26 kg, so m = (1.18 ± 0.09) × 10−26 kg
✓ 1
Allow ECF from (b)(i). Accept ±0.08 to ±0.11 × 10−26 kg.
Part (b)(iii)
m/u = 1.18 × 10−26/1.66 × 10−27 = 7.1 ± 0.6, which includes 7 but not 6: lithium-7
✓ 1
Allow ECF from (b)(ii): the conclusion must use the candidate's uncertainty range.
Answers: (b)(i) (1.51 ± 0.12) × 10−10 V−1 · (b)(ii) (1.18 ± 0.09) × 10−26 kg · (b)(iii) lithium-7 (the remaining parts are explanations — see the table above)
46IBP2H-008-00-09Postulates of special relativityPaper 2Easy4 marks
In a proton therapy unit used to treat tumours, protons are accelerated to a speed of 0.45c relative to the treatment room.
(a)
State the two postulates of special relativity.
(2)
(b)
Calculate the Lorentz factor γ for the protons.
(2)
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Notes
Part (a)
The laws of physics are the same in all inertial reference frames . The speed of light in a vacuum is the same for all inertial observers, independent of the motion of the source or the observer .
47IBP2H-008-01-09Galilean relativityPaper 2Easy6 marks
A cruise ship moves due east at a constant 6.0 m s−1 relative to the shore. A passenger on deck throws a ball due west at 15 m s−1 relative to the ship. Frame S is the shore and frame S′ is the ship. The x-axes of both frames point east.
(a)
State the Galilean transformation equations for the coordinates x′ and t′ of an event in S′, in terms of x, t and the speed v of the ship.
(2)
(b)
Calculate the velocity of the ball relative to the shore.
(2)
(c)
A crew member shines a light pulse due west from the deck. Outline why the Galilean transformation cannot correctly describe this light pulse.
(2)
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Notes
Part (a)
x′ = x − vt
✓ 1
t′ = t
✓ 1
Part (b)
Galilean velocity addition: u = u′ + v = (−15) + 6.0
✓ 1
u = −9.0 m s−1, i.e. 9.0 m s−1 due west
✓ 1
Part (c)
The Galilean transformation predicts that the light's speed relative to the shore is c − 6.0 m s−1, which is different from c
✓ 1
Experiment, and the second postulate, show that light travels at c in every inertial frame, so the Galilean transformation fails for light
✓ 1
Common error: Ignoring direction leads to adding 15 and 6.0 to get 21 m s⁻¹. Choose a positive direction and give each velocity its sign before combining them.
48IBP3H-A5-20r43-09Galilean relativityPaper 2Easy5 marks
A train moves along a straight track at 40 m s−1 relative to the ground. A passenger rolls a ball forward along the aisle at 5.0 m s−1 relative to the train. A second passenger shines a torch beam forward along the aisle. Take c = 3.00 × 108 m s−1.
(a)
Calculate the speed of the ball relative to the ground using Galilean addition of velocities.
(1)
(b)
Outline why Galilean addition of velocities cannot be used to find the speed of the torch light relative to the ground.
(2)
(c)
The relativistic formula for the speed of the ball relative to the ground is u = (u′ + v)/(1 + u′v/c2). Calculate u′v/c2 for the ball and hence comment on the use of Galilean addition for the ball.
(2)
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Notes
Part (a)
A1 45 m s−1
✓ 1
Part (b)
The speed of light in a vacuum has the same value c in all inertial reference frames (second postulate)
✓ 1
Galilean addition would give c + 40 m s−1 relative to the ground, which contradicts this, so the relativistic formula must be used (the speed is c)
49IBP3H-A5-20r43-10Lorentz transformationsPaper 2Easy5 marks
Frame S′ moves at 0.60c in the +x direction relative to frame S. The origins coincide at t = t′ = 0. A pulsar flash is recorded in S at x = 4.0 × 108 m and t = 2.0 s. Use the Lorentz transformations x′ = γ(x − vt) and t′ = γ(t − vx/c2), with c = 3.00 × 108 m s−1.
(a)
Calculate the Lorentz factor γ for S′.
(1)
(b)
Determine the position coordinate x′ of the flash in S′.
(2)
(c)
Determine the time coordinate t′ of the flash in S′.
Common error: Keep v as a speed in m s⁻¹ when multiplying by t, and as a fraction of c when forming vx/c²; mixing the two gives an unreasonably large or small coordinate.
A train moves along a straight horizontal track at a constant velocity of 60.0 m s−1 in the positive x-direction relative to the ground. Frame S is attached to the ground and frame S′ to the train; their origins coincide at t = t′ = 0. The air is still relative to the ground, and the speed of sound in the air is 343 m s−1.
(a)
(i)
State what is meant by an inertial frame of reference.
(1)
(ii)
At t = 15.0 s a lamp beside the track at x = 1850 m flashes. Determine the coordinates of this event in S′.
(2)
(iii)
A car on a road beside the track moves at 35 m s−1 in the positive x-direction. Determine the velocity of the car relative to the train.
(2)
(b)
(i)
The train sounds its horn. Determine the speed, relative to the train, of the sound travelling forwards.
(2)
(ii)
The horn emits a frequency of 400 Hz. Calculate the frequency heard by a person standing beside the track ahead of the train.
(2)
(iii)
Explain, using your answer to (b)(i), why the wavelength of the sound in front of the train is 0.71 m.
(1)
(c)
(i)
The train's headlamp emits light forwards. State the speed of this light relative to the ground predicted by the Galilean transformation, and explain why this prediction must be rejected.
(2)
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Mark
Notes
Part (a)(i)
A frame that is not accelerating, in which Newton's first law holds
✓ 1
Part (a)(ii)
x′ = x − vt = 1850 − 60.0 × 15.0 = 950 m
✓ 1
t′ = t = 15.0 s
✓ 1
Part (a)(iii)
u′ = u − v = 35 − 60.0
✓ 1
= −25 m s−1: 25 m s−1 in the negative x-direction relative to the train
✓ 1
Direction must be stated or shown by the sign.
Part (b)(i)
The speed of sound is fixed relative to the air (the medium), not the source: 343 m s−1 relative to the ground
✓ 1
The hidden step: the source speed is not added.
Relative to the train u′ = 343 − 60.0 = 283 m s−1
✓ 1
403 m s−1 scores [0].
Part (b)(ii)
f′ = fv/(v − us) = 400 × 343/(343 − 60.0)
✓ 1
f′ = 485 Hz
✓ 1
Accept 485 Hz.
