A.5 Galilean and special relativity: IB Physics HL exam-style questions
Galilean and special relativity is an HL-only topic. It starts with reference frames and Galilean transformations, then shows why a constant speed of light for all inertial observers forces the Lorentz transformations, relativistic velocity addition and the invariance of the spacetime interval.
Most marks come from applying time dilation and length contraction to the right observer, so identifying proper time and proper length is the key step. Spacetime diagrams, worldlines and the relativity of simultaneity are tested both graphically and with the Lorentz factor.
50 questions
238 marks
Paper 1A: 30
Paper 1B: 7
Paper 2: 13
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27 practice questions on A.5 Galilean and special relativity
1A-1A-08
Time dilation·A.5 Galilean and special relativity (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark4 steps to full marksDeduce
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.
✓ 1
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.
Syllabus understandingA.5 — proper time interval and proper length; time dilation as given by Δt = γΔt0; the relativity of simultaneity Command term: Deduce
2A-1A-09
Muon decay experiments·A.5 Galilean and special relativity (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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.
✓ 1
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.
Syllabus understandingA.5 — that muon decay experiments provide experimental evidence for time dilation and length contraction; proper time interval and proper length; time dilation as given by Δt = γΔt0; length contraction as given by L = L0/γCommand term: Deduce
3A-1A-10
Spacetime interval·A.5 Galilean and special relativity (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark4 steps to full marksDeduce
In an inertial frame, two events are separated by a distance of 5.0 light-years and a time of 3.0 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.
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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.
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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.
Syllabus understandingA.5 — that the spacetime interval Δs between two events is an invariant quantity as given by (Δs)² = (cΔt)² − (Δx)²; space-like and time-like separations and the relativity of simultaneity Command term: Deduce
4A-1A-14
Galilean relativity·A.5 Galilean and special relativity (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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?
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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.
✓ 1
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.
Syllabus understandingA.5 — that Galilean transformation equations as given by x′ = x − vt relate the coordinates of an event in two inertial reference frames; reference frames Command term: Determine
5A-1A-22
Lorentz transformations·A.5 Galilean and special relativity (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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′?
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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.
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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.
Syllabus understandingA.5 — the Lorentz transformation equations for the coordinates of an event in two inertial reference frames, as given by x′ = γ(x − vt) and t′ = γ(t − vx/c²) Command term: Determine
6A-1A-27
The Lorentz factor·A.5 Galilean and special relativity (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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.
✓ 1
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.
Galilean and Lorentz transformations·A.5 Galilean and special relativity (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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
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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.
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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.
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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.
Syllabus understandingA.5 — Galilean relativity and the Galilean transformation equations; the Lorentz transformation equations; the relativity of simultaneity; that the spacetime interval is invariant Command term: Deduce
8A-1A-52
Postulates of special relativity·A.5 Galilean and special relativity (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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.
✓ 1
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.
Syllabus understandingA.5 — the two postulates of special relativity; reference frames; Newton's laws of motion are the same in all inertial reference frames Command term: Deduce
9A-1A-53
Proper time and proper length·A.5 Galilean and special relativity (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
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.
✓ 1
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.
Syllabus understandingA.5 — proper time interval and proper length; time dilation as given by Δt = γΔt0; length contraction as given by L = L0/γCommand term: Identify
10A-1A-54
Length contraction·A.5 Galilean and special relativity (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A rectangular solar sail has proper dimensions 40 m × 30 m. It moves at 0.80c relative to an observer, in a direction parallel to its 40 m sides.
What area of the sail does the observer measure?
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Step 1γ = 1/√(1 − 0.80²) = 1/0.60 = 5/3.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Only the length parallel to the motion contracts: L = L0/γ = 40 × 0.60 = 24 m. The 30 m sides, perpendicular to the motion, are unchanged.
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Step 3Area = 24 m × 30 m = 720 m².
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis contracts both dimensions (40 × 0.6 × 30 × 0.6 = 432 m²). Lengths perpendicular to the relative velocity are not contracted.
BCorrect: only the 40 m side contracts, to 24 m, so the area is 24 × 30 = 720 m².
CThis is the proper area, ignoring length contraction.
DThis multiplies by γ (1200 × 5/3 = 2000 m²), as though a moving object were lengthened.
Syllabus understandingA.5 — length contraction as given by L = L0/γ; the Lorentz factor Command term: Determine
11A-1A-55
Relativistic velocity addition·A.5 Galilean and special relativity (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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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Notes
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.
AThis uses a plus sign in the denominator: 0.30c/1.40 = 0.21c.
BThis is the Galilean result u − v, which does not apply at these speeds.
CCorrect: 0.30c/0.60 = 0.50c.
DThis adds the velocities relativistically, (0.80 + 0.50)c/(1 + 0.40) = 0.93c, instead of transforming into the spacecraft frame.
Syllabus understandingA.5 — that Lorentz transformation equations lead to the relativistic velocity addition equation as given by u′ = (u − v)/(1 − uv/c²) Command term: Determine
12A-1A-56
Relativity of simultaneity·A.5 Galilean and special relativity (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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?
First flashTime between flashes
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Step 1γ = 1/√(1 − 0.80²) = 5/3. With Δt = 0 and Δx = xQ − xP = 1200 m:
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All 3 steps must be completed — there is no mark for a part-answer.
Syllabus understandingA.5 — the relativity of simultaneity; the Lorentz transformation t′ = γ(t − vx/c²) Command term: Deduce
13A-1A-57
Spacetime diagrams·A.5 Galilean and special relativity (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
The space-time diagram shows the world line of a particle. The world line makes an angle of 37° with the ct axis.
Which row gives the speed of the particle and its Lorentz factor γ?
Space-time diagram in the frame S; the scales on the x and ct axes are the same.
SpeedLorentz factor γ
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Step 1The angle θ between a world line and the ct axis is given by tan θ = v/c, so v = c tan 37° = 0.75c.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2γ = 1/√(1 − 0.75²) = 1/√0.4375 = 1.51 (1.52 if tan 37° = 0.754 is kept unrounded).
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis uses sin 37° = 0.60 instead of tan 37°, and then the Lorentz factor for 0.60c.
BThe speed is right, but γ has been taken as c/v = 1/0.75 instead of 1/√(1 − v²/c²).
CCorrect: v = c tan 37° = 0.75c and γ = 1/√(1 − 0.75²) = 1.51.
DThis uses cos 37° = 0.80 instead of tan 37°, and then the Lorentz factor for 0.80c.
Syllabus understandingA.5 — space–time diagrams; that the angle between the world line of a moving particle and the time axis on a space–time diagram is related to the particle’s speed as given by tan θ = v/c; the Lorentz factor γ = 1/√(1 − v²/c²) Command term: Determine
14A-1A-58
Spacetime diagrams·A.5 Galilean and special relativity (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
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′.
✓ 1
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.
Syllabus understandingA.5 — space–time diagrams; the relativity of simultaneity; time dilation, length contraction and simultaneity visualized using space–time diagrams Command term: Identify
15A-1A-59
Time dilation·A.5 Galilean and special relativity (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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.
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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).
Syllabus understandingA.5 — proper time interval and proper length; time dilation as given by Δt = γΔt0; length contraction as given by L = L0/γCommand term: Determine
16A-1A-72
Length contraction·A.5 Galilean and special relativity (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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.)
✓ 1
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.
Syllabus understandingA.5 — proper time interval and proper length; length contraction as given by L = L0/γ; the relativity of simultaneity Command term: Determine
17A-1A-73
Postulates of special relativity·A.5 Galilean and special relativity (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark4 steps to full marksDetermine
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?
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Step 1In the laboratory the rod is moving, so its length is L0/γ.
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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.
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Step 3Time = (L0/γ)/(c − v).
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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.
Syllabus understandingA.5 — the two postulates of special relativity; length contraction as given by L = L0/γ; the Lorentz transformation equations for the coordinates of an event in two inertial reference frames Command term: Determine
18A-1A-125
Inertial reference frames·A.5 Galilean and special relativity (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark4 steps to full marksIdentify
Three observers each use a frame of reference attached to themselves. Treat a frame fixed to the surface of the Earth as inertial.
