QCAA Physics Alternative Sequence Special relativity

15 sample questions with marking guides and sample answers

Q16
2023
QCAA
Paper 1
1 mark
Q16
1 mark

A train is travelling at relativistic speed and is about to move through a tunnel. An observer on the train measures the train and tunnel to each be 95 m long.

A second observer is stationary relative to the tunnel. They would observe the train to

A

simultaneously enter and exit the tunnel.

B

decrease its mass while in the tunnel.

C

move faster while in the tunnel.

D

be shorter than the tunnel.

Reveal Answer
A

simultaneously enter and exit the tunnel.

Because Observer 2 measures the train to be shorter than the tunnel, the train will fully enter the tunnel before its front exits, meaning these events are not simultaneous.

B

decrease its mass while in the tunnel.

The train's mass is an invariant property (rest mass) or depends only on its speed (relativistic mass), neither of which changes simply by entering the tunnel.

C

move faster while in the tunnel.

The train is traveling at a constant relativistic speed; entering the tunnel does not exert a force to cause it to accelerate.

D

be shorter than the tunnel.

Correct Answer

Observer 1 measures the train's proper length as 95 m and the tunnel's contracted length as 95 m. Therefore, Observer 2 measures the tunnel's proper length (>95> 95 m) and the train's contracted length (<95< 95 m), making the train shorter than the tunnel.

Q10
2023
VCAA
6 marks
Q10

A proton in an accelerator beamline of proper length 4.80 km4.80 \text{ km} has a Lorentz factor, γ\gamma, of 2.002.00.

Q10a
3 marks

Calculate the speed of the proton relative to the beamline in terms of cc, the speed of light in a vacuum. Give your answer to three significant figures.

Reveal Answer

γ=11v2c2\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}

... and after much transposition ...

v=c11γ2v = c \sqrt{1 - \frac{1}{\gamma^2}}

v=c1122v = c \sqrt{1 - \frac{1}{2^2}}

v=c×0.866=0.866cv = c \times 0.866 = 0.866c

Marking Criteria
DescriptorMarks

Identifies the correct formula or correctly transposes the Lorentz factor formula to solve for velocity, e.g., v=c11γ2v = c \sqrt{1 - \frac{1}{\gamma^2}}

1

Correctly substitutes the given values into the formula, e.g., v=c1122v = c \sqrt{1 - \frac{1}{2^2}}

1

Calculates the correct answer to three significant figures, 0.866c0.866c

1
Q10b
1 mark

Calculate the length of the beamline in the reference frame of the proton.

Reveal Answer

L=L0γL = \frac{L_0}{\gamma}

L=4.82.0L = \frac{4.8}{2.0}

L=2.4 kmL = 2.4 \text{ km}

Marking Criteria
DescriptorMarks

Calculates the correct length of the beamline in the reference frame of the proton, 2.4 km2.4 \text{ km}

1
Q10c
2 marks

Calculate the kinetic energy of the proton in joules. Show your working.

Mass of proton =1.67×1027 kg= 1.67 \times 10^{-27} \text{ kg}.

Reveal Answer

Ek=(γ1)mc2E_k = (\gamma - 1)mc^2

Ek=(21)×(1.6×1027)×(3.0×108)2E_k = (2 - 1) \times (1.6 \times 10^{-27}) \times (3.0 \times 10^8)^2

Ek=1.50×1010 JE_k = 1.50 \times 10^{-10} \text{ J}

Marking Criteria
DescriptorMarks

Demonstrates correct working using the relativistic kinetic energy formula, e.g., Ek=(γ1)mc2=(21)×(1.67×1027)×(3.0×108)2E_k = (\gamma - 1)mc^2 = (2 - 1) \times (1.67 \times 10^{-27}) \times (3.0 \times 10^8)^2

1

Calculates the correct kinetic energy, 1.50×1010 J1.50 \times 10^{-10} \text{ J}

1
Q11
2020
VCAA
4 marks
Q11

An astronaut has left Earth and is travelling on a spaceship at 0.800c (γ = 1.67) directly towards the star known as Sirius, which is located 8.61 light-years away from Earth, as measured by observers on Earth.

Q11a
2 marks

How long will the trip take according to a clock that the astronaut is carrying on his spaceship? Show your working.

