A.5 Galilean and special relativity (HL) – IB Physics exam questions with mark scheme

1HardHL only6 marks

Cosmic-ray muons are created high in the atmosphere and travel towards the Earth's surface at speed $0.995c$. The proper lifetime of a muon is $2.2\ \mu\mathrm{s}$. Take $c=3.00\times10^8\ \mathrm{m\,s^{-1}}$.

(a) Calculate the Lorentz factor $\gamma$ for the muon. [2]

(b) Determine the average lifetime of the muon as measured in the Earth frame. [1]

(d) Determine the distance, measured in the muon's frame of reference, that the Earth's atmosphere moves past the muon during its lifetime. [2]

$\gamma = 1/\sqrt{1-v^2/c^2} = 1/\sqrt{1-0.995^2}$(1)
$\gamma=10.0$ (10.0)(1)
$\Delta t=\gamma\Delta t_0 = 22\ \mu\mathrm{s}$ (22 μs)(1)
$d=vt=0.995\times3.00\times10^8\times22.0\times10^{-6} = 6.6$ km (6.6 km)(1)
Length contraction in the muon frame: $L=L_0/\gamma$(1)
$L = 6575/10.01 = 657$ m (about 660 m; allow $0.995c\times2.2\ \mu$s)(1)
Answer: (a) 10.0 (b) 22 μs (c) 6.6 km (d) about 660 m
Examiner tip: Be clear which frame each quantity is measured in: proper time is in the muon frame, proper length is in the Earth frame.

2CoreHL only4 marks

A train moves along a straight track at a constant velocity of $30\ \mathrm{m\,s^{-1}}$ relative to the ground. Frame S is fixed to the ground and frame S′ is fixed to the train. The origins of the two frames coincide at $t = t' = 0$.

(a) State what is meant by an inertial frame of reference. [1]

(b) A firework explodes at $x = 420\ \mathrm{m}$ and $t = 12\ \mathrm{s}$ in frame S. Using a Galilean transformation, calculate the position of the explosion in frame S′. [1]

(c) A car travels in the same direction as the train at $22\ \mathrm{m\,s^{-1}}$ relative to the ground. Determine the velocity of the car in frame S′. [1]

(d) State the time of the explosion in frame S′ according to Galilean relativity. [1]

a frame that is not accelerating / moves at constant velocity OR a frame in which Newton's first law holds (a free body moves at constant velocity); do not accept "a frame at rest" alone(1)
$x' = x - vt = 420 - 30\times12 = 60\ \mathrm{m}$(1)
$u' = u - v = 22 - 30 = -8\ \mathrm{m\,s^{-1}}$ OR $8\ \mathrm{m\,s^{-1}}$ backwards / opposite to the train's motion; sign or direction needed(1)
$t' = 12\ \mathrm{s}$ / the same as in S (time is absolute in Galilean relativity)(1)
Answer: (b) $60\ \mathrm{m}$ (c) $-8\ \mathrm{m\,s^{-1}}$ (8 m s⁻¹ towards the back of the train) (d) $12\ \mathrm{s}$
Examiner tip: Leaving out the minus sign or direction in (c) loses the mark: a velocity in S′ needs its direction relative to the train.

3CoreHL only4 marks

(a) State the two postulates of special relativity. [2]

(b) A spacecraft travelling towards a planet at speed $v$ switches on a forward-facing lamp. Explain why the Galilean velocity transformation is inconsistent with the second postulate in this situation. [2]

the laws of physics are the same in all inertial frames of reference(1)
the speed of light in a vacuum is the same / $c$ for all inertial observers (independent of the motion of the source or observer)(1)
Galilean addition predicts that an observer on the planet measures the light's speed as $c + v$(1)
but the second postulate (and Maxwell's equations / experiment) says it is $c$, so the Galilean transformation cannot be correct at high speeds / for light(1)
Answer: (a) Laws of physics same in all inertial frames; speed of light in vacuum same for all inertial observers. (b) Galilean addition gives $c+v$, which contradicts the measured/postulated $c$.
Examiner tip: "Light always travels at the same speed" loses the mark: say in a vacuum, and for all inertial observers.

