Year 12 Physics Module 7 ⏱ ~45 min 5 MC · 3 Short Answer Lesson 11 of 14

Relativistic Momentum and Energy

On 16 July 1945, the Trinity test at Los Alamos released $8.8 \times 10^{13}$ J from a 21-kiloton nuclear device. Applying $E = mc^2$, the total mass converted was $m = E/c^2 = 8.8 \times 10^{13} / (3 \times 10^8)^2 \approx 0.98\,\text{g}$, the total rest mass of the reacting system fell by less than one gram, and that decrease in mass-energy appeared as the energy of the first nuclear explosion. (Trinity was an implosion device with a plutonium-239 core.) Mass and energy are the same thing measured in different units.

Today's hook: On 16 July 1945, the Trinity nuclear test at Los Alamos produced a yield of 21 kilotons, $8.8 \times 10^{13}$ J of energy. Using $E = mc^2$, calculate the mass that was converted to energy. The answer is less than 1 gram of matter. How does losing less than a gram produce an explosion visible for hundreds of kilometres?
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Orient and predict

Predict what happens to energy as a particle approaches $c$, then set your goals and vocabulary for mass-energy equivalence.

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application & data response Master Mastery tasks Exam Exam-style questions
Before you read, predict

A particle accelerator pushes electrons closer and closer to the speed of light.

  1. What happens to the electron's kinetic energy as its speed approaches $c$? Does it level off or keep growing?
  2. Why can no particle with mass ever reach exactly $c$?
  3. The Sun loses about 4 million tonnes of mass every second. Where does this mass go?

Write your predictions before reading on, you will revisit them at the end.

Warm-up, which statement about relativistic energy is correct?

Beyond the syllabus. Relativistic momentum $p = \gamma m_0 v$, why mass cannot be accelerated to $c$, and $E = mc^2$ energy-release calculations are core. The full energy formalism — $E = \gamma mc^2$, $K = (\gamma - 1)mc^2$, $E^2 = (pc)^2 + (mc^2)^2$ and collision/decay kinematics — is extension for this syllabus: recognise the equations, but the exam assesses the core three.
Learning Intentions
goals

Know, Relativistic Energy

  • $E = \gamma mc^2$ (total energy)
  • $E_k = (\gamma - 1)mc^2$
  • $E^2 = (pc)^2 + (mc^2)^2$

Understand, Mass-Energy Equivalence

  • Rest energy $E_0 = mc^2$
  • Mass is a form of energy
  • Nuclear reactions and particle annihilation

Can Do, Solve Relativistic Dynamics

  • Calculate $\gamma$, total energy and KE from velocity
  • Find velocity from kinetic energy
  • Analyse particle collisions and decays
Scan these before reading
vocab
Rest energy ($E_0$)The energy equivalent of an object's mass at rest: $E_0 = mc^2$. Even a stationary particle carries this energy.
Relativistic kinetic energy$E_k = (\gamma - 1)mc^2$; the energy due to motion, which grows without bound as $v \to c$.
Mass-energy equivalenceThe principle that mass can be converted to energy and vice versa, quantified by $E = mc^2$.
Relativistic momentum$p = \gamma mv$; the correct expression for momentum at any speed, replacing the classical $p = mv$.
Energy-momentum relation$E^2 = (pc)^2 + (mc^2)^2$; works for all particles, including massless photons where $E = pc$.
Cross-lesson links: L13 showed length contraction. L14 reaches the most famous consequence of special relativity: $E = mc^2$. The Trinity nuclear test on 16 July 1945 at Los Alamos, 21 kilotons ($8.8 \times 10^{13}$ J) from less than 1 g of converted mass, is the most visceral demonstration of mass-energy equivalence. L15 explores the conceptual consequences (simultaneity, paradoxes); L20 connects back to the photon's energy-momentum relation $E = pc$ when discussing wave-particle duality.

Core Content

2

Set up relativistic energy and momentum

Total energy, rest energy, kinetic energy and $p = \gamma mv$, and why $c$ is out of reach for anything with mass.

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Relativistic Energy and Momentum
+5 XP

When classical mechanics breaks down

Particle physicists at CERN accelerate protons to $0.9999999896c$, just $10^{-7}$ m/s short of $c$. Each proton has a rest mass energy of $938\,\text{MeV}$, but at that speed its total energy is $6.5\,\text{TeV}$, about 7,000 times larger than its rest energy. Doubling the energy barely changes the speed. Classical kinetic energy $E_k = \frac{1}{2}mv^2$ predicts a finite maximum, but experiments show that accelerating particles to higher speeds requires ever-increasing energy without limit.