Reveal Answer

The distance of 8.61 light-years is the proper length in Earth's frame of reference. The proper time in the Earth's frame of reference is:
t=dv=8.610.8t = \frac{d}{v} = \frac{8.61}{0.8}
t=10.76 yrt = 10.76 \text{ yr}
This time is the dilated time in the astronaut's frame of reference. The proper time as measured by the astronaut is:
t=t0γt = t_0 \gamma
10.76=t0×1.6710.76 = t_0 \times 1.67
t0=6.44 yearst_0 = 6.44 \text{ years}

Marking Criteria
DescriptorMarks

Calculates the time in Earth's frame of reference (10.76 yr10.76 \text{ yr}) OR calculates the contracted length in the astronaut's frame of reference (5.16 ly5.16 \text{ ly})

1

Calculates the correct proper time of 6.44 years6.44 \text{ years}

1
Q11b
2 marks

Is the trip time measured by the astronaut in part a. a proper time? Explain your reasoning.

Reveal Answer

The time measured by the astronaut will be proper time because the clock is stationary in the astronaut's frame of reference.

Marking Criteria
DescriptorMarks

Identifies that the time measured by the astronaut is proper time

1

Explains that this is because the clock is stationary in the astronaut's frame of reference

1
Q18
2022
VCAA
1 mark
Q18
1 mark

Which one of the following is an example of an inertial frame of reference?

A

a bus travelling at constant velocity

B

an express train that is accelerating

C

a car turning a corner at a constant speed

D

a roller-coaster speeding up while heading down a slope

Reveal Answer
A

a bus travelling at constant velocity

Correct Answer

This is correct because an inertial frame of reference is one that is not accelerating. A bus moving at a constant velocity has zero acceleration, making it an inertial frame where Newton's laws of motion hold true.

B

an express train that is accelerating

This is incorrect because an accelerating object represents a non-inertial frame of reference. By definition, an inertial frame must have zero acceleration.

C

a car turning a corner at a constant speed

This is incorrect because turning a corner involves a change in direction, which means the car is experiencing centripetal acceleration. Even at a constant speed, a change in direction makes it a non-inertial frame of reference.

D

a roller-coaster speeding up while heading down a slope

This is incorrect because the roller-coaster is speeding up, meaning its velocity is changing and it is accelerating. Any accelerating environment is a non-inertial frame of reference.

Q1
2023
QCAA
Paper 2
3 marks
Q1
3 marks

Describe the effects of relativistic travel on an object.

Reveal Answer

An object travelling near the speed of light will experience an increase in mass, resulting in increased momentum. It will also experience time slower compared to an observer in another, non-relativistic frame of reference.

Finally, an object moving at relativistic speeds would be observed to decrease in length in the direction of its travel.

Marking Criteria
DescriptorMarks

describes the effect of time dilation

1

describes the effect of length contraction

1

describes the effect of relativistic momentum

1
Q3
2021
QCAA
Paper 1
1 mark
Q3
1 mark

Identify the correct formula for the mass–energy equivalence relationship.

A

E=mc2E = mc^2

B

E=mghE = mgh

C

E=12mc2E = \frac{1}{2}mc^2

D

E=12mv2E = \frac{1}{2}mv^2

Reveal Answer
A

E=mc2E = mc^2

Correct Answer

Correct. This is Einstein's famous mass-energy equivalence formula, stating that energy (EE) equals mass (mm) times the speed of light squared (c2c^2).

B

E=mghE = mgh

Incorrect. This is the formula for gravitational potential energy, where gg is the acceleration due to gravity and hh is height.

C

E=12mc2E = \frac{1}{2}mc^2

Incorrect. This incorrectly combines the structure of the kinetic energy formula with the speed of light (cc), which does not represent any standard physical relationship.

D

E=12mv2E = \frac{1}{2}mv^2

Incorrect. This is the classical formula for kinetic energy, representing the energy of an object in motion with mass mm and velocity vv.

Q11
2022
SCSA
6 marks
Q11

Salman and Priyanka have identical 1.00 m rulers. Priyanka takes her ruler and sets off in a rocket. She travels past Salman at a speed of 0.800c0.800 c. Their metre rulers are aligned in the direction of Priyanka's travel. Each then measures the length of the other's ruler by carefully determining the position of each end of the ruler at the same instant, and measuring the distance between these positions.

Q11a
2 marks

How long does Salman measure Priyanka's ruler to be?

Reveal Answer

l=l010.8002=0.60 ml = l_0 \sqrt{1 - 0.800^2} = 0.60 \text{ m}
Marking Criteria
DescriptorMarks

Uses correct equation and places correct values in correct place

1

Calculates correct answer

1
Q11b
1 mark

How long does Priyanka measure Salman's ruler to be?

Reveal Answer

0.60 m

Marking Criteria
DescriptorMarks

0.60 m

1
Q11c
3 marks

When Priyanka returns, she and Salman compare the results of their measurements. How are they able to explain their seemingly contradictory results?