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4StandardHL only4 marks

Two spacecraft, X and Y, move towards each other along the same straight line. Relative to a space station, X moves at $0.60c$ to the right and Y moves at $0.70c$ to the left.

(a) A student uses Galilean relativity to calculate the speed of Y relative to X. State the value the student obtains and why it cannot be correct. [1]

(b) Calculate the speed of Y as measured by an observer in X. [3]

$1.3c$, which is greater than $c$ (no object can be measured to move faster than light)(1)
correct substitution of signs: $u = -0.70c$ and $v = +0.60c$ (or equivalent with both directions reversed)(1)
$u' = \dfrac{-0.70c - 0.60c}{1 - \dfrac{(-0.70c)(0.60c)}{c^2}} = \dfrac{-1.30c}{1.42}$(1)
speed $= 0.92c$ (allow $0.915c$); direction not needed; $0.10c/0.58 = 0.17c$ from wrong signs scores max 1(1)
Answer: (a) $1.3c$, greater than $c$ (b) $0.92c$
Examiner tip: Most lost marks come from sign errors: take one direction as positive and give the velocity of Y a minus sign before you substitute.

5StandardHL only5 marks

A spacecraft has a proper length of $120\ \mathrm{m}$. It passes a space station at constant speed. An observer on the station measures the length of the spacecraft to be $72\ \mathrm{m}$.

(a) Show that the speed of the spacecraft relative to the station is $0.80c$. [2]

(b) A marker M is fixed on the station. Calculate the time taken for the spacecraft to pass M as measured by the observer on the station. [1]

(d) State which of the two times in (b) and (c) is a proper time. [1]

$\gamma = \dfrac{L_0}{L} = \dfrac{120}{72} = 1.67$ (5/3)(1)
$\dfrac{v}{c} = \sqrt{1 - \dfrac{1}{\gamma^2}} = \sqrt{1-0.36} = 0.80$; answer given, so the working must be seen(1)
$t = \dfrac{72}{0.80\times3.00\times10^8} = 3.0\times10^{-7}\ \mathrm{s}$(1)
$t = \dfrac{120}{0.80\times3.00\times10^8} = 5.0\times10^{-7}\ \mathrm{s}$ OR $\gamma\times3.0\times10^{-7}$; allow ECF from (b)(1)
the station time in (b) / $3.0\times10^{-7}$ s, because the start and end of the passage both happen at M, i.e. at the same place in the station frame(1)
Answer: (b) $3.0\times10^{-7}\ \mathrm{s}$ (c) $5.0\times10^{-7}\ \mathrm{s}$ (d) the station time, (b)
Examiner tip: Proper time belongs to the frame in which both events happen at the same place. Here that is the station, not the crew, so do not assume the moving observer always has the proper time.

6HardHL only8 marks

In frame S, event 1 happens at $x_1 = 0$, $t_1 = 0$ and event 2 happens at $x_2 = 900\ \mathrm{m}$, $t_2 = 2.0\ \mu\mathrm{s}$. Frame S′ moves at $0.80c$ in the $+x$ direction relative to S, and the origins coincide at $t = t' = 0$. Take $c = 3.00\times10^8\ \mathrm{m\,s^{-1}}$.

(a) Calculate $\gamma$ for frame S′ relative to S. [1]

(b) Calculate the coordinates $x_2'$ and $t_2'$ of event 2 in frame S′. [4]

(d) Using your answer to (b), deduce whether event 1 could have caused event 2. [1]