Part (b)(iii)
In one period (1/400 s) a wavefront moves 283 m s−1 × 1/400 s = 0.71 m ahead of the train, which is where the next wavefront is emitted, so λ = 283/400 = 0.71 m
✓ 1
Allow ECF from (b)(i).
Part (c)(i)
Galilean prediction: c + 60.0 m s−1
✓ 1
The second postulate of special relativity (supported by experiment): light travels at c in all inertial frames, whatever the motion of the source; at 60 m s−1γ − 1 ≈ 2 × 10−14, so the Galilean equations remain excellent for the train itself
✓ 1
The postulate is needed for the second mark.
Answers: (a)(ii) x′ = 950 m, t′ = 15.0 s · (a)(iii) −25 m s−1 · (b)(i) 283 m s−1 · (b)(ii) 485 Hz (the remaining parts are explanations — see the table above)
51A-2-32Relativistic velocity additionPaper 2Medium11 marks
In a laboratory, a beam of unstable particles B is selected by passing it between parallel plates 5.0 mm apart with a potential difference of 12 kV between them. A uniform magnetic field of flux density 0.010 T acts perpendicular to both the beam and the electric field. Only particles that pass straight through are used. Beam B travels in the negative x-direction. A second beam of particles A travels at 0.60c in the positive x-direction.
Velocity selector, side view (not to scale)
(a)
(i)
Show that the speed of the particles in beam B is 0.80c.
(2)
(ii)
Calculate the speed of B relative to A according to Galilean relativity, and state why this result cannot be correct.
(1)
(iii)
Determine the velocity of B relative to A according to special relativity.
(2)
(b)
(i)
A bunch of particles in beam B has a length of 50.0 mm in its own rest frame. Determine its length as measured by an observer moving with beam A.
(2)
(ii)
The half-life of the particles at rest is 26 ns. Calculate their half-life as measured in the laboratory.
(2)
(c)
(i)
An observer moving with beam B measures the speed of a light pulse emitted along the negative x-direction in the laboratory. Show, using the relativistic velocity addition equation, that the result is c for any speed v of beam B.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
E = V/d = 12 × 103/5.0 × 10−3 = 2.4 × 106 V m−1
✓ 1
Undeflected: qE = qvB, so v = E/B = 2.4 × 108 m s−1 = 0.80c
✓ 1
Part (a)(ii)
0.80c + 0.60c = 1.4c; nothing (no particle) can move faster than c relative to an inertial observer
✓ 1
Allow ECF from (a)(i).
Part (a)(iii)
u′ = (u − v)/(1 − uv/c²) with u = −0.80c, v = +0.60c: u′ = −1.40c/1.48
✓ 1
Allow ECF from (a)(i).
u′ = −0.95c: 0.95c in the negative x-direction
✓ 1
Accept 0.946c; direction needed.
Part (b)(i)
γ = 1/√(1 − 0.946²) = 3.08
✓ 1
Allow ECF from (a)(iii). Accept 3.0–3.1.
L = 50.0/3.08 = 16 mm
✓ 1
Accept 16–17 mm.
Part (b)(ii)
γ for 0.80c = 1/√(1 − 0.64) = 5/3; the 26 ns is a proper time
✓ 1
Allow ECF from (a)(i).
Laboratory half-life = 5/3 × 26 = 43 ns
✓ 1
Part (c)(i)
In the laboratory u = −c and the frame of B has velocity −v: u′ = (−c + v)/(1 − (−c)(−v)/c²)
✓ 1
Symbolic working required.
= −(c − v)/(1 − v/c) = −c: speed c, as the second postulate requires
✓ 1
Answers: (a)(ii) 1.4c · (a)(iii) 0.95c in the −x direction · (b)(i) 16 mm · (b)(ii) 43 ns (the remaining parts are explanations — see the table above)
52IBP2H-008-00-10Time dilationPaper 2Medium6 marks
A navigation satellite orbits Earth at a speed of 3.87 km s−1. Ignore gravitational effects on time and treat the satellite as an inertial frame over short intervals. Take c = 3.00 × 108 m s−1.
(a)
Show that, for v ≪ c, γ − 1 ≈ v2/2c2.
(2)
(b)
Outline why calculating γ directly on a calculator is not useful here.
(1)
(c)
Calculate, according to an Earth observer, the time by which the satellite clock falls behind an Earth clock in one day (86 400 s).
(2)
(d)
The satellite signals travel at c. Estimate the error in position after one day if this effect were not corrected.
(1)
Show mark scheme
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Mark
Notes
Part (a)
Write γ = (1 − v2/c2)−1/2 and use the binomial expansion (1 − x)−1/2 ≈ 1 + x/2 for small x . With x = v2/c2, γ ≈ 1 + v2/2c2, so γ − 1 ≈ v2/2c2 .
✓ 2
Part (b)
v2/c2 ≈ 1.7 × 10−10, so γ differs from 1 only in about the tenth decimal place. Accept either consequence of the calculator's limited significant figures: γ rounds to 1, or γ − 1 is left with very few (or no) reliable significant figures .
53IBP2H-008-00-11Proper time and proper lengthPaper 2Medium7 marks
Electrons in a linear accelerator travel along a straight evacuated tube that is 3.2 km long in the laboratory frame. At the exit the electrons have a Lorentz factor of 4.0 × 104. Assume this value of γ applies along the whole tube and that the electron speed is effectively c (3.0 × 108 m s−1).
(a)
Define proper length.
(1)
(b)
Calculate the length of the tube in the rest frame of an electron.
(2)
(c)
Calculate the time for an electron to traverse the tube in the laboratory frame.
(1)
(d)
Using your answer to (b), determine the time for the traversal in the electron's frame. Show that your answer is consistent with time dilation.
(3)
Show mark scheme
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Notes
Part (a)
The length of an object (or the distance between two points) measured in the frame in which the object (or the points) is at rest .
✓ 1
Part (b)
L = L0/γ = 3200/(4.0 × 104) . L = 0.080 m .
✓ 2
Part (c)
t = 3200/(3.0 × 108) = 1.07 × 10−5 s (accept 1.1 × 10−5 s) .
✓ 1
Part (d)
t′ = 0.080/(3.0 × 108); allow ECF from (b) . t′ = 2.7 × 10−10 s . Ratio t/t′ = 1.07 × 10−5/2.67 × 10−10 ≈ 4.0 × 104 = γ. The electron measures the proper time because entry and exit happen at the same place in its frame .
✓ 3
Exam tip: Length contraction in one frame and time dilation in the other describe the same physics. Carry unrounded values forward so that the ratio of the times comes out equal to γ.
54IBP2H-008-00-12Spacetime intervalPaper 2Medium7 marks
An astronomer records three events in the Earth frame S. All positions lie along the x-axis.
