QCAA Physics Special relativity

15 sample questions with marking guides and sample answers

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
Q13
2023
NESA
1 mark
Q13
1 mark

Nucleus X has a greater binding energy than nucleus Y.

What can be deduced about X and Y?

A

X is more stable than Y.

B

Y is more stable than X.

C

X has a greater mass defect than Y.

D

Y has a greater mass defect than X.

Reveal Answer
A

X is more stable than Y.

Stability is determined by binding energy per nucleon, not total binding energy, so we cannot determine which nucleus is more stable without knowing their nucleon numbers.

B

Y is more stable than X.

Stability depends on binding energy per nucleon, which cannot be determined from total binding energy alone.

C

X has a greater mass defect than Y.

Correct Answer

Binding energy is directly proportional to mass defect according to the mass-energy equivalence equation E=Δmc2E = \Delta m c^2. Therefore, a greater binding energy means a greater mass defect.

D

Y has a greater mass defect than X.

Since nucleus X has a greater binding energy, it must have a greater mass defect than Y, not the other way around.

Q32
2025
NESA
8 marks
Q32
8 marks

Analyse the consequences of the theory of special relativity in relation to length, time and motion. Support your answer with reference to experimental evidence.

Reveal Answer

The two postulates of special relativity are: the speed of light in a vacuum is an absolute constant, and all inertial frames of reference are equivalent.

The consequences of these postulates are time dilation, length contraction and relativistic momentum dilation. Time dilation means that the measured time between events increases as the speed of an observer increases. Length contraction means that the length of an object decreases as the speed of an observer increases, and relativistic momentum dilation means that objects (with mass) cannot travel at c as it takes increasing amounts of energy to accelerate objects as they approach the speed of light.

Length contraction and time dilation can both be seen in observations of muons at Earth's surface. The short half-life of muons suggests that there should be very few of them that arrive at Earth's surface from their point of origin high in Earth's atmosphere, but the muon flux is greater than suggested by half-life and distance travelled alone. However, when the distance travelled is considered from the muon's frame of reference, it is length-contracted according to l=l0(1v2c2)l = l_0 \sqrt{\left(1 - \frac{v^2}{c^2}\right)}, giving an effective shorter distance for them to travel. From the perspective of an observer on Earth's surface, the time taken for them to decay is dilated, giving an effective greater amount of time for them to travel before decaying.

The effects of relativistic momentum can be seen in particle accelerators in terms of the increasing amounts of energy to required to accelerate the particles as they approach the speed of light – it is though the objects are much more massive than they are.

Marking Criteria
DescriptorMarks
  • Provides a comprehensive analysis of the consequences of special relativity on time, length and motion
  • Refers to applicable experimental evidence
8
  • Provides a thorough analysis of the consequences of special relativity on time, length and motion
  • Refers to applicable experimental evidence
7

The student response meets all criteria of the 5-mark band, and additionally meets the majority of criteria in the 7-mark band.

6
  • Provides analysis of the consequences of special relativity on time and/or length and/or motion
  • Refers to some relevant experimental evidence
5

The student response meets all criteria of the 3-mark band, and additionally meets the majority of criteria in the 5-mark band.

4

Makes some reference to special relativity and/or related experimental evidence

3

The student response meets all criteria of the 1-mark band, and additionally meets the majority of criteria in the 3-mark band.

2

Provides some relevant information

1

None of the above

0
Q19
2021
QCAA
Paper 1
1 mark
Q19
1 mark

A spaceship with a velocity of 9.0×1079.0 \times 10^7 m s1^{-1} is measured to be 125 m in length by an observer at rest.
Calculate the length of the spaceship as measured by somebody on board the spaceship.

A

119 m

B

131 m

C

137 m

D

178 m

Reveal Answer
A

119 m

This calculation incorrectly treats the given 125 m as the proper length (L0L_0) and solves for the contracted length. Since the observer on the spaceship is at rest relative to it, they measure the proper length, which must be longer than the contracted length measured by the outside observer.

B

131 m

Correct Answer

The observer on board measures the proper length (L0L_0). Using the length contraction formula L=L01v2c2L = L_0 \sqrt{1 - \frac{v^2}{c^2}}, with L=125L=125 m and v=0.3cv=0.3c, we rearrange to find L0=12510.09131L_0 = \frac{125}{\sqrt{1 - 0.09}} \approx 131 m.

C

137 m

This answer results from omitting the square root in the Lorentz factor calculation, dividing 125 by (10.32)(1 - 0.3^2) instead of 10.32\sqrt{1 - 0.3^2}.