$\gamma = \dfrac{1}{\sqrt{1-0.80^2}} = 1.67$ (5/3)(1)
$x_2' = \gamma(x - vt) = 1.67\,(900 - 0.80\times3.00\times10^8\times2.0\times10^{-6})$(1)
$x_2' = 700\ \mathrm{m}$(1)
$t_2' = \gamma\left(t - \dfrac{vx}{c^2}\right) = 1.67\left(2.0\times10^{-6} - \dfrac{0.80\times900}{3.00\times10^8}\right)$(1)
$t_2' = -0.67\ \mu\mathrm{s}$ / $-6.7\times10^{-7}\ \mathrm{s}$; negative sign required(1)
in S: $(ct)^2 - x^2 = 600^2 - 900^2 = -4.5\times10^5\ \mathrm{m^2}$(1)
in S′: $(ct')^2 - x'^2 = 200^2 - 700^2 = -4.5\times10^5\ \mathrm{m^2}$, so equal; allow ECF from (b); accept the positive version $x^2-(ct)^2$ if used consistently(1)
no: the order of the events is reversed in S′ (event 2 happens before event 1) OR light could only travel 600 m in 2.0 μs, less than 900 m (interval is space-like), so no signal could link them(1)
Answer: (a) 1.67 (b) $x_2' = 700\ \mathrm{m}$, $t_2' = -0.67\ \mu\mathrm{s}$ (c) both $-4.5\times10^5\ \mathrm{m^2}$ (d) no, the events are not causally connected
Examiner tip: A negative $t'$ is not an error to fix: it is the physics. Students who drop the sign lose the mark in (b) and cannot reach the conclusion in (d).

7HardHL only7 marks

The spacetime diagram shows the axes of frame S (x, ct) and the axes of frame S′ (x′, ct′). Frame S′ moves at $0.50c$ relative to S. Distances are in light-years (ly). Events A and B and the worldline of a particle P, which passes through the origin O, are shown.

(a) Calculate the angle between the ct′ axis and the ct axis. [1]

(b) Using the diagram, determine the speed of P in frame S. [1]

(d) Event B lies on the x′ axis. State what this tells you about events O and B in frame S′. [1]

(e) Events A and B are simultaneous in frame S. Determine, with a reason, which of A and B happens first in frame S′. [2]

$\theta = \tan^{-1}(0.50) = 27°$ (allow 26.6°)(1)
P passes through $x = 3$ ly, $ct = 4$ ly, so speed $= \dfrac{3}{4}c = 0.75c$; allow any correct pair of points read from the grid(1)
$u' = \dfrac{0.75c - 0.50c}{1 - 0.75\times0.50}$(1)
$u' = 0.40c$; allow ECF from (b)(1)
O and B are simultaneous in S′ / B happens at $ct' = 0$(1)
lines of simultaneity in S′ are parallel to the x′ axis; the line through A parallel to x′ meets the ct′ axis above the one through B (A lies on the ct′ axis at $ct' > 0$) OR $ct'_A = \gamma(ct - \tfrac{v}{c}x) = 1.15\times(2 - 0.5) = 1.7$ ly and $ct'_B = 0$(1)
B happens first (before A) in S′(1)
Answer: (a) 27° (b) $0.75c$ (c) $0.40c$ (d) simultaneous in S′ (e) B first ($ct'_B = 0$, $ct'_A \approx 1.7$ ly)
Examiner tip: To compare times in S′, draw lines parallel to the x′ axis, not horizontal lines: horizontal lines are simultaneity in S.

8StandardHL only6 marks

Muons are produced at the top of the atmosphere. In an experiment, a detector on a mountain at height $2000\ \mathrm{m}$ above sea level records 560 muons per hour moving downwards at $0.994c$. An identical detector at sea level records 410 muons per hour with the same speed. The half-life of a muon in its rest frame is $1.52\ \mu\mathrm{s}$.

(a) Calculate the time taken for a muon to travel from the mountain-top detector to sea level, as measured in the Earth frame. [1]

(b) Using no relativistic effects, predict the count rate at sea level. [2]

(d) Explain what the results show about special relativity. [1]

$t = \dfrac{2000}{0.994\times3.00\times10^8} = 6.7\ \mu\mathrm{s}$(1)
number of half-lives $= \dfrac{6.71}{1.52} = 4.4$(1)
$560\times0.5^{4.4} \approx 26$ per hour (allow 25–28)(1)
$\gamma = 9.1$ so dilated half-life $= 9.14\times1.52 = 13.9\ \mu\mathrm{s}$, giving $0.48$ half-lives OR contracted height $= 2000/9.14 = 219\ \mathrm{m}$ in the muon frame(1)
$560\times0.5^{0.48} \approx 400$ per hour (allow 395–405)(1)
the measured 410 per hour agrees with the relativistic prediction and is far above the non-relativistic 26, so the data support time dilation / special relativity(1)
Answer: (a) $6.7\ \mu\mathrm{s}$ (b) about 26 per hour (c) about 400 per hour (d) agrees with relativity
Examiner tip: Use the half-life with $0.5^{t/T}$, not $t/T$ as a fraction lost: dividing 560 by 4.4 is a common wrong method.

