IB DP Physics: Galilean and special relativity: Lorentz transformations and space-time diagrams (DP Physics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover sub-topic A.5, Galilean and special relativity, from the IB Physics guide, first assessment 2025. All of A.5 is additional higher level content, so every question is HL only. Use c = 3.00 × 10⁸ m/s. Read the study guide first and condense with the revision notes. Course links: IB DP Physics hub and printable checklist.

Questions

1. Frame S′ moves at 8.0 m/s along +x relative to frame S. An event has x = 120 m and t = 5.0 s in S.

(a) Use the Galilean transformations to find x′ and t′. [2] (b) A runner moves at 11.0 m/s along +x in S. Find the runner’s velocity in S′. [1]

2. (a) State what is meant by Galilean relativity. [1] (b) State the two postulates of special relativity. [2] (c) A lamp in S emits a light pulse along +x. S′ moves at 0.50c along +x. State the speed of the pulse in S′ according to Galilean addition and according to special relativity, and say which agrees with the second postulate. [2]

3. An unstable particle has a proper mean lifetime of 26 ns and moves at 0.92c in the laboratory.

(a) Calculate γ. [1] (b) Find the mean lifetime measured in the laboratory. [1] (c) Find the distance it travels in the laboratory frame in that time. [1] (d) State which of the two lifetimes is a proper time interval and why. [1]

4. A rocket has proper length 150 m and passes a station at 0.85c.

(a) Find the length measured from the station. [2] (b) State the length of the rocket measured by the pilot. [1]

5. A star is 6.0 light-years from Earth in the Earth frame. A ship travels there at a constant 0.80c.

(a) Find the journey time in the Earth frame. [1] (b) Find the journey time measured on the ship, and identify which time is the proper time interval. [2] (c) Find the Earth-star distance in the ship frame. [2] (d) Show that your answers to (b) and (c) are consistent. [1]

6. Frame S′ moves at 0.40c along +x relative to S. An event has x = 6.0 × 10⁸ m and t = 3.0 s in S.

(a) Calculate γ. [1] (b) Find x′. [2] (c) Find t′. [2] (d) Show that (cΔt)² − (Δx)² takes the same value in both frames, measuring from the origin event. [2]

7. Relative to Earth, spacecraft A moves at 0.75c along +x and spacecraft B moves at 0.60c along −x.

(a) Find the velocity of B in the frame of A. [3] (b) Explain why Galilean addition gives an impossible answer. [1]

8. In frame S, two events are 8.0 μs apart in time and 1440 m apart in space.

(a) Calculate the space-time interval Δs. [3] (b) Find the proper time between the events. [1] (c) Find the speed of the frame in which the events happen at the same place. [1] (d) Explain why no frame exists in which these events are simultaneous. [1]

9. Two lightning strikes are simultaneous in S and 450 m apart along x. Frame S′ moves at 0.80c along +x.

(a) Find Δt′, the time of the larger-x strike minus the time of the smaller-x strike, giving its sign. [3] (b) State which strike S′ observers say happened first, and explain the answer. [2]

10. (a) A particle moves at 0.80c. Find the angle between its world line and the ct axis. [2] (b) A world line makes 30.0° with the ct axis. Find the speed as a fraction of c. [2] (c) Describe how the ct′ and x′ axes of a frame moving at 0.80c are drawn relative to ct and x. [2] (d) State why the scales along ct′ and x′ differ from those along ct and x. [1]

11. Muons form 12 km above the ground (Earth frame) and move at 0.995c. Their proper mean lifetime is 2.2 μs.

(a) Find the distance a muon would travel in one proper lifetime if there were no time dilation. [2] (b) Find γ, the lifetime in the Earth frame, and the corresponding distance. [3] (c) In the muon frame, find the contracted thickness of the 12 km layer and the time to cross it. [2] (d) Explain why (b) and (c) support special relativity. [1]

Answers

1. (a) x′ = x − vt = 120 − 8.0 × 5.0 [1] = 80 m, and t′ = t = 5.0 s. [1] (b) u′ = u − v = 11.0 − 8.0 = 3.0 m/s. [1] [3] Examiner insight: Give both coordinates in (a); a candidate who finds only x′ has answered half the question.

2. (a) Newton’s laws of motion hold in the same form for observers in every inertial frame. [1] (b) The laws of physics take the same form in every inertial frame. [1] The speed of light in a vacuum is the same for all inertial observers, regardless of source or observer motion. [1] (c) Galilean: c − 0.50c = 0.50c. Special relativity: c. [1] The relativistic result agrees with the second postulate. [1] [5] Examiner insight: Postulate 2 loses credit if it omits “all inertial observers” or “vacuum”; say who measures the speed.

3. (a) γ = 1/√(1 − 0.92²) = 2.55. [1] (b) Δt = γΔt₀ = 2.55 × 26 = 66 ns. [1] (c) d = 0.92 × 3.00 × 10⁸ × 66.3 × 10⁻⁹ = 18 m. [1] (d) The 26 ns is proper, because it is measured by a clock at rest with the particle, so both decay events happen at one place. [1] [4] Examiner insight: Justify “proper” by the events being at the same place in that frame; naming the smaller number alone is weaker.