D

178 m

This value is incorrect and does not align with the standard length contraction formula. It likely results from a calculation error or misapplication of the Lorentz factor γ\gamma.

Q15
2024
VCAA
1 mark
Q15
1 mark

At the Australian Synchrotron, electrons are accelerated to close to the speed of light, cc, by the linear accelerator.

Consider electrons that enter the Australian Synchrotron booster ring with a Lorentz factor γ=200\gamma = 200, and as a result of further acceleration in that ring reach a much higher value, with a Lorentz factor γ=2000\gamma = 2000.

Which one of the following best describes the effect of the booster ring on each electron's speed and energy?

A

The electron's speed increases by a large amount and its energy increases by a large amount.

B

The electron's speed increases by a small amount and its energy increases by a large amount.

C

The electron's speed increases by a large amount and its energy increases by a small amount.

D

The electron's speed increases by a small amount and its energy increases by a small amount.

Reveal Answer
A

The electron's speed increases by a large amount and its energy increases by a large amount.

Incorrect. Since the electron is already traveling close to the speed of light when γ=200\gamma = 200, its speed can only increase by a very small fraction to approach cc.

B

The electron's speed increases by a small amount and its energy increases by a large amount.

Correct Answer

Correct. The electron's speed is already very close to cc, so it only increases by a small amount, but its total energy, given by E=γmc2E = \gamma mc^2, increases by a factor of 10, which is a large amount.

C

The electron's speed increases by a large amount and its energy increases by a small amount.

Incorrect. The speed only increases by a small amount because it cannot exceed the speed of light, while the energy increases by a large amount.

D

The electron's speed increases by a small amount and its energy increases by a small amount.

Incorrect. The energy increases by a large amount because it is directly proportional to the Lorentz factor (E=γmc2E = \gamma mc^2), which increases tenfold from 200 to 2000.

Q2
2024
QCAA
Paper 1
1 mark
Q2
1 mark

Which property of light is described by the postulates of special relativity?

A

The energy of light is greater when the frequency of the photons decreases.

B

The wavelength of light decreases as the velocity of the source increases.

C

The velocity of light remains constant in all inertial frames of reference.

D

The frequency of light changes depending on media.

Reveal Answer
A

The energy of light is greater when the frequency of the photons decreases.

This statement is factually incorrect and unrelated to special relativity. According to the quantum relation E=hfE=hf, energy is directly proportional to frequency, so energy decreases as frequency decreases.

B

The wavelength of light decreases as the velocity of the source increases.

This describes the Doppler effect rather than a fundamental postulate. While relativity influences the Doppler shift, the postulates specifically define the behavior of the speed of light, not the change in wavelength due to source motion.

C

The velocity of light remains constant in all inertial frames of reference.

Correct Answer

This is the second postulate of special relativity. It states that the speed of light in a vacuum (cc) is constant and independent of the motion of the light source or the observer in all inertial frames.

D

The frequency of light changes depending on media.

This is incorrect regarding wave mechanics; frequency remains constant when light changes media, while speed and wavelength change. Furthermore, this is a concept of optics, not a postulate of special relativity.

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
Q24
2020
QCAA
Paper 1
4 marks
Q24
4 marks

A spaceship travelled from Planet A to Planet B at a speed of 0.90c0.90c. An observer that was stationary relative to both planets measured the time taken for the trip to be 4.0 years.
Calculate the time taken for the trip as measured by an observer on the spaceship. (years to 1 decimal place)

Reveal Answer

t=t01v2c2t = \frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}

4.0=t01(0.90c)2c24.0 = \frac{t_0}{\sqrt{1-\frac{(0.90c)^2}{c^2}}}

t0=1.7t_0 = 1.7 years

Time = 1.7 years (to 1 decimal place)

Marking Criteria
DescriptorMarks

Indicates an understanding of the physical scenario in relation to time dilation (or other relevant physical concept/s).

1

Indicates an understanding that the time provided in the question represents relativistic (dilated) time. (If proper time and relativistic time are confused, a maximum of 2 marks can be awarded overall.)

1

Provides pertinent mathematical operation/s correctly performed.

1

Determines the time correctly (accept 1.7 years to 1.8 years inclusive).

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 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 occurs along the direction of motion, not necessarily the height, and it is a consequence of relativistic speed rather than a barrier preventing acceleration.

B

time dilation will decrease the velocity of the object.

Time dilation describes the relative slowing of time in a moving frame; it does not actively decrease velocity or physically prevent an object from accelerating.

C

the object will gain infinite momentum.