Event | x / ly | t / years A | 0 | 0 B | 3.0 | 5.0 C | 6.0 | 2.0
Use (Δs)2 = (cΔt)2 − (Δx)2.
(a)
Calculate (Δs)2 for the pair A–B and for the pair A–C.
(2)
(b)
Using your answers to (a), deduce whether event A could be the cause of event B, and whether event A could be the cause of event C.
(2)
(c)
Calculate the proper time between A and B.
(1)
(d)
Determine the speed, relative to S, of the inertial frame in which A and B occur at the same position. Show that this is consistent with your answer to (c).
A–B: (Δs)2 > 0 (timelike) and B occurs after A, so a signal travelling at ≤ c could reach B from A; A could cause B . A–C: (Δs)2 < 0 (spacelike). A signal would need to travel faster than c (6.0 ly in 2.0 years, i.e. 3c), so A cannot cause C. Allow ECF from (a) .
✓ 2
Part (c)
Δτ = √16 = 4.0 years. Allow ECF from (a) .
✓ 1
Part (d)
v = Δx/Δt = 3.0 ly / 5.0 years = 0.60c . γ = 1/√(1 − 0.602) = 1.25, so Δτ = 5.0/1.25 = 4.0 years, which agrees with (c) .
✓ 2
Exam tip: With this sign convention, a positive (Δs)² means the interval is timelike: a proper time exists and the earlier event can cause the later one. A negative value means no causal link is possible.
55IBP2H-008-00-15Spacetime diagramsPaper 2Medium6 marks
Electrons in a hospital radiotherapy linear accelerator are accelerated until their Lorentz factor γ is 2.50. Take c = 3.00 × 108 m s−1.
(a)
Describe how γ varies with v/c for 0 ≤ v/c < 1.
(2)
(b)
Calculate the speed of the electrons when γ = 2.50.
(2)
(c)
Outline, with reference to γ, why the electrons can never reach the speed of light.
(2)
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Notes
Part (a)
γ = 1 when v = 0 and stays close to 1 for small v/c, so relativistic effects are negligible at everyday speeds . γ increases ever more rapidly and tends to infinity as v/c → 1 .
✓ 2
Part (b)
v = c√(1 − 1/γ2) = c√(1 − 0.16) = c√0.84 . v = 0.917c = 2.75 × 108 m s−1 .
✓ 2
Part (c)
The total energy is E = γm0c2 (or the kinetic energy is (γ − 1)m0c2), and γ → ∞ as v → c . An infinite amount of work or energy would therefore be needed to accelerate a particle with mass to c, which is impossible, so v < c always .
✓ 2
Exam tip: Rearrange to v = c√(1 − 1/γ²) before you substitute. This avoids most algebra slips when γ is given.
56IBP2H-008-01-10Time dilation and length contractionPaper 2Medium8 marks
Students analyse data from a cosmic-ray experiment. Detector A is on a mountain summit and detector B is at sea level. Both detectors count only muons moving vertically downward at the speed given in the table.
Quantity — Value Height of A above B — 3.20 km Speed of the muons — 0.980c Half-life of muons at rest — 1.52 μs Count rate at A — 600 per hour
Take c = 3.00×108 m s−1.
Not to scale
(a)
Calculate the Lorentz factor γ of the muons.
(1)
(b)
Calculate the time taken for the muons to travel from A to B, as measured in the Earth frame.
(1)
(c)
Determine the count rate expected at B if relativistic effects are ignored.
(2)
(d)
Determine the count rate expected at B when relativistic effects are included.
(2)
(e)
Explain how an observer moving with the muons accounts for the count rate at B.
(2)
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Notes
Part (a)
γ = 1/√(1 − 0.9802) = 5.03
✓ 1
Part (b)
t = 3200/(0.980 × 3.00×108) = 1.09×10−5 s (10.9 μs)
✓ 1
Part (c)
Number of half-lives = 10.9/1.52 = 7.16
✓ 1
Rate = 600 × 2−7.16 ≈ 4.2 per hour
✓ 1
Part (d)
Proper time = 10.9 μs/5.03 = 2.17 μs, which is 1.42 half-lives (ECF from (a) and (b))
57IBP2H-008-01-11Relativistic velocity additionPaper 2Medium8 marks
In a particle physics laboratory, a radioactive nucleus moves at 0.70c relative to the laboratory. It emits an electron.
(a)
The electron is emitted forward at 0.90c relative to the nucleus. State the Galilean prediction for the electron's speed in the laboratory and explain why it must be wrong.
(2)
(b)
Calculate the relativistic speed of this electron in the laboratory.
(2)
(c)
A second, identical nucleus moving at 0.70c emits an electron backwards at 0.90c relative to the nucleus. Calculate the velocity of this electron in the laboratory.
(2)
(d)
A third nucleus moves at 0.70c in the laboratory. It emits an electron forward, and the laboratory measures the electron's speed as 0.95c. Determine the speed of this electron relative to the nucleus.
(2)
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Notes
Part (a)
Galilean prediction: 0.70c + 0.90c = 1.60c
✓ 1
This is greater than c; no massive particle can reach c, and the result is inconsistent with the invariance of c
✓ 1
Part (b)
u = (0.90 + 0.70)c/(1 + 0.90 × 0.70) = 1.60c/1.63
✓ 1
u = 0.982c
✓ 1
Part (c)
u = (−0.90 + 0.70)c/(1 + (−0.90)(0.70)) = −0.20c/0.37
✓ 1
u = −0.541c, i.e. 0.541c opposite to the direction of the nucleus
58IBP2H-008-01-13Spacetime diagramsPaper 2Medium7 marks
A spacetime diagram is drawn for the Earth frame S. It has axes ct and x, both in kilometres and drawn to equal scale. A rocket passes the origin at t = 0, moving at 0.50c in the +x direction. Event E1 is at the origin. Event E2 is at x = 4.0 km, ct = 2.0 km.
(a)
Calculate the angle between the rocket's worldline and the ct axis.
(1)
(b)
Describe how the rocket's x′ axis is drawn on the diagram and explain its position.
(2)
(c)
Show that E2 is simultaneous with E1 in the rocket frame.
(2)
(d)
Calculate the distance between E1 and E2 in the rocket frame.
