Free A.5 Galilean & special relativity questions for IB Physics (SL & HL), HL only, part of A Space, time & motion. Each question has the answer and the marking points an examiner would look for.
Generate a full A.5 Galilean & Special Relativity worksheet →All A Space, time & motion questionsTwo spaceships move directly towards each other. Relative to the Earth, one moves at 0.8c and the other at 0.6c.
What is the speed of one spaceship relative to the other?
Answer: C
Spaceship A moves to the right at 0.60c relative to the Earth. Spaceship B moves to the left at 0.80c relative to the Earth. The proper length of A is 100 m.
State why the Galilean transformation cannot give the speed of B relative to A.
Calculate the speed of B relative to A.
Calculate the length of A measured by an Earth observer.
Two lamps at the front and back of A flash at the same time in the Earth frame. Explain whether the flashes are simultaneous for an observer on A.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | gives 1.4c > c it gives 1.4c, faster than light ✓ | 1 |
| (b) | 0.95c u′ = (0.80 + 0.60)/(1 + 0.48) ✓ = 0.946c ✓ | 2 |
| (c) | 80 m γ = 1.25 ✓ L = 100/1.25 = 80 m ✓ | 2 |
| (d) | not simultaneous simultaneity is relative — events at different places simultaneous in one frame are not in another ✓ for A the front flash happens first ✓ | 2 |
Muons are produced at a height of 10 km above the ground and travel vertically down at 0.99c. The half-life of muons in their rest frame is 1.5 μs.
Calculate the Lorentz factor.
Calculate the time taken to reach the ground in the Earth frame.
Calculate the fraction of muons that reach the ground (i) without relativity and (ii) with time dilation.
Explain the result from the muons’ frame of reference.
State which observer measures the proper time between the muon’s creation and its arrival at the ground, and explain why.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 7.09 γ = 1√1 − 0.992 = 7.09 ✓ | 1 |
| (b) | 33.7 μs t = 100000.99 × 3.00 × 108 = 33.7 μs ✓ | 1 |
| (c) | 1.7 × 10−7; 0.111 (i) number of half-lives = 22.4 ✓ fraction = 1.7 × 10−7 ✓ (ii) proper time = 33.77.09 = 4.75 μs ✓ fraction = 0.111 ✓ | 4 |
| (d) | length contraction in the muon frame the distance to the ground is length contracted ✓ to 1411 m, so they reach it in a shorter proper time ✓ | 2 |
| (e) | muon frame an observer moving with the muon ✓ both events occur at the same place in that frame ✓ | 2 |
A detector on a mountain at a height of 1910 m counts 563 muons per hour moving vertically downwards at a constant speed. An identical detector at sea level, directly below, counts 408 per hour. The half-life of muons in their rest frame is 1.52 μs. Assume no muons are created or absorbed between the detectors. c = 3.00 × 108 m s−1.
Determine the speed of the muons.
Calculate the distance between the detectors in the rest frame of the muons.
Estimate the count rate expected at sea level if time dilation did not occur, and comment on the result.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 0.994c (γβ ≈ 9.0) 408/563 = (½)n ✓ n = 0.465 half-lives ✓ proper time = 0.465 × 1.52 μs = 0.706 μs ✓ in the ground frame t = γ × 0.706 μs and 1910 m = vt ✓ so γβc × 0.706 μs = 1910 m ✓ γβ = 9.02 ✓ β = 9.02/√(1 + 9.02²) = 0.994 ✓ | 7 |
| (b) | ≈ 211 m γ = √(1 + 9.02²) = 9.07 ✓ L = 1910/9.07 ≈ 211 m ✓ | 2 |
| (c) | ≈ 30 per hour t = 1910/(0.994c) = 6.4 μs = 4.2 half-lives ✓ expected 563 × 2−4.2 ≈ 30 per hour ✓ the observed 408 per hour is far larger — evidence for time dilation ✓ | 3 |
A spacecraft moves at 0.80c relative to Earth. A clock on the spacecraft records 6.0 years for a journey.
Calculate the Lorentz factor.
Calculate the time for the journey measured on Earth.
Calculate the distance travelled according to the spacecraft observers.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 1.67 γ = 1/√(1 − 0.64) = 1.67 ✓ | 1 |
| (b) | 10 years Δt = γΔt₀ ✓ 10 years ✓ | 2 |
| (c) | 4.8 ly 0.80c × 6.0 years ✓ 4.8 light-years ✓ | 2 |
Two events P and Q are observed in inertial frame S. Q occurs 5 ly from P and 6 y after P (so cΔt = 6 ly). Frame S′ moves at 0.8c relative to S in the direction from P to Q.
Calculate the space–time interval (Δs)2 between P and Q.
Deduce whether P could have caused Q.
Determine the time separation cΔt′ and distance Δx′ between the events in S′.
Show that your answers give the same space–time interval.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 11 ly2 (Δs)2 = (cΔt)2 − Δx2 ✓ = 11 ly2 ✓ | 2 |
| (b) | yes, timelike interval is positive / timelike ✓ a signal travelling slower than light can connect them ✓ | 2 |
| (c) | cΔt′ = 3.3 ly, Δx′ = 0.33 ly γ = 1.67 ✓ cΔt′ = γ(cΔt − βΔx) = 3.3 ly ✓ Δx′ = γ(Δx − βcΔt) = 0.33 ly ✓ | 3 |
| (d) | 11 (3.3)2 − (0.33)2 ≈ 11 ly2 ✓ | 1 |
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Meet Riyaz →Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in