4. (a) γ = 1.898 [1], L = 150/1.898 = 79 m. [1] (b) The pilot is at rest relative to the rocket, so measures the proper length 150 m. [1] [3] Examiner insight: Part (b) rewards recognising that the proper length is read in the rest frame, not computing again.

5. (a) t = 6.0/0.80 = 7.5 years. [1] (b) The ship’s clock is present at both events, so its time is proper: t₀ = 7.5/γ with γ = 1.667 [1], giving 4.5 years. [1] (c) L = 6.0/1.667 [1] = 3.6 light-years. [1] (d) In the ship frame, 3.6 ly at 0.80c takes 3.6/0.80 = 4.5 years, equal to (b). [1] [6] Examiner insight: “Show that” needs the check written out with numbers, not just a statement that they agree.

6. (a) γ = 1/√(1 − 0.16) = 1.09. [1] (b) vt = 0.40 × 3.00 × 10⁸ × 3.0 = 3.6 × 10⁸ m [1]; x′ = 1.091 × (6.0 − 3.6) × 10⁸ = 2.6 × 10⁸ m. [1] (c) vx/c² = 0.40 × 2.0 s = 0.80 s [1]; t′ = 1.091 × (3.0 − 0.80) = 2.4 s. [1] (d) In S: (9.0 × 10⁸)² − (6.0 × 10⁸)² = 4.5 × 10¹⁷ m². [1] In S′: (7.2 × 10⁸)² − (2.62 × 10⁸)² = 5.19 × 10¹⁷ − 0.69 × 10¹⁷ = 4.5 × 10¹⁷ m², the same. [1] [7] Examiner insight: Carry unrounded values into (d); early rounding of γ can make the two intervals differ in the third figure.

7. (a) Take S′ as A’s frame, so v = +0.75c, and u = −0.60c. [1] u′ = (−0.60 − 0.75)c/(1 − (−0.60)(0.75)) = −1.35c/1.45 [1] = −0.93c, so B approaches A at 0.93c. [1] (b) Galilean addition gives 1.35c, which exceeds the speed of light. [1] [4] Examiner insight: Show the signed substitution into the equation; a wrong sign for u gives a speed that is easy to challenge, and the working shows where it went wrong.

8. (a) cΔt = 3.00 × 10⁸ × 8.0 × 10⁻⁶ = 2400 m. [1] Δs² = 2400² − 1440² = 3.69 × 10⁶ m². [1] Δs = 1920 m. [1] (b) Δt₀ = Δs/c = 1920/3.00 × 10⁸ = 6.4 μs. [1] (c) v = Δx/Δt = 1440/2400 = 0.60c. [1] (d) Δs² is positive and invariant. In a frame where the events were simultaneous, Δs² would equal −(Δx′)², which cannot be positive. [1] [6] Examiner insight: Part (d) needs the invariance link; saying “nothing can go faster than light” alone does not use the interval.

9. (a) γ = 1.667 [1]; vΔx/c² = 0.80 × 450/(3.00 × 10⁸) = 1.2 × 10⁻⁶ s [1]; Δt′ = −γvΔx/c² = −2.0 μs. [1] (b) The negative sign means the larger-x strike occurs first in S′. [1] Time depends on position in t′ = γ(t − vx/c²), so equal t in S gives different t′. [1] [5] Examiner insight: State which event is first and tie it to the sign; a number without direction is incomplete.

10. (a) tan θ = v/c = 0.80 [1], θ = 38.7°. [1] (b) tan θ = v/c [1], so v/c = tan 30.0° = 0.577. [1] (c) ct′ is the world line of the S′ origin, at 38.7° to ct. [1] x′ is tilted up by the same angle from x, towards the 45° light line. [1] (d) The scales are fixed by lines of constant space-time interval. [1] [7] Examiner insight: Measure θ from the ct axis, not x; the speed is greater than c if tan is read from the wrong axis.

11. (a) d = 0.995 × 3.00 × 10⁸ × 2.2 × 10⁻⁶ [1] = 660 m. [1] (b) γ = 10.0 [1]; Δt = 22 μs [1]; d = 0.995c × 22.0 μs ≈ 6.6 km. [1] (c) L = 12 km/10.0 = 1.2 km [1], crossed in 1.2 km/(0.995c) = 4.0 μs. [1] (d) Without relativity a muon would cover only 660 m, yet muons reach the ground; time dilation (Earth frame) and length contraction (muon frame) both explain this, so the observations support both effects. [1] [8] Examiner insight: Keep frames separate: dilation belongs in the Earth frame and contraction in the muon frame, not both in one argument.

Where marks are usually lost

• Using Galilean addition once the speeds are a sizeable fraction of c.
• Swapping which frame gives the proper time or proper length.
• Dropping the sign of v or u in the velocity addition and Lorentz equations.
• Rounding γ early, then failing to reproduce an invariant interval.
• Writing “time slows” without naming the clock and the observer.
• Treating x′ as perpendicular to ct′ on a space-time diagram.
• Measuring the world-line angle from the x axis.
• Leaving out units or mixing light-years, metres and seconds in one line.

Next steps

• Revise with the revision notes
• Reread the study guide
• Open the IB DP Physics hub and printable checklist
• Take all free 10-minute diagnostics
• Book a free trial class

Official syllabus

IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization. Sub-topic A.5, Galilean and special relativity (additional higher level).