I. An astronaut in a space probe coasting with its engines off, far from any planet or star II. A passenger in a car travelling round a roundabout at a constant speed III. A passenger in a lift that is moving upwards at a constant speed
Which observers are in an inertial frame of reference?
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Step 1An inertial frame of reference is one that is not accelerating; in it, a body with no resultant force on it moves with constant velocity (Newton's first law holds).
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All 4 steps must be completed — there is no mark for a part-answer.
Step 2I: with no engine thrust and no gravitational field the probe has constant velocity, so its frame is inertial.
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Step 3II: the car moves at constant speed but its direction changes continuously, so it has a centripetal acceleration; its frame is not inertial.
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Step 4III: the lift moves at constant velocity relative to the Earth, which is treated as inertial, so the lift frame is also inertial.
✓ 1
Answer B
Answer: B · 4 stages of work, one mark
Every option, and why
AThis accepts II, taking constant speed to mean no acceleration. A body moving in a circle accelerates towards the centre, so the car frame is not inertial.
BCorrect: the coasting probe and the lift moving at constant velocity are non-accelerating frames; the car on the roundabout accelerates towards the centre.
CThis rejects I, as if a frame had to be "at rest" to be inertial, and accepts II, confusing constant speed with constant velocity.
DThis includes II. A change in the direction of the velocity is an acceleration, so the passenger in the car is in a non-inertial frame.
Syllabus understandingA.5 — reference frames (guidance: an inertial reference frame is a non-accelerating frame of reference); that Newton's laws of motion are the same in all inertial reference frames; A.2 — circular motion requires a centripetal acceleration Command term: Identify
19A-1A-126
Galilean transformation·A.5 Galilean and special relativity (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
The graph shows how the position x of a cyclist varies with time t in the frame S of the road. Frame S′ moves at a constant velocity of 10 m s−1 in the positive x-direction relative to S, and the origins of S and S′ coincide at t = 0.
The graph of the position x′ of the cyclist in S′ against t is a straight line. Which row gives the gradient and the intercept on the position axis of this line?
Position of the cyclist against time in frame S (drawn to scale).
GradientIntercept on the position axis
Show mark scheme
Marking point
Mark
Notes
Step 1From the graph, in S: x = 40 + 6.0t (gradient (100 − 40)/10 = 6.0 m s−1).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 3Gradient = u′ = u − v = −4.0 m s−1; at t = 0 the origins coincide, so the intercept is unchanged at 40 m.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: x′ = 40 − 4.0t: the cyclist moves backwards at 4.0 m s−1 relative to S′, starting 40 m from its origin.
BThis writes the transformation the wrong way round, x′ = vt − x = 4.0t − 40, which reverses both the gradient and the intercept.
CThis assumes that the velocity of the cyclist is the same in every frame. Only the acceleration is the same in all inertial frames; the velocity changes by v.
DThis adds the velocity of the frame, x′ = x + vt, as if S′ moved in the negative x-direction.
Syllabus understandingA.5 — that Galilean transformation equations as given by x′ = x − vt and t′ = t relate the coordinates of an event in two inertial reference frames; that Galilean transformation equations lead to the velocity addition equation u′ = u − v; A.1 — motion graphs Command term: Determine
20A-1A-127
Galilean velocity addition·A.5 Galilean and special relativity (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
In frame S a ball moves with velocity u in the positive x-direction. Frame S′ moves relative to S with speed v in the negativex-direction. Both speeds are much less than c.
What is the velocity of the ball in S′, taking the positive x′-direction to be the same as the positive x-direction?
Show mark scheme
Marking point
Mark
Notes
Step 1Galilean velocity transformation: u′ = u − vframe, where vframe is the velocity of S′ relative to S.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2S′ moves in the negative direction, so vframe = −v.
—
Step 3u′ = u − (−v) = u + v. Check: an observer moving backwards sees the ball moving forwards faster.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis uses the speed v as the velocity of S′ and ignores its direction: the frame moves in the negative direction, so its velocity is −v.
BThis subtracts the wrong way round (velocity of the frame minus velocity of the ball) and also ignores the direction of S′.
CThis gives the right size but the wrong sign: the ball moves in the positive direction in S′, faster than in S.
DCorrect: u′ = u − (−v) = u + v.
Syllabus understandingA.5 — that Galilean transformation equations lead to the velocity addition equation as given by u′ = u − v; reference frames Command term: Deduce
21A-1A-128
Proper time and the space-time interval·A.5 Galilean and special relativity (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
In an inertial frame S, two events are separated by a distance Δx and a time Δt, where Δx < cΔt. An inertial observer travels in a straight line at constant velocity and is present at both events.
What time interval does this observer measure between the two events?
Show mark scheme
Marking point
Mark
Notes
Step 1The observer is present at both events, so in the observer's frame the events happen at the same place: Δx′ = 0, and the observer measures the proper time Δt0.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 3Δt0 = √(Δt² − Δx²/c²), which is less than Δt, as expected for a proper time.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis subtracts the light travel time Δx/c instead of using the invariant interval: the squares of the separations, not the separations, combine.
BThis adds the squares, treating the interval like a Euclidean distance. The space-time interval has a minus sign, and the proper time is the shortest time between the events.
CCorrect: from the invariant interval with Δx′ = 0, the proper time is √(Δt² − Δx²/c²).
DThis is the interval Δs itself, which is a distance (unit m); it must be divided by c to give a time.
Syllabus understandingA.5 — that the space–time interval Δs between two events is an invariant quantity as given by (Δs)² = (cΔt)² − Δx²; proper time interval and proper length Command term: Determine
22A-1A-129
Length contraction·A.5 Galilean and special relativity (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
Two spacecraft P and Q have the same proper length. They pass an observer, moving parallel to their lengths, at speeds of 0.60c (P) and 0.80c (Q).
What is (length of P measured by the observer)/(length of Q measured by the observer)?
Show mark scheme
Marking point
Mark
Notes
Step 1Measured length = L0/γ = L0√(1 − v²/c²).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 3Ratio = 0.80/0.60 = 4/3: the slower spacecraft is contracted less, so it appears longer.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis inverts the ratio, as if the faster spacecraft were contracted less (or uses the ratio of the speeds, 0.60/0.80).
BThis assumes that both spacecraft have their proper length for the observer; length contraction depends on the speed of the object relative to the observer.
Syllabus understandingA.5 — length contraction as given by L = L0/γ; the Lorentz factor γ = 1/√(1 − v²/c²); proper length Command term: Determine
23A-1A-130
Proper time on a space-time diagram·A.5 Galilean and special relativity (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
The space-time diagram, drawn in frame S, shows the world lines of three inertial observers P, Q and R. Event E1 is at the origin and event E2 is on the world line of Q.
Which observer measures the shortest time interval between E1 and E2?
Space-time diagram in frame S. The scales on the two axes are the same.Show mark scheme
Marking point
Mark
Notes
Step 1Q's world line passes through both E1 and E2, so Q is present at both events and measures the proper time between them with a single clock.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2P and R are at rest in S; for them the events occur at different places, so they measure the S time interval, 4 years (ct = 4 ly).
—
Step 3The proper time is the shortest time between two events: √(4² − 2²) = 3.5 years for Q, less than 4 years.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AP is present at E1 only and is at rest in S, so P measures the dilated S time of 4 years. Being "at rest" does not make a time interval proper.
BCorrect: Q is present at both events, so Q measures the proper time, which is the shortest (3.5 years).
CR is at rest in S and is present at neither event; R measures the same S time interval as P.
DTime intervals depend on the frame: P and R (frame S) measure 4 years, Q measures 3.5 years.
Syllabus understandingA.5 — proper time interval and proper length; time dilation as given by Δt = γΔt0; space–time diagrams Command term: Identify
24A-1A-131
Invariance of the space-time interval·A.5 Galilean and special relativity (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
In frame S, event Y occurs 5.0 years after event X and 3.0 ly from it. In another inertial frame S′ the two events are 4.5 ly apart.
All 3 steps must be completed — there is no mark for a part-answer.