(2)
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Notes
Part (a)
θ = tan−1(v/c) = tan−1(0.50) = 26.6°
✓ 1
Part (b)
It is a line through the origin at 26.6° above the x axis, tilted towards the 45° light line
✓ 1
The x′ axis is the set of events with t′ = 0, which is the line ct = (v/c)x = 0.50x
✓ 1
Part (c)
ct′ = γ(ct − (v/c)x) = γ(2.0 − 0.50 × 4.0) km
✓ 1
This gives ct′ = 0, so E2 lies on the x′ axis and occurs at the same t′ as E1 (accept: E2 satisfies ct = 0.50x, so it lies on the x′ axis)
Common error: The x′ axis tilts upward from the x axis, towards the light line, not downward. Check that the ct′ and x′ axes are symmetric about the 45° line.
59IBP2H-008-01-16Galilean relativityPaper 2Medium7 marks
A maglev train moves along a straight track at 125 m s−1. Frame S is the ground and frame S′ is the train. The origins coincide at t = t′ = 0. A warning light flashes at x = 2500 m, t = 12.0 s in the ground frame. Take c = 3.00×108 m s−1.
(a)
Use the Galilean transformation to calculate x′ and t′ for the flash.
(2)
(b)
Show that γ − 1 is approximately 8.7×10−14 for the train.
(2)
(c)
Use the Lorentz transformation to determine t′ − t for the flash.
(2)
(d)
Outline why engineers can use Galilean transformations when designing the train's control systems.
t′ − t = 1.04×10−12 − 3.47×10−12 ≈ −2.4×10−12 s (ECF from (b))
✓ 1
Part (d)
The relativistic corrections are of order picoseconds, far smaller than any timing precision relevant to the train, so the Galilean results are effectively exact
✓ 1
Common error: At low speeds, the vx/c² term can be larger than the time-dilation term (γ − 1)t. Include both terms rather than assuming time dilation is the only correction.
60IBP3H-A5-20r43-13Relativistic velocity additionPaper 2Medium6 marks
A spacecraft moves directly away from Earth at 0.70c, as measured from Earth. It launches two probes along its line of motion, each at 0.50c relative to the spacecraft: probe P is launched forwards (away from Earth) and probe Q is launched backwards (towards Earth). Use u = (u′ + v)/(1 + u′v/c2).
(a)
Calculate the speed of probe P relative to Earth.
(3)
(b)
Probe P emits a pulse of light forwards. State the speed of this pulse relative to Earth.
(1)
(c)
Calculate the velocity of probe Q relative to Earth, stating its direction.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 use of u = (u′ + v)/(1 + u′v/c2) with u′ = 0.50c and v = 0.70c
M1 u′ = −0.50c for the backward launch, so u = (−0.50 + 0.70)c/(1 − 0.50 × 0.70) = 0.20c/0.65
✓ 1
A1 0.308c directed away from Earth (probe Q still moves away from Earth)
✓ 1
Exam tip: For an object launched backwards, put a negative sign on u′ before using the formula, then interpret the sign of the result as the direction of motion.
61IBP3H-A5-20r43-14Simultaneity and the Lorentz transformationPaper 2Medium8 marks
Two flares occur in a distant gas cloud. In frame S, flare X occurs at x = 0, t = 0 and flare Y occurs at x = 5.0 × 108 m, t = 1.0 s. Use c = 3.00 × 108 m s−1, Δt′ = γ(Δt − vΔx/c2) and Δx′ = γ(Δx − vΔt) for a frame S′ moving at speed v along +x.
(a)
Show that flare X cannot have caused flare Y.
(2)
(b)
Determine the speed of the frame S′ in which the two flares are simultaneous.
(3)
(c)
Calculate the distance between the flares in S′.
(3)
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Mark
Notes
Part (a)
M1 light travels cΔt = 3.00 × 108 m in 1.0 s
✓ 1
A1 this is less than the separation 5.0 × 108 m, so no signal (which cannot exceed c) could connect X to Y
Exam tip: If the events are separated by more distance than light can travel in the time between them, a frame exists where they are simultaneous; setting the transformed time difference to zero gives its speed.
62IBP3H-A5-20r43-15Spacetime diagramsPaper 2Medium7 marks
On a spacetime diagram for frame S, with ct vertical and x horizontal in the same units (metres), event P is at the origin and event Q has coordinates x = 3.0 m and ct = 5.0 m. Use c = 3.00 × 108 m s−1.
(a)
Describe, in words, how the ct′ axis of a frame S′ moving at +0.60c relative to S appears on this diagram.
(2)
(b)
Show that an object could travel from P to Q, and state its speed.
(2)
(c)
Calculate the time between P and Q measured by a clock carried by this object. Give your answer to three significant figures.
(3)
Show mark scheme
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Notes
Part (a)
It is a straight line through the origin (the worldline of the origin of S′)
✓ 1
It is tilted from the ct axis towards the light line, with gradient c/v = 1/0.60 ≈ 1.7
✓ 1
Part (b)
M1 speed = Δx/Δt = c × (3.0/5.0) = 0.60c
✓ 1
A1 this is less than c, so P could cause Q
✓ 1
Part (c)
M1 use of the invariant interval, (cΔτ)2 = (cΔt)2 − Δx2, or Δτ = Δt/γ
✓ 1
A1 cΔτ = √(5.02 − 3.02) = 4.0 m
✓ 1
A1 Δτ = 4.0/(3.00 × 108) = 1.33 × 10−8 s
✓ 1
Common error: The time on the moving object's own clock is shorter than the time in frame S; use the invariant interval or divide by γ, never multiply by γ.
63A-2-03Time dilation and length contractionPaper 2Hard12 marks
A beam of pions travels at 0.980c along a straight 40.0 m beam line in a laboratory. Detector 1 at the start of the beam line and detector 2 at the end count the pions passing them. The half-life of pions at rest is 18.0 ns.
Laboratory frame; not to scale
(a)
(i)
Show that the Lorentz factor γ for the pions is about 5.0.
(1)
(ii)
Determine, for the laboratory frame, the fraction of the pions passing detector 1 that also reach detector 2.
(3)
(iii)
Calculate the fraction that would reach detector 2 if there were no time dilation.
(1)
(b)
(i)
In the rest frame of the pions, determine the length of the beam line and the time for which a pion is between the detectors.
(2)
(ii)
Show that an observer in the pion frame predicts the same fraction as in (a)(ii).
(1)
(c)
(i)
Show that the fraction f reaching detector 2 can be written f = 2−L/(γvT½), where L is the laboratory length of the beam line. Hence state what happens to f as v approaches c.
(2)
(ii)
Determine the speed that the pions would need for exactly half of those passing detector 1 to reach detector 2.
(2)
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Notes
Part (a)(i)
γ = 1/√(1 − 0.980²) = 1/√0.0396 = 5.03
✓ 1
Full substitution or 5.03 needed.