9HardHL only6 marks

A railway carriage moves at a constant high speed $v$ past a platform. A lamp fixed at the exact centre of the carriage emits a single flash. The doors R and F at each end of the carriage open automatically when light from the flash reaches them. Observer M is at rest in the carriage and observer Q is at rest on the platform.

Discuss whether the two doors open at the same time, according to M and according to Q. Your answer should explain why both observers are correct. [6]

in M's frame the lamp is at the midpoint and both doors are at rest(1)
light travels at $c$ in both directions for M, over equal distances, so M finds the doors open simultaneously(1)
for Q, the light also travels at $c$ in both directions (second postulate), not $c \pm v$(1)
for Q, door R moves towards the point where the flash was emitted and door F moves away from it(1)
so the light travels a shorter distance to R than to F: Q finds that R (the rear door) opens first(1)
simultaneity is relative: events at different places that are simultaneous in one frame are not simultaneous in another frame moving relative to it(1)
both are correct since all inertial frames are equally valid (first postulate); there is no absolute time(1)
the opening of each door is a single event that both agree happens; only the time order / interval differs; the events cannot cause each other so there is no contradiction(1)

Level	Marks	What it looks like
Level 3	5–6	Clear, logical discussion of both frames. States M sees simultaneous opening and Q sees the rear door first, with a reason based on the constant speed of light and the motion of the doors in Q's frame. Links this to the relativity of simultaneity and why neither observer is wrong.
Level 2	3–4	Correct outcome for both observers with some reasoning, e.g. constant $c$ or the doors moving in Q's frame, but the explanation of why both are correct is missing or weak.
Level 1	1–2	Simple statements, e.g. that M sees the doors open together or that Q does not; little or no correct reasoning. Wrong door named for Q limits to this level.
No marks	0	Nothing worthy of credit.

Answer: M: doors open simultaneously. Q: rear door R opens first. Both correct, since simultaneity of separated events depends on the frame.
Examiner tip: Many answers say the light "catches up" because it moves at $c+v$ for Q: that is Galilean thinking and caps the answer, as Q also measures $c$ both ways.

10StretchHL only7 marks

A spacecraft leaves Earth and travels at constant speed $0.60c$ to a star that is $9.0\ \mathrm{ly}$ from Earth in the Earth frame. Ignore the time taken to accelerate. Treat the Earth and star as at rest relative to each other.

(a) Calculate the time for the journey in the Earth frame. [1]

(b) Using the invariance of the spacetime interval, determine the time for the journey measured by a clock on the spacecraft. [2]

(c) Calculate the distance between Earth and the star in the spacecraft frame and show that it is consistent with your answer to (b). [2]

(d) A second spacecraft makes the same journey and its clock records $6.0$ years. Determine its speed relative to Earth. [2]

$t = \dfrac{9.0}{0.60} = 15$ years(1)
$(c\tau)^2 = (ct)^2 - x^2 = 15^2 - 9.0^2$ (in ly); in the spacecraft frame both events are at the same place so $x' = 0$(1)
$\tau = 12$ years(1)
$\gamma = 1.25$ so $L = \dfrac{9.0}{1.25} = 7.2\ \mathrm{ly}$(1)
$\dfrac{7.2}{0.60} = 12$ years, which matches (b)(1)
$\left(\dfrac{9.0}{\beta}\right)^2 - 9.0^2 = 6.0^2$ (or $\gamma = \dfrac{t}{\tau}$ with $t = 9.0/\beta$)(1)
$\beta^2 = \dfrac{81}{117}$ so $v = 0.83c$(1)
Answer: (a) 15 years (b) 12 years (c) 7.2 ly; 7.2/0.60 = 12 years (d) $0.83c$
Examiner tip: In (d) the speed is unknown in both $t$ and $\gamma$: write $t = 9.0/\beta$ first and solve one equation; guessing $\gamma$ from 15/6 uses the wrong Earth time.

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