Step 2In S: 5.0² − 3.0² = 16 ly².
—
Step 3In S′: (cΔt′)² = 16 + 4.5² = 36.25 ly², so Δt′ = 6.02 years. The events are further apart in S′, so the time between them is longer as well.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis uses a plus sign in the interval, 5.0² + 3.0² = 34, and then subtracts 4.5²: √(34 − 20.25) = 3.7 years.
BThis is the proper time, √(5.0² − 3.0²) = 4.0 years, measured in the frame where the events are at the same place; in S′ they are 4.5 ly apart.
CThis assumes that time intervals are the same in all frames (Galilean t′ = t).
DCorrect: (cΔt′)² = 16 + 20.25 = 36.25 ly², so Δt′ = 6.0 years.
Syllabus understandingA.5 — that the space–time interval Δs between two events is an invariant quantity as given by (Δs)² = (cΔt)² − Δx² Command term: Determine
25A-1A-132
Space-like and time-like separations·A.5 Galilean and special relativity (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
The space-time diagram shows event O and four other events E1, E2, E3 and E4. The dashed lines are the world lines of light rays through O. The scales on the two axes are the same.
Which event could occur at the same place as O in some inertial frame, and occurs after O in every inertial frame?
Space-time diagram; the dashed lines are at 45° to the axes.Show mark scheme
Marking point
Mark
Notes
Step 1Two events can occur at the same place in some frame only if a body moving slower than light could be present at both: |Δx| < cΔt (time-like separation, inside the light cone of O).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2E3: |Δx| = 1 ly, cΔt = 3 ly, so the separation is time-like and E3 is later than O. For time-like separations the time order is the same in every frame.
—
Step 3E1 is outside the light cone (space-like); E2 is on it (light-like: only light connects O and E2); E4 is inside the cone but earlier than O.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AE1 is space-like separated from O (3 ly apart, 1 year later): no observer can be present at both, and its time order relative to O depends on the frame.
BE2 lies on the light world line: only a light signal connects O and E2, and no inertial observer moving slower than light can be present at both.
CCorrect: E3 is inside the future light cone of O, so an observer moving at 1/3 c could be present at both, and E3 is after O in every frame.
DE4 is time-like separated from O, so the events can happen at the same place, but E4 is in the past light cone: it is before O in every frame.
Syllabus understandingA.5 — space–time diagrams; that the space–time interval is an invariant quantity, (Δs)² = (cΔt)² − Δx²; the relativity of simultaneity; world lines of light; proper time interval Command term: Deduce
26A-1A-133
Relativistic velocity addition·A.5 Galilean and special relativity (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
The relativistic velocity addition equation is u′ = (u − v)/(1 − uv/c²), where v is the velocity of frame S′ relative to frame S.
Which statements are correct?
I. If u = c, then u′ = c for every value of v less than c. II. If u and v are both much less than c, then u′ ≈ u − v. III. Two spacecraft approach each other, each moving at 0.60c relative to a space station. The station observer measures the distance between them decreasing at a rate less than c.
Show mark scheme
Marking point
Mark
Notes
Step 1I: u′ = (c − v)/(1 − v/c) = c(c − v)/(c − v) = c, in agreement with the second postulate. Correct.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2II: when uv/c² ≪ 1 the denominator is ≈ 1, giving the Galilean result u − v. Correct.
—
Step 3III: in the station frame each spacecraft moves at 0.60c, so the gap closes at 0.60c + 0.60c = 1.2c. This rate is not the velocity of any object, so it may exceed c. The speed of one spacecraft relative to the other, measured by its crew, is 1.2c/1.36 = 0.88c. III is wrong.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: I and II follow from the equation; III is wrong because a closing rate measured in a third frame is not a velocity of an object and may exceed c.
BThis rejects II, but the denominator 1 − uv/c² tends to 1 at low speeds, so the Galilean equation is recovered; it also accepts III, confusing the closing rate with a relative velocity.
CThis rejects I, but substituting u = c gives u′ = c exactly; it also accepts III.
DThis accepts III: the relativistic addition equation applies to the velocity of one object measured in the rest frame of another, not to the rate at which the gap between two objects closes in a third frame (1.2c).
Syllabus understandingA.5 — that Lorentz transformation equations lead to the relativistic velocity addition equation as given by u′ = (u − v)/(1 − uv/c²); the two postulates of special relativity; that Galilean transformation equations lead to u′ = u − vCommand term: Deduce
27A-1A-134
Lorentz transformation of an event·A.5 Galilean and special relativity (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
In frame S an event occurs at x = 900 m and t = 2.0 μs. Frame S′ moves at 0.80c in the positive x-direction relative to S, and the origins of S and S′ coincide at t = t′ = 0.
Which row gives the coordinates of the event in S′?
All 3 steps must be completed — there is no mark for a part-answer.
Step 2x′ = γ(x − vt) = (5/3)(900 − 480) = 700 m.
—
Step 3t′ = γ(t − vx/c²) = (5/3)(2.0 − 2.4) μs = −0.67 μs: in S′ the event happens before the origins pass each other.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: x′ = 700 m and t′ = (5/3)(2.0 − 2.4) μs = −0.67 μs.
BThe position is right but the sign of the time is wrong: 2.0 μs − 2.4 μs is negative. A negative t′ is allowed because the event is space-like separated from the origin (ct = 600 m < x = 900 m).
CThis uses the Galilean transformation: x′ = 900 − 480 = 420 m and t′ = t.
DThis transforms the time with the time-dilation formula, t′ = γt = 3.3 μs, ignoring the vx/c² term; 2.0 μs is not a proper time because the event is not at the origin of S.
Syllabus understandingA.5 — the Lorentz transformation equations for the coordinates of an event in two inertial reference frames, as given by x′ = γ(x − vt) and t′ = γ(t − vx/c²); the relativity of simultaneity Command term: Determine
28A-1A-135
Relativistic velocity addition on a space-time diagram·A.5 Galilean and special relativity (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
The space-time diagram, drawn in the frame S of a space station, shows the world lines of two spacecraft P and Q that move towards each other along the x-axis. The scales on the two axes are the same.
What is the speed of Q measured by an observer on P?
Space-time diagram in the station frame S (drawn to scale).Show mark scheme
Marking point
Mark
Notes
Step 1From the diagram the speed is Δx/Δ(ct) × c (not the gradient Δ(ct)/Δx): P moves 5 ly in ct = 10 ly, so vP = +0.50c; Q moves −2.5 ly in 10 ly, so uQ = −0.25c.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Relativistic velocity addition with S′ the frame of P: u′ = (u − v)/(1 − uv/c²) = (−0.25 − 0.50)c/(1 − (−0.25)(0.50)) = −0.75c/1.125.
—
Step 3u′ = −0.67c: Q approaches P at 0.67c.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis subtracts the speeds, 0.50c − 0.25c, as if the two spacecraft moved in the same direction; Q moves in the negative x-direction.
BCorrect: u′ = −0.75c/1.125 = −0.67c.
CThis is the Galilean result u − v = −0.75c, which ignores the denominator of the relativistic equation.
DThis uses the wrong sign in the denominator: 0.75c/(1 − 0.125) = 0.86c. With u and v in opposite directions, −uv/c² is positive and the denominator is greater than 1.
Syllabus understandingA.5 — space–time diagrams; that the angle between the world line of a moving particle and the time axis is related to the particle's speed as given by tan θ = v/c; the relativistic velocity addition equation u′ = (u − v)/(1 − uv/c²) Command term: Determine
29A-1A-136
Relativity of simultaneity and time order·A.5 Galilean and special relativity (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
In frame S, event B occurs 2.0 μs after event A at a point 900 m from A in the positive x-direction.
Which inertial frames, moving in the positive x-direction relative to S, measure event B to occur before event A?
Show mark scheme
Marking point
Mark
Notes
Step 1ct = 3.00 × 108 × 2.0 × 10−6 = 600 m < 900 m, so the separation is space-like: no signal can link A and B, and their order can depend on the frame.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Δt′ = γ(Δt − vΔx/c²) is negative when v > c²Δt/Δx = c × 600/900.