Part (a)(ii)
Time in the laboratory t = 40.0/(0.980 × 3.00 × 108) = 136.1 ns
✓ 1
Dilated half-life = γ × 18.0 = 90.5 ns
✓ 1
Allow ECF from (a)(i); with γ = 5.0, 90.0 ns.
Fraction = (½)136.1/90.5 = 0.35
✓ 1
Accept 0.35. Using e−λt with λ = ln 2/T½ is equivalent.
Part (a)(iii)
(½)136.1/18.0 = 5.3 × 10−3
✓ 1
Allow ECF from (a)(ii) for the time of flight. Accept 5 × 10−3.
Part (b)(i)
Length = 40.0/γ = 7.96 m
✓ 1
Allow ECF from (a)(i). Accept 7.9–8.0 m.
Time = 7.96/(0.980 × 3.00 × 108) = 27.1 ns
✓ 1
Or t/γ. Accept 27 ns.
Part (b)(ii)
Proper half-life applies: (½)27.1/18.0 = (½)1.50 = 0.35, the same as (a)(ii)
✓ 1
Allow ECF from (b)(i).
Part (c)(i)
Laboratory time L/v and dilated half-life γT½, so the number of half-lives is L/(γvT½)
✓ 1
Symbolic working required.
As v → c, γv → ∞, so the exponent → 0 and f → 1 (almost no pions decay)
✓ 1
Do not accept "f increases" without the limit.
Part (c)(ii)
For f = ½ the exponent in (c)(i) must equal 1: γv = L/T½ = 40.0/(18.0 × 10−9) = 2.22 × 109 m s−1, i.e. γv/c = 7.41
✓ 1
Allow ECF from (c)(i). Note that γv may exceed c.
(v/c)²/(1 − v²/c²) = 54.9, so v = 0.991c (2.97 × 108 m s−1)
✓ 1
Accept 0.990–0.992c. A trial-and-improvement solution is acceptable. A value ≥ c scores 0 for this mark.
Answers: (a)(ii) 0.35 · (a)(iii) 5.3 × 10−3 · (b)(i) 7.96 m, 27.1 ns · (c)(ii) 0.991c(the remaining parts are explanations — see the table above)
64A-2-06Simultaneity and the Lorentz transformationPaper 2Hard11 marks
A spacecraft passes the Earth at 0.60c, travelling directly towards the Moon. Take the Earth and the Moon to be at rest in frame S, with the Earth at x = 0 and the Moon on the positive x-axis. Frame S′ is the spacecraft frame. As the spacecraft passes the Earth, at t = t′ = 0, a laser pulse is fired from the Earth towards the Moon (event E1). The pulse arrives at the Moon (event E2).
The Earth–Moon distance in S is 3.84 × 108 m.
Frame S; not to scale
(a)
(i)
Show that the Lorentz factor γ for S′ relative to S is 1.25.
(1)
(ii)
Calculate the time between E1 and E2 in frame S.
(1)
(b)
(i)
Use the Lorentz transformation to determine the coordinates t′ and x′ of E2 in S′.
(3)
(ii)
Explain, using length contraction, why the pulse reaches the Moon in less time in S′ than in S.
(2)
(c)
(i)
Show that the space-time interval between E1 and E2 is zero both in S and in S′.
(2)
(ii)
Explain why there is no inertial frame in which E1 and E2 are simultaneous.
(2)
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Notes
Part (a)(i)
γ = 1/√(1 − 0.60²) = 1/√0.64 = 1/0.80 = 1.25
✓ 1
Full substitution needed.
Part (a)(ii)
Δt = 3.84 × 108/3.00 × 108 = 1.28 s
✓ 1
Part (b)(i)
For E2 in S: x = 3.84 × 108 m, t = 1.28 s; vx/c² = 0.60 × 1.28 = 0.768 s and vt = 0.60 × 3.00 × 108 × 1.28 = 2.30 × 108 m
In S′ the Earth–Moon distance is contracted to 3.84 × 108/1.25 = 3.07 × 108 m
✓ 1
Allow ECF from (a)(i).
and the Moon moves towards the pulse at 0.60c while the pulse moves at c, so they meet after 3.07 × 108/(1.6 × 3.00 × 108) = 0.640 s, less than 1.28 s
✓ 1
Closing speed 1.6c is a separation rate, not a speed of any object: no conflict with the postulates.
Part (c)(i)
In S: cΔt = 3.00 × 108 × 1.28 = 3.84 × 108 m = Δx, so (Δs)² = (cΔt)² − (Δx)² = 0
✓ 1
Allow ECF from (a)(ii).
In S′: cΔt′ = 3.00 × 108 × 0.640 = 1.92 × 108 m = Δx′, so (Δs′)² = 0: the interval is the same (zero) in both frames
✓ 1
Allow ECF from (b)(i).
Part (c)(ii)
Simultaneity in a frame moving at v needs Δt′ = γ(Δt − vΔx/c²) = 0, i.e. v = c²Δt/Δx = c, because Δx = cΔt
✓ 1
Or: if Δt′ = 0 then (Δs)² = −(Δx′)² = 0 forces Δx′ = 0 as well, so E1 and E2 would be the same event.
No inertial observer can move at c relative to S, so no such frame exists: E1 and E2 are connected by a light signal (light-like separation) and occur in the same order in every frame
✓ 1
The link to the speed of light (second postulate) is needed for this mark.
Answers: (a)(ii) 1.28 s · (b)(i) t′ = 0.640 s, x′ = 1.92 × 108 m (the remaining parts are explanations — see the table above)
A nearby star has a parallax angle of 0.817 arc-seconds measured from the Earth. A spacecraft travels from the Earth to the star at a constant speed of 0.800c relative to the Earth. Treat the Earth and the star as at rest relative to each other, and consider only the outward journey.
(a)
(i)
Show that the distance from the Earth to the star is about 4.0 light-years.
(2)
(ii)
Outline why this method cannot be used to find the distance to a star 4000 light-years away.
(1)
(b)
(i)
Calculate the time taken for the journey as measured in the Earth frame.
(1)
(ii)
Determine, for the crew, the distance between the Earth and the star and the duration of the journey.
(3)
(iii)
Each observer considers the other's clocks to run slow. Explain why both nevertheless agree that the crew's clock records less time for the journey.
(2)
(c)
(i)
On reaching the star the spacecraft immediately turns round and returns to the Earth at the same speed. A member of the crew left her twin on the Earth. Determine the difference between their ages when they meet again.
(2)
(ii)
The crew member argues that, in her frame, it was the Earth that moved away and came back, so her twin should be the younger. Explain why this argument is not valid.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
d = 1/p = 1/0.817 = 1.224 pc
✓ 1
d = 1.224 × 3.09 × 1016/9.46 × 1015 = 4.00 ly
✓ 1
Must see 4.00 or full substitution.