Correct Answer

As an object with mass approaches the speed of light, the Lorentz factor approaches infinity, meaning its relativistic momentum (p=γmvp = \gamma mv) and the energy required to accelerate it further become infinite.

D

the inertia of the object will decrease.

As an object approaches the speed of light, its effective inertia (resistance to acceleration) increases towards infinity, making it harder to accelerate, not easier.

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

This is Albert Einstein's famous equation for mass–energy equivalence, stating that energy (EE) equals mass (mm) multiplied by the speed of light squared (c2c^2).

B

E=mghE = mgh

This formula represents gravitational potential energy, where gg is the acceleration due to gravity and hh is the height above a reference point.

C

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

This is an incorrect variation that conflates the kinetic energy coefficient (12\frac{1}{2}) with the speed of light (cc).

D

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

This is the formula for classical kinetic energy, describing the energy of an object in motion with velocity vv, rather than the energy equivalent of mass itself.

Q13
2020
VCAA
1 mark
Q13
1 mark

Matter is converted to energy by nuclear fusion in stars.

If the star Alpha Centauri converts mass to energy at the rate of 6.6 × 10^9 kg s−1, then the power generated is closest to

A

2.0 × 10^18 W

B

2.0 × 10^18 J

C

6.0 × 10^26 W

D

6.0 × 10^26 J

Reveal Answer
A

2.0 × 10^18 W

Incorrect. This value incorrectly calculates power by multiplying the mass rate by the speed of light (cc) instead of c2c^2.

B

2.0 × 10^18 J

Incorrect. This option uses the incorrect formula (mcmc instead of mc2mc^2) and the wrong unit for power, which should be Watts, not Joules.

C

6.0 × 10^26 W

Correct Answer

Correct. Power is the rate of energy generation, calculated using P=dmdtc2P = \frac{dm}{dt}c^2. Multiplying 6.6×109 kg/s6.6 \times 10^9 \text{ kg/s} by (3.0×108 m/s)2(3.0 \times 10^8 \text{ m/s})^2 gives approximately 6.0×1026 W6.0 \times 10^{26} \text{ W}.

D

6.0 × 10^26 J

Incorrect. Although the numerical calculation is correct, the unit for power is Watts (Joules per second), not Joules.

Q6
2020
QCAA
Paper 1
1 mark
Q6
1 mark

The definition of relativistic momentum is the

A

momentum of an object when measured in a Newtonian frame of reference.

B

momentum of an object when measured regardless of its frame of reference.

C

momentum of an object when measured in the frame of reference in which the object is in motion.

D

momentum of an object when measured in the frame of reference in which the object is stationary.

Reveal Answer
A

momentum of an object when measured in a Newtonian frame of reference.

This is incorrect because relativistic momentum is a concept in special relativity, not Newtonian mechanics. In a Newtonian framework, momentum is simply defined as p=mvp=mv without the Lorentz factor.

B

momentum of an object when measured regardless of its frame of reference.

Momentum is a frame-dependent quantity, meaning its value changes depending on the observer's velocity relative to the object. It cannot be defined regardless of the frame.

C

momentum of an object when measured in the frame of reference in which the object is in motion.

Correct Answer

Relativistic momentum, defined as p=γmvp = \gamma mv, describes the momentum of an object moving with velocity vv relative to an observer. It is measured in the frame where the object has non-zero velocity.

D

momentum of an object when measured in the frame of reference in which the object is stationary.

In the frame where the object is stationary (its rest frame), the velocity is zero, and therefore the momentum is zero. This does not define the general concept of relativistic momentum.

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
Q17
2024
QCAA
Paper 1
1 mark
Q17
1 mark

A person on Earth experiences a time period of 15 years.

Approximately how much time will have passed for a passenger on a spaceship travelling at 0.7 c relative to Earth during that time?

A

8 years

B

11 years

C

21 years

D

27 years

Reveal Answer
A

8 years

This value is lower than the calculated proper time derived from the time dilation formula.

B

11 years

Correct Answer

Using the time dilation formula Δt0=Δt1v2/c2\Delta t_0 = \Delta t \sqrt{1 - v^2/c^2}, the passenger's time is 1510.7215×0.71410.715\sqrt{1 - 0.7^2} \approx 15 \times 0.714 \approx 10.7 years, which rounds to 11 years.

C

21 years

This result incorrectly divides by the Lorentz factor (15/10.722115 / \sqrt{1-0.7^2} \approx 21), implying time passes faster on the spaceship, whereas special relativity states moving clocks run slower.

D

27 years

This value suggests much more time passes on the spaceship than on Earth, contradicting the principle of time dilation where the moving observer experiences less time.

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