—
Step 3So B precedes A in every frame moving in the positive direction at more than 0.67c (B and A are simultaneous at exactly 0.67c).
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AOnly time-like (or light-like) separated events have a fixed order. A and B are space-like separated (600 m < 900 m), so their order is frame-dependent.
BThis reverses the inequality: for v < 0.67c the term vΔx/c² is smaller than Δt, so Δt′ stays positive and B is still after A.
CThis squares the ratio: (600/900)² = 0.44. The condition is linear in v: v/c > cΔt/Δx.
DCorrect: Δt′ < 0 when v > c²Δt/Δx = 0.67c.
Syllabus understandingA.5 — the relativity of simultaneity; the Lorentz transformation equations t′ = γ(t − vx/c²); that the space–time interval is invariant (space-like and time-like separations) Command term: Deduce
30A-1A-137
Space-time diagrams·A.5 Galilean and special relativity (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
The diagram shows the axes of an inertial frame S and those of a frame S′ that moves at speed v relative to S, drawn on a space-time diagram of S. The angle between the ct′ axis and the x′ axis is 50°.
What is v?
Axes of S (black) and S′ (gold) on a space-time diagram of S (not to scale).Show mark scheme
Marking point
Mark
Notes
Step 1The ct′ axis makes an angle θ with the ct axis, and the x′ axis makes the same angle θ with the x axis, where tan θ = v/c.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2The angle between the two S′ axes is therefore 90° − 2θ = 50°, so θ = 20°.
—
Step 3v = c tan 20° = 0.36c.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis uses sin 20° = 0.34 instead of tan 20°.
BCorrect: θ = (90° − 50°)/2 = 20° and v = c tan 20° = 0.36c.
CThis halves the 50° angle, taking θ = 25°: the 50° is the angle between the S′ axes, not twice the tilt.
DThis takes the tilt of one axis to be 90° − 50° = 40°, forgetting that both S′ axes tilt by θ towards each other.
Syllabus understandingA.5 — space–time diagrams; that the angle between the world line of a moving particle and the time axis on a space–time diagram is related to the particle's speed as given by tan θ = v/cCommand term: Determine
31A-1B-02
Galilean velocity addition·A.5 Galilean and special relativity (HL)
Paper 1BMedium7 marks
Data-based question7 steps to full marksDetermine
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
(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)
Show mark scheme
Marking point
Mark
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)
Syllabus understandingA.5 — that Galilean transformation equations lead to the velocity addition equation as given by u′ = u − v; reference frames; C.2 — the nature of sound waves; Tool 3 — uncertainties of a difference, zero (systematic) error; Inquiry 3 — evaluate hypotheses Command term: Determine
32A-1B-20
Time dilation and length contraction·A.5 Galilean and special relativity (HL)
Paper 1BMedium7 marks
Data-based question7 steps to full marksDetermine
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)
Show mark scheme
Marking point
Mark
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)
Syllabus understandingA.5 — proper time interval; time dilation as given by Δt = γΔt0; that muon decay experiments provide experimental evidence for time dilation and length contraction; E.3 — the radioactive decay law N = N0e−λt; Tool 3 — linearize graphs using logarithms, gradient with units, systematic errors Command term: Determine
33A-1B-21
Time dilation·A.5 Galilean and special relativity (HL)
Paper 1BHard7 marks
Data-based question7 steps to full marksDetermine
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)
Show mark scheme
Marking point
Mark
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)
Syllabus understandingA.5 — proper time interval; time dilation as given by Δt = γΔt0; D.2 — the work done in moving a charge q in an electric field as given by W = qΔVe; E.3 — nuclear masses expressed in (unified) atomic mass units u; Tool 3 — error bars, lines of maximum and minimum gradient, uncertainty in a gradient Command term: Determine
34A-1B-23
Postulates of special relativity·A.5 Galilean and special relativity (HL)
Paper 1BEasy7 marks
Data-based question7 steps to full marksDetermine
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)
Show mark scheme
Marking point
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)
Syllabus understandingA.5 — the two postulates of special relativity; Galilean velocity addition u′ = u − v; Tools 3 — gradient with units, extrapolating to an intercept, systematic error from a fixed delay, comparing a result with two predictions Command term: Determine
35A-1B-39
Testing the Galilean transformation·A.5 Galilean and special relativity (HL)
Paper 1BEasy7 marks
Data-based question7 steps to full marksDetermine
A student tests the Galilean transformation. A ball rolls in a straight line along a long horizontal bench, slowing down. Camera 1 is fixed to the bench (frame S). Camera 2 is fixed to a motorised trolley that moves along a rail beside the bench at a constant velocity in the positive x-direction (frame S′). Video analysis gives the position x of the ball measured from a mark on the bench and its position x′ measured from a mark on the trolley, at the same times t. At t = 0 the trolley mark is a distance d ahead of the bench mark. Each position reading has an uncertainty of ±0.01 m; the uncertainty in t is negligible.
The Galilean transformation predicts that x − x′ = vt + d, where v is the speed of the trolley. The graph shows the data, with the line of best fit.
t / s
x / m
x′ / m
(x − x′) / m
0.0
0.10
−0.20
0.30
0.4
0.71
0.12
0.59
0.8
1.25
0.35
0.90
1.2
1.73
0.53
1.20
1.6
2.14
0.65
1.49
2.0
2.49
0.70
1.79
Graph of (x − x′) against t with the line of best fit (drawn to scale).
(a)
(i)
State the absolute uncertainty in each value of x − x′.
(1)
(b)
(i)
Determine the speed of the trolley. Give its unit.
(2)
(ii)
Determine the distance d.
(1)
(c)
(i)
Galilean relativity predicts that the ball has the same acceleration in S and S′. Test this prediction by calculating the change in the average velocity of the ball between the intervals t = 0 to 0.4 s and t = 1.6 s to 2.0 s, in each frame.
(2)
(ii)
Explain, using the Galilean velocity addition equation, why the acceleration of the ball is the same in both frames.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
±0.02 m: the absolute uncertainties of the two readings add
✓ 1
Award this mark only for ±0.02 m; ±0.01 m scores 0.
Part (b)(i)
Gradient of the line from two well-separated points on it, e.g. (1.79 − 0.30)/(2.0 − 0.0) = 0.745
✓ 1
Points on the line (or the end points through which it passes), not adjacent data points.
v = 0.75 m s−1
✓ 1
Accept 0.72–0.77 m s−1. Unit required.
Part (b)(ii)
d = intercept on the vertical axis = 0.30 m
✓ 1
Accept 0.28–0.32 m. Allow ECF from (b)(i) if the intercept is calculated from the gradient and a point.
Part (c)(i)
S: (0.71 − 0.10)/0.4 = 1.525 and (2.49 − 2.14)/0.4 = 0.875 m s−1, change = −0.65 m s−1; S′: (0.12 + 0.20)/0.4 = 0.80 and (0.70 − 0.65)/0.4 = 0.125 m s−1, change = −0.675 m s−1
✓ 1
Both changes needed.
Each average velocity is uncertain by ±0.02/0.4 = ±0.05 m s−1, so each change is uncertain by about ±0.1 m s−1; the changes in velocity (and so the accelerations) agree within this uncertainty, as predicted
✓ 1
The conclusion must refer to the uncertainty. Accept the argument that each velocity in S′ is about 0.75 m s−1 less than in S, so the changes are equal.
Part (c)(ii)
u′ = u − v with v constant, so Δu′ = Δu in any time interval, and Δt′ = Δt: the accelerations Δu′/Δt and Δu/Δt are equal
✓ 1
The constancy of v (an inertial frame) must be used.