Reveal Answer

To successfully measure the length of the ruler moving relative to them, they determine the position of the ends of the ruler at the same time and measure the distance between these two positions. Each thought the other's measurements were not made simultaneously. Therefore they both measure the other's ruler as a different length to 1.00 m.

Marking Criteria
DescriptorMarks

Explains that to successfully measure the length of the ruler moving relative to them, they determine the position of the ends of the ruler at the same time and measure the distance between these two positions.

1

States that each thought the other's measurements were not made simultaneously.

1

Concludes that therefore they both measure the other's ruler as a different length to 1.00 m.

1
Q12
2020
VCAA
1 mark
Q12
1 mark

A high-energy proton is travelling through space at a constant velocity of 2.50 × 10^8 m s−1.

The Lorentz factor, γ, for this proton would be closest to

A

1.81

B

2.44

C

3.27

D

3.39

Reveal Answer
A

1.81

Correct Answer

The Lorentz factor is calculated using γ=11v2/c2\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}. Substituting v=2.50×108 m s1v = 2.50 \times 10^8 \text{ m s}^{-1} and c=3.00×108 m s1c = 3.00 \times 10^8 \text{ m s}^{-1} yields γ1.81\gamma \approx 1.81.

B

2.44

This result comes from incorrectly omitting the squares on the velocities, calculating 11v/c\frac{1}{\sqrt{1 - v/c}} instead of the proper Lorentz factor formula.

C

3.27

This is the value of γ2\gamma^2 (approximately 3.273.27), which occurs if you forget to take the square root of the denominator in the Lorentz factor formula.

D

3.39

This is an incorrect value resulting from a miscalculation. The correct Lorentz factor must be found using the formula γ=11v2/c2\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}.

Q1
2023
QCAA
Paper 1
1 mark
Q1
1 mark

An object is unable to accelerate to the speed of light because

A

length contraction will change the height of the object.

B

time dilation will decrease the velocity of the object.

C

the object will gain infinite momentum.

D

the inertia of the object will decrease.

Reveal Answer
A

length contraction will change the height of the object.

Length contraction only occurs parallel to the direction of motion, so the object's height would remain unchanged. Furthermore, length contraction is a consequence of relativistic speeds, not the limiting factor preventing an object from reaching the speed of light.

B

time dilation will decrease the velocity of the object.

Time dilation means time passes slower for the moving object relative to a stationary observer, but it does not act as a physical drag force that decreases the object's velocity.

C

the object will gain infinite momentum.

Correct Answer

As an object's velocity approaches the speed of light (cc), its relativistic factor (γ\gamma) approaches infinity. This causes its relativistic momentum (p=γmvp = \gamma mv) to become infinite, meaning it would require an infinite amount of energy to accelerate it to cc.

D

the inertia of the object will decrease.

As an object approaches the speed of light, its resistance to acceleration (effective inertia) actually increases toward infinity, making it harder to accelerate, rather than decreasing.

Q17
2023
QCAA
Paper 1
1 mark
Q17
1 mark

The half-life of an unstable subatomic particle is measured by a stationary detector to be longer when its velocity approaches the speed of light. This is because the particle

A

is moving relative to its frame of reference.

B

is in the same frame of reference as the detector.

C

experiences time differently relative to the detector.

D

cannot be accurately observed at relativistic speeds.

Reveal Answer
A

is moving relative to its frame of reference.

An object cannot move relative to its own frame of reference; by definition, it is always stationary within its own frame.

B

is in the same frame of reference as the detector.

The particle is moving at relativistic speeds relative to the stationary detector, meaning they are in completely different frames of reference.

C

experiences time differently relative to the detector.

Correct Answer

According to special relativity, time dilation causes time to pass more slowly for the moving particle relative to the stationary detector, resulting in a longer measured half-life.

D

cannot be accurately observed at relativistic speeds.

The longer half-life is a real, measurable physical consequence of time dilation, not an error or limitation in observation.

Q10
2021
QCAA
Paper 1
1 mark
Q10
1 mark

Proper length is the length measured in the frame of reference where the object is

A

at rest.

B

in motion.

C

accelerating.

D

in motion but not accelerating.

Reveal Answer
A

at rest.

Correct Answer

Proper length is defined in special relativity as the length of an object measured by an observer who is at rest relative to the object.

B

in motion.

If the object is in motion relative to the observer's frame of reference, the measured length will be shorter than the proper length due to length contraction.

C

accelerating.

Proper length is specifically defined in the object's rest frame. Measurements taken from an accelerating frame do not yield the proper length.

D

in motion but not accelerating.

Even if the object is moving at a constant velocity, an observer in a different frame will measure a contracted length rather than the proper length.