Part (a)(ii)
The parallax angle would be about 1000 times smaller (≈ 8 × 10−4 arc-seconds), too small to measure with the available precision
✓ 1
A numerical estimate is not required.
Part (b)(i)
t = 4.00 ly/0.800c = 5.00 years
✓ 1
Allow ECF from (a)(i).
Part (b)(ii)
γ = 1/√(1 − 0.800²) = 5/3 (1.67)
✓ 1
Distance = 4.00/γ = 2.40 ly
✓ 1
Allow ECF from (a)(i).
Duration = 2.40/0.800 = 3.00 years (= 5.00/γ)
✓ 1
Allow ECF from (b)(i).
Part (b)(iii)
Departure and arrival occur at the same place in the spacecraft frame, so the crew's clock measures the proper time, the shortest time between the two events
✓ 1
The Earth-frame time uses two clocks (at the Earth and at the star) that are synchronised in the Earth frame but not in the spacecraft frame, so the situation is not symmetric (relativity of simultaneity)
The crew member is 4.0 years younger than her twin
✓ 1
Allow ECF. The direction (who is younger) is needed.
Part (c)(ii)
The twin on the Earth stays in one inertial frame, but the crew member does not: she changes from the outward to the return inertial frame (she accelerates) at the turnaround, so the two situations are not symmetric
✓ 1
Do not accept "because she is the one moving".
Time dilation may be applied to the crew’s clock in the Earth frame for the whole journey; in the crew’s description the Earth-clock reading that is simultaneous with her own jumps forward when she changes frame (relativity of simultaneity), so she too concludes that her twin is older
✓ 1
Award for a correct link to the relativity of simultaneity or to the change of inertial frame making the Earth-frame calculation the valid one.
Answers: (b)(i) 5.00 years · (b)(ii) 2.40 ly, 3.00 years · (c)(i) 4.0 years; the crew member is younger (the remaining parts are explanations — see the table above)
66A-2-14Lorentz transformation and the invariant intervalPaper 2Hard14 marks
In the inertial frame S of a laboratory, event 1 occurs at x = 0, t = 0 and event 2 occurs at x = 1500 m, t = 8.00 μs. Frame S′ moves at 0.500c in the positive x-direction relative to S, and the origins coincide at t = t′ = 0.
(a)
(i)
Calculate the space-time interval (Δs)² between the events, and state whether their separation is time-like or space-like.
(2)
(ii)
Deduce whether event 1 could have caused event 2.
(1)
(b)
(i)
Show that the Lorentz factor of S′ relative to S is 1.15.
(1)
(ii)
Determine the coordinates x′ and t′ of event 2 in S′.
(3)
(iii)
Show that the space-time interval is the same in S′ as in S.
(2)
(c)
(i)
Determine the proper time between the two events.
(2)
(ii)
An inertial observer is present at both events. Determine the speed of this observer relative to S, and use the time dilation formula to confirm your answer to (c)(i).
(2)
(d)
(i)
Calculate x′ for event 2 using the Galilean transformation and comment on your answer.
Clock P is at rest at the origin of frame S, the rest frame of a space station. An identical clock Q is carried by a probe that moves at 0.60c in the positive x-direction relative to S. The probe passes the station at event O, when both clocks read zero. The probe frame is S′.
The space-time diagram, drawn in S, shows the world line of the probe (the ct′ axis) and the x′ axis. Event A is clock P reading 5.0 μs. Event B is clock Q reading 4.0 μs. Take c = 3.00 × 108 m s−1.
Space-time diagram drawn to scale in the station frame S. The gold lines are the axes of the probe frame S′; the ct′ axis is the world line of the probe.
(a)
(i)
Show that the Lorentz factor of the probe relative to the station is 1.25.
(1)
(ii)
Show that, in frame S, event B occurs at x = 900 m and ct = 1500 m.
(2)
(iii)
State and explain which clock the station observer concludes is running slow.
(1)
(b)
(i)
Event C is on the world line of clock P and is simultaneous with event B in the probe frame S′. Use the Lorentz transformation to determine the reading of clock P at event C.
(2)
(ii)
Show that, on the diagram, events C and B lie on a line parallel to the x′ axis.
(2)
(iii)
Explain how the station observer and the probe observer can each conclude that the other’s clock runs slow.
(2)
(c)
(i)
Determine the coordinates of event B in S according to the Galilean transformation, and state what Galilean relativity predicts about the readings of the two clocks.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
γ = 1/√(1 − 0.60²) = 1/√0.64 = 1/0.80 = 1.25
✓ 1
Full substitution needed.
Part (a)(ii)
Q is present at both O and B, so 4.0 μs is a proper time; in S, Δt = γΔt0 = 1.25 × 4.0 = 5.0 μs, so ct = 3.00 × 108 × 5.0 × 10−6 = 1500 m
✓ 1
Allow ECF from (a)(i).
x = vt = 0.60 × 3.00 × 108 × 5.0 × 10−6 = 900 m
✓ 1
Both coordinates must be shown.
Part (a)(iii)
Clock Q: A and B have the same ct (1500 m), so they are simultaneous in S; at that instant P reads 5.0 μs but Q reads only 4.0 μs
✓ 1
The reference to A and B being simultaneous in S is needed.
Part (b)(i)
In S′, t′ of B is 4.0 μs, because Q is at rest at the origin of S′ (or t′ = γ(t − vx/c²) = 1.25 × (5.0 − 1.8) μs = 4.0 μs)
✓ 1
Allow ECF from (a)(i) and (a)(ii).
C has x = 0, so t′ = γt: t = 4.0/1.25 = 3.2 μs, so P reads 3.2 μs
✓ 1
5.0 μs (simultaneity in S used) scores [0].
Part (b)(ii)
C is at x = 0, ct = 3.00 × 108 × 3.2 × 10−6 = 960 m
✓ 1
Allow ECF from (b)(i).
Gradient of CB = Δ(ct)/Δx = (1500 − 960)/900 = 0.60, the same as the gradient of the x′ axis (ct = 0.60x), so CB is a line of constant t′
✓ 1
Allow ECF from the candidate’s coordinates.
Part (b)(iii)
Each observer compares the two clocks at events that are simultaneous in their own frame: S uses A and B (a horizontal line on the diagram), S′ uses C and B (a line parallel to the x′ axis)
✓ 1
A and C are different events because simultaneity is relative: in S′, P reads only 3.2 μs when Q reads 4.0 μs, so the two conclusions do not contradict each other
✓ 1
Do not accept "it is only an illusion" or "because light takes time to arrive".