Answers: (a)(i) ±0.02 m · (b)(i) 0.75 m s−1 · (b)(ii) 0.30 m · (c)(i) S: −0.65 m s−1; S′: −0.675 m s−1; consistent (the remaining parts are explanations — see the table above)
Syllabus understandingA.5 — that Galilean transformation equations as given by x′ = x − vt and t′ = t relate the coordinates of an event in two inertial reference frames; the velocity addition equation u′ = u − v; that Newton's laws of motion are the same in all inertial reference frames; A.1 — average velocity and acceleration; Tool 3 — uncertainty of a difference, gradient and intercept of a linear graph with units; Inquiry 3 — evaluate a prediction Command term: Determine
36A-1B-40
Testing time dilation with unstable particles·A.5 Galilean and special relativity (HL)
Paper 1BMedium7 marks
Data-based question7 steps to full marksDetermine
Unstable particles are produced when a proton beam strikes a target. Particles moving at a chosen speed v are selected and sent along an evacuated beam pipe. For each speed, the mean lifetime τ of the particles in the laboratory frame is found from the decrease in the number of particles along the pipe. Each value of τ has an uncertainty of ±1.0 ns; the uncertainty in v is negligible.
Special relativity predicts that τ = γτ0, where τ0 is the mean lifetime of the particles in their rest frame. The graph shows τ against γ with error bars, for all six speeds.
v/c
γ
τ / ns ± 1.0 ns
0.500
1.15
14.5
0.700
1.40
17.7
0.800
1.67
20.8
0.900
27.9
0.950
3.20
40.1
0.980
5.03
62.5
Mean lifetime in the laboratory against the Lorentz factor (drawn to scale).
(a)
(i)
Calculate the missing value of γ for v = 0.900c.
(1)
(b)
(i)
Draw the line of best fit and the lines of maximum and minimum gradient that pass through all the error bars. Hence determine τ0 and its absolute uncertainty.
(3)
(c)
(i)
Deduce whether the data support the prediction τ = γτ0.
(1)
(ii)
Galilean relativity predicts that the mean lifetime does not depend on the speed of the particles. State how the graph would look if this prediction were correct, and compare it with the data.
(1)
(iii)
Explain why τ is plotted against γ rather than against v.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
γ = 1/√(1 − 0.900²) = 2.29
✓ 1
Accept 2.29 or 2.3.
Part (b)(i)
Best-fit gradient = 12.4 ns, so τ0 = 12.4 ns
✓ 1
Accept 12.0–12.8 ns. Gradient must be of a line, not a single ratio τ/γ.
Maximum gradient ≈ (63.5 − 13.5)/(5.03 − 1.15) = 12.9 ns (through the bottom of the first bar and the top of the last); minimum gradient ≈ (61.5 − 28.9)/(5.03 − 2.29) = 11.9 ns (through the top of the bar at γ = 2.29 and the bottom of the last bar)
✓ 1
Accept other extreme lines that pass through all the error bars. A minimum line through (1.15, 15.5) and (5.03, 61.5) misses the bar at γ = 2.29.
τ0 = (12.4 ± 0.5) ns
✓ 1
Accept an uncertainty of ±0.3 to ±0.7 ns. Allow ECF from the candidate's gradients.
Part (c)(i)
Yes: a straight line passes through all the error bars, and its extension passes through the origin (intercept ≈ 0.1 ns, zero within the uncertainty), so τ is proportional to γ
✓ 1
Both linearity and the origin are needed. Allow ECF from (b)(i).
Part (c)(ii)
A horizontal line (constant τ ≈ τ0); the measured τ increases more than fourfold from 14.5 ns to 62.5 ns, far beyond the ±1.0 ns uncertainty, so the Galilean prediction is rejected
✓ 1
Both the predicted shape and the comparison are needed.
Part (c)(iii)
The prediction τ = γτ0 gives a straight line through the origin with gradient τ0 when τ is plotted against γ; a graph of τ against v would be a curve, from which τ0 and the proportionality are much harder to test
✓ 1
Linearisation idea needed.
Answers: (a)(i) 2.29 · (b)(i) (12.4 ± 0.5) ns · (c)(i) supported (the remaining parts are explanations — see the table above)
Syllabus understandingA.5 — proper time interval; time dilation as given by Δt = γΔt0; the Lorentz factor γ = 1/√(1 − v²/c²); that in Galilean relativity t′ = t; Tool 3 — error bars, lines of best fit and of maximum and minimum gradient, uncertainty in a gradient, linearising a relationship; Inquiry 3 — evaluate a hypothesis Command term: Determine
37A-1B-41
Muon decay as evidence for time dilation·A.5 Galilean and special relativity (HL)
Paper 1BHard7 marks
Data-based question7 steps to full marksDetermine
Cosmic-ray muons that come to rest in a large block of plastic scintillator later decay. For each stopped muon, the detector records the time t between its arrival and its decay. The table gives the number of decays N recorded in time bins 1.0 μs wide, centred on t, for six selected bins. Random coincidences add a constant background of B = 40 counts to every bin, found from bins at very long times. For the decays, N − B = N0e−t/τ0, where τ0 is the mean lifetime of muons at rest.
The graph shows ln(N − B) against t with the line of best fit.
t / μs
N
ln(N − B)
0.5
2045
7.60
2.0
1017
6.88
3.5
551
5.0
309
5.59
6.5
185
4.98
8.0
101
4.11
Graph of ln(N − B) against decay time with the line of best fit (drawn to scale).
(a)
(i)
Explain why the background B must be subtracted before the logarithm is taken.
(1)
(ii)
Calculate the missing value of ln(N − B) for t = 3.5 μs.
(1)
(b)
(i)
Determine τ0.
(2)
(c)
(i)
In a separate experiment, muons moving vertically downwards at 0.995c are counted at a mountain station 1900 m above sea level and at sea level, with identical detectors. The counting rates are 570 h−1 at the mountain station and 425 h−1 at sea level. Assume that the speed of the muons is constant and that they are lost only by decay. Determine the Lorentz factor implied by these rates.
(2)
(ii)
Discuss whether the result of (c)(i) is evidence for time dilation.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
ln(N0e−t/τ0 + B) is not a linear function of t: only for N − B does taking logs give a straight line of gradient −1/τ0 (ln N would curve and level off as the decays approach the background, giving too large a τ0)
✓ 1
Reference to linearity of the log graph is needed.
Part (a)(ii)
ln(551 − 40) = ln 511 = 6.24
✓ 1
ln 551 = 6.31 scores 0.
Part (b)(i)
Gradient from two well-separated points on the line, e.g. (4.11 − 7.60)/(8.0 − 0.5) = −0.47 μs−1; best line −0.453 μs−1
✓ 1
Accept −0.44 to −0.47 μs−1.
τ0 = −1/gradient = 2.2 μs
✓ 1
Accept 2.1–2.3 μs. Allow ECF from the candidate's gradient.
Part (c)(i)
Time in the muon frame (proper time) for the descent: t0 = τ0 ln(570/425) = 2.2 × 0.294 = 0.646 μs
✓ 1
Allow ECF from (b)(i).
Time in the Earth frame = 1900/(0.995 × 3.00 × 108) = 6.37 μs, so γ = 6.37/0.646 = 9.9
✓ 1
Accept 9.3–10.5. Allow ECF from the candidate's proper time.
Part (c)(ii)
Without time dilation the expected sea-level rate is 570 e−6.37/2.2 ≈ 32 h−1, far below the measured 425 h−1; and γ for 0.995c is 1/√(1 − 0.995²) = 10.0, which agrees with (c)(i) within the uncertainty of τ0, so the data are evidence for time dilation
✓ 1
Allow ECF from (b)(i) and (c)(i). Both the comparison of rates (or of γ) and a conclusion are needed.
Answers: (a)(ii) 6.24 · (b)(i) 2.2 μs · (c)(i) ≈ 9.9 · (c)(ii) yes: γ ≈ 10 as predicted (the remaining parts are explanations — see the table above)
Syllabus understandingA.5 — that muon decay experiments provide experimental evidence for time dilation and length contraction; proper time interval; time dilation as given by Δt = γΔt0; E.3 — the exponential nature of radioactive decay; Tool 3 — systematic error (background), linearising a relationship using logarithms, gradient Command term: Determine
38A-2-03
Time dilation and length contraction·A.5 Galilean and special relativity (HL)
Paper 2Hard12 marks
Short answer & extended response7 steps to full marksDetermine
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.