Q16
2025
VCAA
1 mark
Q16
1 mark

Protons with a Lorentz factor, γ\gamma, equal to 2.102.10 are injected into a 100 m100 \text{ m} long beamline in a particle physics laboratory.

In the reference frame of the protons, which one of the following is closest to the length of the beamline?

A

17.4 m17.4 \text{ m}

B

47.6 m47.6 \text{ m}

C

100 m100 \text{ m}

D

210 m210 \text{ m}

Reveal Answer
A

17.4 m17.4 \text{ m}

This value is incorrect and does not follow the proper length contraction formula. The length must be divided by the Lorentz factor, not calculated using an incorrect relativistic relation.

B

47.6 m47.6 \text{ m}

Correct Answer

In the protons' reference frame, the beamline is moving and undergoes length contraction. The contracted length is calculated as L=L0/γ=100/2.1047.6 mL = L_0 / \gamma = 100 / 2.10 \approx 47.6 \text{ m}.

C

100 m100 \text{ m}

This is the proper length of the beamline measured in the laboratory frame where it is at rest. The protons will observe a shorter, contracted length due to their relative motion.

D

210 m210 \text{ m}

This result comes from incorrectly multiplying the proper length by the Lorentz factor (L=γL0L = \gamma L_0), which would represent length expansion rather than length contraction.

Q26
2024
NESA
3 marks
Q26
3 marks

Muons are unstable particles produced when cosmic rays strike atoms high in the atmosphere. The muons travel downward, perpendicular to Earth's surface, at almost the speed of light.

Classical physics predicts that these muons will decay before they have time to reach Earth's surface.

Explain qualitatively why these muons can reach Earth's surface, regardless of whether their motion is considered from either the muon's frame of reference or the Earth's frame of reference.

Reveal Answer

In the muon's reference frame, the distance travelled is less than that observed by a stationery observer on Earth's surface, due the effects of length contraction. This shortened distance means that they will get further than would be expected by an observer observing the rest length.

From the Earth's frame of reference, the time dilation means that the half-life of the muon is dilated compared to the half-life measured in the rest frame of the muon. This greater time allows more muons to reach the ground than would otherwise be expected.

Marking Criteria
DescriptorMarks

Explains why muons reach the Earth's surface with reference to relativistic effects in both frames of reference

3
  • Outlines some relativistic effects that apply to the muon
    OR
  • Explains one relativistic effect that applies to the muon
2

Provides some relevant information

1

None of the above

0
Q25
2023
QCAA
Paper 1
3 marks
Q25
3 marks

An observer who is stationary relative to a moving spaceship measures the velocity of the spaceship to be 2.0×108 m s12.0 \times 10^8 \text{ m s}^{-1}.

Calculate the length of the spaceship if the observer records it as 18 m. Show your working.

Reveal Answer

L=L01v2c218=L01(2.0×108)2(3.0×108)2L0=24 m\begin{align*} L &= L_0 \sqrt{1 - \frac{v^2}{c^2}}\\ 18 &= L_0 \sqrt{1 - \frac{(2.0 \times 10^8)^2}{(3.0 \times 10^8)^2}}\\ L_0 &= 24 \text{ m} \end{align*}

Length = 24 m24 \text{ m} (to two significant figures)

Marking Criteria
DescriptorMarks

recognises the scenario relates to an object experiencing length contraction

1

correctly substitutes for relativistic length

1

calculates the length of the spaceship

1
Q20
2021
VCAA
1 mark
Q20
1 mark

One of Einstein's postulates for special relativity is that the laws of physics are the same in all inertial frames of reference.

Which one of the following best describes a property of an inertial frame of reference?

A

It is travelling at a constant speed.

B

It is travelling at a speed much slower than cc.

C

Its movement is consistent with the expansion of the universe.

D

No observer in the frame can detect any acceleration of the frame.

Reveal Answer
A

It is travelling at a constant speed.

Incorrect. Travelling at a constant speed is insufficient because the direction could be changing (such as in circular motion), which implies acceleration. An inertial frame must travel at a constant velocity.

B

It is travelling at a speed much slower than cc.

Incorrect. The definition of an inertial frame does not depend on its speed relative to the speed of light, cc. A frame moving at a constant velocity close to cc is still a valid inertial frame.

C

Its movement is consistent with the expansion of the universe.

Incorrect. The expansion of the universe is a cosmological phenomenon and does not define an inertial frame. Inertial frames are defined strictly by the absence of acceleration.

D

No observer in the frame can detect any acceleration of the frame.

Correct Answer

Correct. An inertial frame of reference is defined as one that is not accelerating. Therefore, an observer within this frame will not experience any fictitious forces and cannot detect any acceleration.

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