Part (c)(i)
t = t′ = 4.0 μs, so ct = 1200 m, and x = vt = 0.60 × 3.00 × 108 × 4.0 × 10−6 = 720 m
✓ 1
Both coordinates needed.
Time is absolute: both clocks read 4.0 μs at B and at the event on P’s world line simultaneous with it, in either frame, so neither clock runs slow (contradicted by experiment at speeds comparable with c)
✓ 1
Answers: (a)(iii) clock Q · (b)(i) 3.2 μs · (c)(i) x = 720 m, ct = 1200 m (the remaining parts are explanations — see the table above)
A courier ship passes a space station at event O and continues at a constant speed of 0.50c in the positive x-direction. Frame S is the station frame and frame S′ the ship frame; O is the origin of both. The space-time diagram, drawn in S, shows the ct′ and x′ axes of the ship frame.
Event E1 is the flash of a beacon at rest in S at x = 900 m, ct = 450 m. Event E2 is an alarm on the ship at x = 400 m, ct = 800 m. Take c = 3.00 × 108 m s−1.
Space-time diagram drawn to scale in the station frame S. The gold lines are the axes of the ship frame S′.
(a)
The ship frame.
(i)
Show that the angle θ between the ct′ axis and the ct axis is about 27°.
(1)
(b)
Event E1.
(i)
Explain, with reference to the diagram, why E1 and O are simultaneous for the ship but not for the station.
(2)
(ii)
Calculate the x′ coordinate of E1.
(2)
(c)
Event E2.
(i)
State the x′ coordinate of E2, giving a reason.
(1)
(ii)
Use the Lorentz transformation to determine the time of E2 recorded by the ship’s clock.
(2)
(d)
E1 and E2.
(i)
Show that E1 cannot have caused E2.
(2)
(ii)
Determine the velocity, relative to the station, of an inertial frame in which E1 and E2 are simultaneous.
(2)
(e)
Scales on the axes.
(i)
The event on the ship’s world line at which the ship’s clock reads ct′ = 800 m is P. Determine the coordinates of P in S and explain why the scale on the ct′ axis differs from that on the ct axis.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
tan θ = v/c = 0.50, so θ = 26.6° ≈ 27°
✓ 1
Part (b)(i)
E1 lies on the x′ axis, which is the line t′ = 0: all events on it (including O) are simultaneous in S′
✓ 1
In S, simultaneous events lie on a horizontal line (ct = constant); E1 is at ct = 450 m, not 0, so it occurs after O in S
Accept 2.3 μs. Equivalent: t′ = γ(t − vx/c²) with t = 2.67 μs.
Part (d)(i)
Δx = 400 − 900 = −500 m and cΔt = 800 − 450 = 350 m
✓ 1
|Δx| > cΔt, i.e. (Δs)² = 350² − 500² < 0 (space-like): a signal would have to travel faster than light, so E1 cannot cause E2
✓ 1
Or: E2 lies outside the light cone of E1.
Part (d)(ii)
Δt′ = γ(Δt − vΔx/c²) = 0 gives v = c²Δt/Δx = c × 350/(−500)
✓ 1
v = −0.70c: 0.70c in the negative x-direction
✓ 1
Direction required.
Part (e)(i)
Along the world line x = 0.50ct and (ct)² − x² = 800²: ct = γ × 800 = 924 m, x = 462 m
✓ 1
Or by inverse Lorentz transformation with x′ = 0.
P is about 1030 m from O on the diagram although ct′ = 800 m: the scales differ because they are set by the invariant interval (hyperbolae of constant (ct)² − x²), not by distance on the page
✓ 1
Answers: (b)(ii) 7.8 × 102 m · (c)(ii) 2.31 μs · (d)(ii) 0.70c in the −x direction · (e)(i) x = 462 m, ct = 924 m (the remaining parts are explanations — see the table above)
69IBP2H-008-00-13Time dilation and length contractionPaper 2Hard8 marks
A beam of pions is produced for a medical research facility. In their rest frame, pions have a half-life of 18 ns. The pions travel at 0.990c along a beamline 25 m long, measured in the laboratory. Take c = 3.00 × 108 m s−1.
Laboratory frame; not to scale
(a)
Calculate the Lorentz factor of the pions.
(1)
(b)
Calculate the time for the pions to travel along the beamline in the laboratory frame.
(1)
(c)
Determine, using the laboratory frame, the fraction of pions that reach the end of the beamline without decaying.
(3)
(d)
Show, using the pion rest frame, that the same fraction is obtained.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
γ = 1/√(1 − 0.9902) = 7.09 .
✓ 1
Part (b)
t = 25/(0.990 × 3.00 × 108) = 8.42 × 10−8 s (84.2 ns) .
✓ 1
Part (c)
Dilated half-life = 7.09 × 18 = 128 ns, ECF from (a) . Number of half-lives = 84.2/128 = 0.66, ECF from (b) . Fraction = 2−0.66 = 0.63 .
✓ 3
Part (d)
Contracted beamline length = 25/7.09 = 3.53 m . Time in pion frame = 3.53/(0.990 × 3.00 × 108) = 11.9 ns . 11.9/18 = 0.66 half-lives, giving fraction 2−0.66 = 0.63, the same as (c). ECF from (a) .
✓ 3
Common error: Do not dilate the half-life and contract the length in the same calculation. Choose one frame and apply only the effect relevant to that frame.
70IBP2H-008-00-14Spacetime diagramsPaper 2Hard8 marks
Two beacons, P at the rear and Q at the front, are fixed 600 m apart at opposite ends of a space station. They flash simultaneously in the station frame. A ship moves past the station at 0.80c, travelling in the direction from P to Q. Take c = 3.0 × 108 m s−1.
Station frame; not to scale
(a)
Explain, with reference to the second postulate, why the flashes need not be simultaneous for the ship's crew.
(2)
(b)
Calculate γ for the ship relative to the station.
(1)
(c)
Calculate the time interval between the flashes in the ship frame.
(2)
(d)
State which beacon flashes first in the ship frame.
(1)
(e)
Calculate the distance between the two flash events in the ship frame, and explain why this is not the contracted length of the station.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Light from each flash travels at c in every frame, so simultaneity is judged by correcting for light travel time at speed c in that frame . In the ship frame the station is moving, so the corrected emission times are different and the two flashes are not simultaneous .
Δt′ is negative, so Q (the front beacon, ahead in the direction of the ship's motion) flashes first .
✓ 1
Part (e)
Δx′ = γ(Δx − vΔt) = 1.67 × 600 = 1.0 × 103 m, ECF from (b) . The contracted length (360 m) requires the positions of the two ends to be measured simultaneously in the ship frame; these flashes are not simultaneous in that frame .