(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)
Show mark scheme
Marking point
Mark
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)
Syllabus understandingA.5 — time dilation as given by Δt = γΔt0; length contraction as given by L = L0/γ; proper time interval and proper length; that muon decay experiments provide experimental evidence for time dilation and length contraction; E.3 — half-life and the exponential nature of radioactive decay Command term: Determine
39A-2-06
Simultaneity and the Lorentz transformation·A.5 Galilean and special relativity (HL)
Paper 2Hard11 marks
Short answer & extended response7 steps to full marksDetermine
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.
(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)
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)
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)
Syllabus understandingA.5 — the Lorentz transformation equations; length contraction; the two postulates of special relativity; that the space-time interval is an invariant quantity, (Δs)² = (cΔt)² − (Δx)²; the relativity of simultaneity Command term: Determine
40A-2-11
Time dilation·A.5 Galilean and special relativity (HL)
Paper 2Hard13 marks
Short answer & extended response9 steps to full marksDetermine
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)
Syllabus understandingA.5 — reference frames (an inertial reference frame is non-accelerating); proper time interval and proper length; time dilation as given by Δt = γΔt0; length contraction as given by L = L0/γ; the relativity of simultaneity; E.5 — the use of stellar parallax as a method to determine the distance to celestial bodies; the conversion between light years and parsecs Command term: Determine
41A-2-14
Lorentz transformation and the invariant interval·A.5 Galilean and special relativity (HL)
Paper 2Hard14 marks
Short answer & extended response8 steps to full marksDetermine
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.
Δt0 = Δt/γ = 8.00/1.281 = 6.24 μs, the same as the value from the interval in (c)(i)
✓ 1
Allow ECF from (c)(i). Accept 6.2–6.3 μs.
Part (d)(i)
x′ = 1500 − 1200 = 300 m, about 13 % smaller than the Lorentz value (346 m): the Galilean transformation is not valid at 0.500c
✓ 1
Allow ECF from (b)(ii).
Answers: (a)(i) 3.51 × 106 m², time-like · (a)(ii) yes · (b)(ii) x′ = 346 m, t′ = 6.35 μs · (c)(i) 6.24 μs · (c)(ii) 0.625c; 6.24 μs · (d)(i) 300 m (the remaining parts are explanations — see the table above)
Syllabus understandingA.5 — the Lorentz transformation equations for the coordinates of an event in two inertial reference frames; that the space–time interval Δs between two events is an invariant quantity as given by (Δs)² = (cΔt)² − (Δx)²; proper time interval; time dilation as given by Δt = γΔt0; that in Galilean relativity x′ = x − vtCommand term: Determine
42A-2-16
Spacetime diagrams·A.5 Galilean and special relativity (HL)
Paper 2Hard12 marks
Short answer & extended response8 steps to full marksDetermine
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)
Syllabus understandingA.5 — space–time diagrams; time dilation as given by Δt = γΔt0; the relativity of simultaneity; the Lorentz transformation equations for the coordinates of an event in two inertial reference frames; that in Galilean relativity x′ = x − vt and t′ = t (guidance: time dilation, length contraction and simultaneity can be visualized using space–time diagrams) Command term: Determine
43A-2-30
Galilean relativity·A.5 Galilean and special relativity (HL)
Paper 2Medium12 marks
Short answer & extended response8 steps to full marksExplain
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)
Show mark scheme
Marking point
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)
Syllabus understandingA.5 — reference frames; that Newton’s laws of motion are the same in all inertial reference frames (Galilean relativity); that in Galilean relativity x′ = x − vt and t′ = t; the velocity addition equation u′ = u − v; the two postulates of special relativity; C.2 — the nature of sound waves; C.5 — the observed frequency for a moving source, f′ = fv/(v ± us) Command term: Explain
44A-2-31
Spacetime diagrams·A.5 Galilean and special relativity (HL)
Paper 2Hard14 marks
Short answer & extended response10 steps to full marksDeduce
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)
Syllabus understandingA.5 — space–time diagrams; tan θ = v/c; the relativity of simultaneity; the Lorentz transformation equations; that the space–time interval is an invariant quantity, (Δs)² = (cΔt)² − (Δx)²; the scales on the time axes ct and ct′ are defined by lines of constant space–time interval Command term: Deduce
45A-2-32
Relativistic velocity addition·A.5 Galilean and special relativity (HL)
Paper 2Medium11 marks
Short answer & extended response7 steps to full marksDetermine
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.
(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)
Syllabus understandingA.5 — relativistic velocity addition u′ = (u − v)/(1 − uv/c²); the two postulates; length contraction and time dilation; D.2/D.3 — the motion of a charged particle in perpendicular uniform electric and magnetic fields; E = V/d; E.3 — half-life Command term: Determine
46A-2-60
A collision in two inertial frames·A.5 Galilean and special relativity (HL)
Paper 2Easy13 marks
Short answer & extended response9 steps to full marksDetermine
Two gliders move towards each other along a straight horizontal air track. Frame S is attached to the track. In S, glider A of mass 0.40 kg moves at +0.60 m s−1 and glider B of mass 0.20 kg moves at −0.30 m s−1. The gliders collide and stick together. Friction and air resistance are negligible.
(a)
(i)
Calculate the velocity of the gliders after the collision, in S.
(2)
(ii)
Determine the kinetic energy transferred to other forms in the collision, in S.
(2)
(b)
(i)
A camera moves along a rail beside the track at a constant velocity of +0.30 m s−1 relative to the track. Frame S′ is attached to the camera. State why S′ is an inertial frame of reference.
(1)
(ii)
Determine, in S′, the velocities of A and B before the collision and of the combined gliders after it.
(2)
(iii)
Deduce whether momentum is conserved in S′. Determine the kinetic energy transferred in the collision according to an observer in S′, and compare your answer with (a)(ii).
(3)
(c)
(i)
Show that, if the total momentum of a system of bodies is conserved in S, it is also conserved in any frame S′ that moves at a constant velocity v relative to S. Assume that v is much less than c.
(2)
(ii)
The camera is now accelerated along the rail during the collision. Outline why an observer using the camera's frame would find that Newton's laws do not apply to the gliders.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Total momentum: 0.40 × 0.60 + 0.20 × (−0.30) = 0.18 kg m s−1
✓ 1
Award [0] for this mark if the sign of the velocity of B is ignored (0.30 kg m s−1).
Velocity = 0.18/0.60 = +0.30 m s−1
✓ 1
Direction (positive) needed.
Part (a)(ii)
Before: ½ × 0.40 × 0.60² + ½ × 0.20 × 0.30² = 0.072 + 0.009 = 0.081 J
✓ 1
After: ½ × 0.60 × 0.30² = 0.027 J, so 0.054 J is transferred
✓ 1
Allow ECF from (a)(i).
Part (b)(i)
S′ moves at constant velocity relative to S (an inertial frame), so S′ is not accelerating
✓ 1
Do not accept "it moves at constant speed" without reference to velocity or acceleration.
Part (b)(ii)
Using u′ = u − v: A: 0.60 − 0.30 = +0.30 m s−1; B: −0.30 − 0.30 = −0.60 m s−1
✓ 1
After: 0.30 − 0.30 = 0: the combined gliders are at rest in S′
✓ 1
Allow ECF from (a)(i).
Part (b)(iii)
Momentum in S′: 0.40 × 0.30 + 0.20 × (−0.60) = 0 before and 0 after, so it is conserved
✓ 1
Allow ECF from (b)(ii).
Kinetic energy before = ½ × 0.40 × 0.30² + ½ × 0.20 × 0.60² = 0.018 + 0.036 = 0.054 J; after = 0
✓ 1
Allow ECF from (b)(ii).
Transferred: 0.054 J, the same as in S, although the total kinetic energies differ between the frames
✓ 1
Allow ECF from (a)(ii). The comparison is needed.
Part (c)(i)
In S′ each velocity is ui − v, so the total momentum is Σmi(ui − v) = Σmiui − vΣmi
✓ 1
Symbolic working required.
Σmiui is the same before and after (conserved in S) and the total mass and v do not change, so the total momentum in S′ is also the same before and after
✓ 1
The constancy of both the total mass and v must be stated.
Part (c)(ii)
In this frame each glider has an extra acceleration equal and opposite to that of the camera, so the gliders would appear to accelerate (and their total momentum to change) although no resultant external force acts on them: the frame is not inertial
✓ 1
Reference to acceleration without a force (or momentum change without an external force) is needed.
Answers: (a)(i) +0.30 m s−1 · (a)(ii) 0.054 J · (b)(ii) A: +0.30 m s−1; B: −0.60 m s−1; after: 0 · (b)(iii) 0.054 J in both frames (the remaining parts are explanations — see the table above)
Syllabus understandingA.5 — reference frames (an inertial reference frame is non-accelerating); that Newton's laws of motion are the same in all inertial reference frames; the Galilean velocity addition equation u′ = u − v; A.2 — the law of conservation of linear momentum; A.3 — kinetic energy; energy transfers in inelastic collisions Command term: Determine
47A-2-61
Time dilation of an orbiting clock·A.5 Galilean and special relativity (HL)
Paper 2Medium10 marks
Short answer & extended response9 steps to full marksDetermine
A navigation satellite moves in a circular orbit of radius 2.66 × 107 m around the Earth. An atomic clock on the satellite is compared with an identical clock at rest in a frame centred on the Earth, which may be treated as inertial. Consider only the effect of the motion of the satellite (special-relativistic time dilation).
Mass of the Earth = 5.97 × 1024 kg. For v ≪ c, γ ≈ 1 + v²/(2c²).
(a)
(i)
Show that the orbital speed of the satellite is about 3.9 × 103 m s−1.
(2)
(b)
(i)
Show that γ − 1 for the satellite is about 8 × 10−11.
(1)
(ii)
Determine the time by which the satellite clock falls behind the Earth clock during one day (8.64 × 104 s) of Earth time.
(2)
(iii)
A receiver finds its distance from the satellite from the travel time of a radio signal timed by the satellite clock. Estimate the error in this distance after one day if the time dilation is not corrected.
(1)
(c)
(i)
A second satellite moves in a circular orbit of radius 6.78 × 106 m. Deduce, without calculating its speed, the ratio (time lost per day by the clock of the second satellite)/(time lost per day by the clock of the first satellite).
(2)
(ii)
The frame of the satellite is not an inertial frame. Explain why, and outline why the time dilation formula can nevertheless be used in the analysis above.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gravitational force provides the centripetal force: GMm/r² = mv²/r, so v = √(GM/r)
✓ 1
v = √(6.67 × 10−11 × 5.97 × 1024/2.66 × 107) = 3.869 × 103 m s−1
✓ 1
An answer to at least 3 s.f. or the full substitution is needed.
Allow ECF from (a)(i) for the relation v = √(GM/r).
Ratio = 2.66 × 107/6.78 × 106 = 3.92
✓ 1
Accept 3.9. The square root of this ratio (2.0) scores [1 max].
Part (c)(ii)
The satellite moves in a circle, so its velocity changes direction: it has a centripetal acceleration and its frame is accelerating (not inertial)
✓ 1
Do not accept "it moves" or "it is in orbit" alone.
The calculation is made in the Earth-centred frame, which is inertial; in that frame the satellite clock moves at a constant speed, so the same γ applies to its ticking at every instant
✓ 1
Reference to the analysis being done in the inertial (Earth) frame is needed.
Answers: (b)(ii) 7.2 μs · (b)(iii) ≈ 2 km · (c)(i) 3.92 (the remaining parts are explanations — see the table above)
Syllabus understandingA.5 — reference frames (an inertial reference frame is non-accelerating); proper time interval; time dilation as given by Δt = γΔt0; the Lorentz factor; D.1 — orbital motion of a satellite: gravitational force provides the centripetal force, v = √(GM/r); A.2 — circular motion Command term: Determine
48A-2-62
Time dilation and length contraction for a moving radioactive source·A.5 Galilean and special relativity (HL)
Paper 2Medium13 marks
Short answer & extended response9 steps to full marksDetermine
A probe passes the Earth and travels away from it in a straight line at a constant speed of 0.80c relative to the Earth. It carries a sample of a radioactive isotope. The graph shows how the activity A of the sample, measured by instruments on the probe, varies with the time t shown by the probe clock. Both the probe clock and an Earth clock read zero as the probe passes the Earth.
1 year = 3.16 × 107 s.
Activity of the sample measured on the probe against probe time (drawn to scale).
(a)
(i)
Use the graph to determine the half-life of the isotope.
(1)
(ii)
Determine the number of radioactive nuclei in the sample when the probe passes the Earth.
(3)
(b)
(i)
Explain why the half-life in (a)(i) is a proper time interval.
(1)
(ii)
Calculate the half-life of the sample measured in the frame of the Earth.
(2)
(c)
(i)
The activity measured on the probe falls to A0/16. Determine, in the frame of the Earth, the time of this event and the distance of the probe from the Earth.
(2)
(ii)
A marker buoy is at rest relative to the Earth at the position of the probe at the event in (c)(i). Determine the distance between the Earth and the buoy in the frame of the probe, and show that it is consistent with the motion of the Earth in that frame.
(2)
(d)
(i)
Deduce whether observers on the Earth agree that the number of radioactive nuclei in the sample at the event in (c)(i) is N0/16.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Time for A to halve, e.g. 8.0 → 4.0 at 2.5 years (or 4.0 → 2.0 between 2.5 and 5.0 years): T½ = 2.5 years
✓ 1
Accept 2.4–2.6 years.
Part (a)(ii)
T½ = 2.5 × 3.16 × 107 = 7.9 × 107 s
✓ 1
Allow ECF from (a)(i).
λ = ln 2/T½ = 8.8 × 10−9 s−1
✓ 1
N0 = A0/λ = 8.0 × 1010/8.8 × 10−9 = 9.1 × 1018
✓ 1
Accept 9.0–9.3 × 1018. Using T½ in years scores [2 max].
Part (b)(i)
The two events (activity A0 and activity A0/2) both occur at the sample, and the probe clock is at rest relative to the sample, so a single clock present at both events measures the interval
✓ 1
Do not accept "it is measured on the probe" without reference to both events at the same place / one clock.
Part (b)(ii)
γ = 1/√(1 − 0.80²) = 5/3 (1.67)
✓ 1
Half-life = γ × 2.5 = 4.2 years
✓ 1
Allow ECF from (a)(i).
Part (c)(i)
Probe time = 4 half-lives = 10 years, so Earth time = γ × 10 = 16.7 years
✓ 1
Allow ECF from (a)(i) and (b)(ii) (4 × the Earth-frame half-life).
Distance = 0.80c × 16.7 years = 13.3 ly
✓ 1
Allow ECF from the candidate's Earth time.
Part (c)(ii)
The Earth–buoy distance is a proper length in the Earth frame, contracted in the probe frame: 13.3/γ = 8.0 ly
✓ 1
Allow ECF from (c)(i).
In the probe frame the Earth moves away at 0.80c for 10 years of probe time: 0.80 × 10 = 8.0 ly, the same distance
✓ 1
The link to the probe time of 10 years is needed.
Part (d)(i)
Earth frame: 16.7 years/4.17 years = 4.0 half-lives, so N = N0/24 = N0/16
✓ 1
Allow ECF from (b)(ii) and (c)(i).
Yes: the number of undecayed nuclei at a given event is a count that all observers agree on; time dilation of both the elapsed time and the half-life by the same factor γ gives the same result
✓ 1
A conclusion with a reason is needed.
Answers: (a)(i) 2.5 years · (a)(ii) 9.1 × 1018 · (b)(ii) 4.2 years · (c)(i) 16.7 years; 13.3 ly · (c)(ii) 8.0 ly · (d)(i) yes (the remaining parts are explanations — see the table above)
Syllabus understandingA.5 — proper time interval and proper length; time dilation as given by Δt = γΔt0; length contraction as given by L = L0/γ; reference frames; E.3 — half-life; activity A = λN; T½ = ln 2/λCommand term: Determine
49A-2-63
Synchronised clocks and the relativity of simultaneity·A.5 Galilean and special relativity (HL)
Paper 2Hard13 marks
Short answer & extended response10 steps to full marksDeduce
Three beacons B0, B1 and B2 are at rest in the frame S of a space station, at x = 0, 600 m and 1200 m. Each beacon carries a clock; the clocks are synchronised in S. A shuttle moves at 0.60c in the positive x-direction relative to S; its frame is S′. The shuttle clock and the clock of B0 both read zero as the shuttle passes B0 (event O).