✓ 2
Common error: The separation of two events is not automatically a length. A length must be measured from simultaneous position readings in the observer's own frame.
71IBP2H-008-01-12Simultaneity and the Lorentz transformationPaper 2Hard10 marks
A space station traffic system defines frame S. Beacon 1 at x = 0 flashes at t = 0 (event E1). Beacon 2 at x = 900 m flashes at t = 2.0 μs (event E2). A spacecraft, frame S′, passes at 0.60c in the +x direction. The origins of S and S′ coincide at E1. Take c = 3.00×108 m s−1.
(a)
Calculate the time between E1 and E2 in S′.
(2)
(b)
Calculate the distance between E1 and E2 in S′.
(2)
(c)
Show that the spacetime interval between E1 and E2 is the same in S and S′.
(2)
(d)
Explain whether E1 could be the cause of E2.
(2)
(e)
Determine the speed of the frame in which E1 and E2 are simultaneous.
In S′: (cΔt′)2 − Δx′2 = 752 − 6752 = −4.5×105 m2, the same value (ECF from (a) and (b))
✓ 1
Part (d)
In 2.0 μs light travels only 600 m, which is less than 900 m; equivalently, the interval is negative, so the events are spacelike
✓ 1
No signal can travel from E1 to E2, so E1 cannot cause E2
✓ 1
Part (e)
Δt′ = 0 requires Δt = vΔx/c2, so v = c2Δt/Δx = c(600/900)
✓ 1
v = 0.667c in the +x direction
✓ 1
Exam tip: Convert time to distance by working in cΔt (here 600 m and 75 m). This keeps the interval calculation in metres and makes the spacelike or timelike decision obvious.
72IBP2H-008-01-14Postulates of special relativityPaper 2Hard7 marks
A ground station tracks two deep-space probes launched in opposite directions. Probe P moves east at 0.75c and probe Q moves west at 0.60c, both relative to Earth. Q carries a straight antenna rod of proper length 12.0 m, aligned with its direction of motion.
Earth frame; not to scale
(a)
Calculate the rate at which the separation of the probes increases, as measured on Earth. Explain why this result does not contradict special relativity.
(2)
(b)
Calculate the speed of Q relative to P.
(2)
(c)
P sends a radio signal towards Q. State the speed of the signal measured by Q.
(1)
(d)
Determine the length of Q's antenna rod as measured in P's frame.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Rate = 0.75c + 0.60c = 1.35c
✓ 1
No object or signal moves faster than c in any frame; this is the rate of change of a distance between two objects, measured in one frame
✓ 1
Part (b)
u = (0.75 + 0.60)c/(1 + 0.75 × 0.60) = 1.35c/1.45
✓ 1
u = 0.931c
✓ 1
Part (c)
c, because the speed of light is the same in all inertial frames
✓ 1
Part (d)
γ = 1/√(1 − 0.9312) = 2.74 (ECF from (b))
✓ 1
L = 12.0/2.74 = 4.38 m
✓ 1
Exam tip: Length contraction needs the speed of the rod relative to the observer. Find the relative velocity of Q and P first; do not use either probe's speed relative to Earth.
73IBP2H-008-01-15Muon decay experimentsPaper 2Hard8 marks
In a research accelerator, a muon beam is found to travel a mean distance of 1.98 km in the laboratory before the muons decay. The mean lifetime of a muon at rest is 2.20 μs. Take c = 3.00×108 m s−1.
(a)
Show that γβ = d/(cτ0), where d is the mean distance, τ0 is the proper mean lifetime and β = v/c.
(2)
(b)
Determine the speed of the muons.
(2)
(c)
Calculate the distance the laboratory moves past a typical muon during the muon's mean lifetime, in the muon's frame. Comment on your answer.
(2)
(d)
The beam passes through a shielding block of proper thickness 5.0 m. Calculate the time a muon spends inside the block, measured in the muon's frame.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The mean lifetime in the laboratory is dilated to γτ0
✓ 1
d = v × γτ0 = γβcτ0, so γβ = d/(cτ0)
✓ 1
Part (b)
γβ = 1980/(3.00×108 × 2.20×10−6) = 3.00
✓ 1
β2/(1 − β2) = 9, so β2 = 0.90 and v = 0.949c
✓ 1
Part (c)
Distance = 0.949 × 3.00×108 × 2.20×10−6 = 626 m (ECF from (b))
✓ 1
This equals 1980/γ = 1980/3.16 m, the length-contracted distance, so both frames agree that the muon decays at the same point in the laboratory
✓ 1
Part (d)
Contracted thickness = 5.0/3.16 = 1.58 m
✓ 1
t = 1.58/(0.949 × 3.00×108) = 5.56×10−9 s, i.e. 5.56 ns (ECF)
✓ 1
Exam tip: When the product γβ is given or is easy to find, solve for β by squaring: β² = (γβ)²/(1 + (γβ)²). This avoids trial and error.
74IBP3H-A5-20r43-16Simultaneity and the Lorentz transformationPaper 2Hard10 marks
Frame S′ moves at constant speed v along +x relative to frame S. The Lorentz transformations from S′ to S are x = γ(x′ + vt′) and t = γ(t′ + vx′/c2). An object moves along the x-axis at constant velocity u′ in S′ and u in S.
(a)
Derive the relation u = (u′ + v)/(1 + u′v/c2).
(4)
(b)
Show that if u′ = c then u = c for any v.
(2)
(c)
A rocket moves at 0.80c relative to Earth and fires an object forwards at 0.90c relative to the rocket. Calculate the speed of the object relative to Earth.
(2)
(d)
Determine the Lorentz factor of the object relative to Earth. Give your answer to three significant figures and use the unrounded value of u from part (c).
M1 divide numerator and denominator by Δt′ and use u′ = Δx′/Δt′
✓ 1
A1 u = (u′ + v)/(1 + u′v/c2)
✓ 1
Part (b)
M1 substitute u′ = c: u = (c + v)/(1 + v/c)
✓ 1
A1 = c(c + v)/(c + v) = c
✓ 1
Part (c)
M1 u = (0.90 + 0.80)c/(1 + 0.90 × 0.80)
✓ 1
A1 u = 1.70c/1.72 = 0.988c
✓ 1
Part (d)
M1 γ = 1/√(1 − u2/c2) with u/c = 1.70/1.72
✓ 1
A1 γ = 6.58
✓ 1
Common error: Rounding u to three significant figures before finding γ gives a noticeably wrong answer, because γ is very sensitive to speeds close to c; keep the unrounded value.
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