The space-time diagram, drawn in S, shows the world lines of the beacons and the axes of S′. Take c = 3.00 × 108 m s−1.
Space-time diagram in the station frame S (drawn to scale). The gold lines are the axes of the shuttle frame S′.
(a)
(i)
Show that the Lorentz factor for the shuttle relative to the station is 1.25.
(1)
(ii)
Calculate the reading of the clock of B2 and the reading of the shuttle clock when the shuttle passes B2.
(2)
(b)
(i)
Use the Lorentz transformation to show that, in S′, at the instant at which the shuttle passes B0, the clock of B2 reads vx/c², where x = 1200 m. Calculate this reading.
(2)
(ii)
Explain how the space-time diagram confirms your answer to (b)(i).
(1)
(c)
(i)
In S′ the beacons move at 0.60c in the negative x′-direction. Determine, in S′, the time interval recorded by the clock of B2 between the instant in (b)(i) and the instant at which B2 reaches the shuttle.
(3)
(ii)
Hence show that the description in S′ gives the same reading of the clock of B2 when it reaches the shuttle as found in (a)(ii).
(1)
(d)
(i)
When they meet, the clock of B2 reads more than the shuttle clock. Explain why this does not contradict the shuttle observer's claim that the station clocks run slow.
(2)
(e)
(i)
State the reading of the clock of B2 at the instant in (b)(i) according to Galilean relativity.
(1)
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)
B2: t = 1200/(0.60 × 3.00 × 108) = 6.67 μs
✓ 1
Shuttle (proper time, present at both O and the passing): 6.67/1.25 = 5.33 μs
✓ 1
Allow ECF from (a)(i).
Part (b)(i)
Events simultaneous with O in S′ have t′ = 0: γ(t − vx/c²) = 0, so t = vx/c² at the position of B2
✓ 1
Symbolic working required.
Reading = 0.60 × 1200/(3.00 × 108) = 2.4 μs
✓ 1
Part (b)(ii)
Events simultaneous with O in S′ lie on the x′ axis; it meets the world line of B2 at ct = 720 m (= 0.60 × 1200 m), i.e. t = 720/(3.00 × 108) = 2.4 μs
✓ 1
Allow ECF from (b)(i). Accept 700–740 m read from the diagram.
Part (c)(i)
In S′ the distance B0–B2 is contracted: 1200/1.25 = 960 m
✓ 1
Allow ECF from (a)(i).
Time in S′ for B2 to reach the shuttle = 960/(0.60 × 3.00 × 108) = 5.33 μs
✓ 1
The clock of B2 moves in S′, so it runs slow: 5.33/1.25 = 4.27 μs
✓ 1
Allow ECF from the candidate's time in S′.
Part (c)(ii)
2.40 + 4.27 = 6.67 μs, the same as (a)(ii)
✓ 1
Allow ECF from (b)(i) and (c)(i).
Part (d)(i)
In S′ the station clocks are not synchronised: at t′ = 0 the clock of B2 already reads 2.4 μs ahead of B0 (relativity of simultaneity)
✓ 1
Allow ECF from (b)(i).
During the 5.33 μs of shuttle time the clock of B2 advances only 4.27 μs (it runs slow in S′); its head start makes it read more than the shuttle clock at the meeting
✓ 1
Allow ECF from (c)(i). Do not accept "it is an illusion" or answers based on the light travel time.
Part (e)(i)
Zero: in Galilean relativity t′ = t, so simultaneity is absolute and all the station clocks read zero when the shuttle passes B0
✓ 1
Answers: (a)(ii) B2: 6.67 μs; shuttle: 5.33 μs · (b)(i) 2.4 μs · (c)(i) 4.27 μs · (e)(i) 0 (the remaining parts are explanations — see the table above)
Syllabus understandingA.5 — the relativity of simultaneity; the Lorentz transformation equations t′ = γ(t − vx/c²); time dilation as given by Δt = γΔt0; length contraction as given by L = L0/γ; proper time interval; space–time diagrams; that in Galilean relativity t′ = tCommand term: Deduce
50A-2-64
Relativistic velocity addition·A.5 Galilean and special relativity (HL)
Paper 2Medium11 marks
Short answer & extended response8 steps to full marksDetermine
A carrier spacecraft moves at a constant velocity relative to a space station. It launches probes straight ahead with a speed u′ relative to the carrier. The graph shows how the speed u of a probe relative to the station varies with u′, according to special relativity. The dashed line marks u = c.
Speed of a probe relative to the station against its speed relative to the carrier (drawn to scale).
(a)
(i)
State the speed of the carrier relative to the station, giving a reason.
(1)
(ii)
Show, using the relativistic velocity addition equation, that a probe launched at u′ = 0.60c moves at about 0.85c relative to the station.
(2)
(iii)
Draw, on the graph, the line that Galilean relativity predicts. Hence explain why the relativistic curve, unlike the Galilean line, can never cross the dashed line.
(2)
(b)
(i)
A probe of proper length 15.0 m is launched at u′ = 0.60c. Determine the length of the probe measured by an observer on the station.
(3)
(ii)
Calculate the length of the same probe measured by an observer on the carrier.
(1)
(c)
(i)
A probe must move at 0.90c relative to the station. Determine the speed with which it must be launched relative to the carrier.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
0.50c: a probe with u′ = 0 moves with the carrier, and the graph gives u = 0.50c at u′ = 0
✓ 1
Reason needed.
Part (a)(ii)
With S′ the carrier frame, u′ = (u − v)/(1 − uv/c²) rearranges to u = (u′ + v)/(1 + u′v/c²) = (0.60 + 0.50)c/(1 + 0.30)
✓ 1
Allow ECF from (a)(i) for v.
u = 1.10c/1.30 = 0.846c
✓ 1
An answer to at least 3 s.f. (0.846c) is needed.
Part (a)(iii)
Galilean: u = u′ + 0.50c, a straight line of gradient 1 from (0, 0.50) that reaches u = c at u′ = 0.50c
✓ 1
Allow ECF from (a)(i).
Second postulate: light (u′ = c) has speed c in every inertial frame, so u = c only when u′ = c; for u′ < c the relativistic equation always gives u < c
✓ 1
Reference to the postulate (or to substituting u′ = c in the equation) is needed.
Part (b)(i)
Speed relative to the station = 0.85c (from the graph or (a)(ii))
✓ 1
Allow ECF from (a)(ii).
γ = 1/√(1 − 0.846²) = 1.88
✓ 1
Accept 1.88–1.90 from 0.85c.
Length = 15.0/1.88 = 8.0 m
✓ 1
Accept 7.9–8.0 m. Allow ECF from the candidate's γ. Using γ for 0.60c scores [1 max].
Part (b)(ii)
15.0 × √(1 − 0.60²) = 12.0 m
✓ 1
Part (c)(i)
From the graph, u = 0.90c at u′ ≈ 0.73c, or u′ = (0.90 − 0.50)c/(1 − 0.90 × 0.50)
✓ 1
Allow ECF from (a)(i) for v.
u′ = 0.40c/0.55 = 0.73c
✓ 1
Accept 0.70–0.75c from the graph. The Galilean answer 0.40c scores [0].
Answers: (a)(i) 0.50c · (b)(i) 8.0 m · (b)(ii) 12.0 m · (c)(i) 0.73c(the remaining parts are explanations — see the table above)
Syllabus understandingA.5 — that Lorentz transformation equations lead to the relativistic velocity addition equation as given by u′ = (u − v)/(1 − uv/c²); that Galilean transformation equations lead to u′ = u − v; the two postulates of special relativity; length contraction as given by L = L0/γ; proper length Command term: Determine
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