Einstein showed that the correct relativistic expressions are:

Relativistic energy and momentum

$$E = \gamma mc^2 \quad \text{(total energy)}$$

$$E_k = (\gamma - 1)mc^2 \quad \text{(kinetic energy)}$$

$$p = \gamma mv \quad \text{(relativistic momentum)}$$

These equations have profound properties:

  • At rest ($v = 0$, $\gamma = 1$): $E = mc^2$. Even a stationary particle has energy, its rest energy.
  • As $v \to c$: $\gamma \to \infty$, so $E_k \to \infty$. Infinite energy would be required to reach $c$, therefore, massive objects cannot reach the speed of light.
  • For massless particles ($m = 0$): $E = pc$ and $v = c$ exactly. Photons are massless and always travel at $c$.
v/c Ek/(mc²) Classical: ½v²/c² Relativistic: γ − 1 v = c 0 → ∞

Figure 1, Kinetic energy vs velocity: the classical expression increases as $\tfrac12mv^2$ and would reach the finite value $\tfrac12mc^2$ if incorrectly extrapolated to $c$; relativistic kinetic energy diverges as $v \to c$. No massive object can reach the speed of light.

Stop & Check

An electron is accelerated through 1 MV. Calculate its kinetic energy in joules and in MeV. Calculate its Lorentz factor and hence its speed. ($m_e = 9.11\times10^{-31}$ kg)

Three energy equations: rest energy $E_0 = mc^2$; total energy $E = \gamma mc^2$; kinetic energy $E_k = (\gamma-1)mc^2$. Relativistic momentum: $p = \gamma mv$. As $v \to c$, $E_k \to \infty$, massive objects cannot reach $c$. Massless photons ($m = 0$): $E = pc$ and naturally travel at $c$.

Write all three energy equations and note the difference between rest energy and kinetic energy.

A particle has $\gamma = 3$. Its kinetic energy equals:

3

Read mass-energy equivalence in the real world

Fission, fusion, annihilation and pair creation, plus the energy-momentum relation that also covers photons.

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Mass-Energy Equivalence
+5 XP

The world's most famous equation

We just saw that relativistic energy is $E = \gamma mc^2$ and that even a stationary particle has rest energy $mc^2$. That raises a question: what does mass-energy equivalence actually mean physically, and where does it show up in the real world? This card answers it → nuclear fission, fusion, annihilation and the energy-momentum relation.

$E = mc^2$ tells us that mass is a form of energy. The conversion factor is $c^2$, an enormous number: 1 kg of mass equals $9\times10^{16}$ J of energy, roughly the output of a large power station running for several years.

Nuclear fission: A heavy nucleus (like uranium-235) splits into smaller fragments. The total mass of the fragments is slightly less than the original nucleus. This "missing mass" (mass defect) is converted to kinetic energy of the fragments, about 200 MeV per fission.

Nuclear fusion: Light nuclei combine to form a heavier nucleus with less total mass. The Sun converts about 4 million tonnes of mass to energy every second through fusion. About 0.7% of the mass of hydrogen fused is converted to energy, carried away as photons and neutrinos.

Particle-antiparticle annihilation: When an electron meets a positron, they annihilate, converting their entire rest mass into two gamma-ray photons: $e^- + e^+ \to \gamma + \gamma$. Each photon carries energy $mc^2 = 511$ keV.

Particle creation: The reverse process also occurs. High-energy photons or collisions can create particle-antiparticle pairs, converting energy into mass.

Relativistic Energy Equations, Summary

$E = \gamma mc^2$  , Total relativistic energy

$E_0 = mc^2$  , Rest energy

$E_k = (\gamma - 1)mc^2$  , Relativistic kinetic energy

$E^2 = (pc)^2 + (mc^2)^2$  , Energy-momentum relation

$p = \gamma mv$  , Relativistic momentum

Nuclear Fission U-235 → Mass defect → ~200 MeV E = Δmc² Nuclear Fusion H H → He 0.7% mass → energy Powers the Sun 4×10⁶ t/s Annihilation e⁻ e⁺ → γ γ Each γ: 511 keV E = mec²

Figure 2, Three applications of mass-energy equivalence: nuclear fission (~200 MeV per event), nuclear fusion (powers the Sun), and electron-positron annihilation (each photon carries 511 keV)

Stop & Check

A proton has total energy 5.0 GeV. Its rest mass is 938 MeV/$c^2$. Calculate its kinetic energy, momentum, and speed.

Applications of $E = \Delta mc^2$: nuclear fission releases ~200 MeV per U-235 event (mass defect); nuclear fusion powers the Sun ($4\times10^6$ t/s converted); $e^- + e^+ \to \gamma + \gamma$ produces two 511 keV photons. Energy-momentum relation: $E^2 = (pc)^2 + (mc^2)^2$; for photons ($m=0$): $E = pc = hf$.

Record the three real-world applications and the energy-momentum relation.

In a nuclear reaction, the total mass of the products is slightly less than the reactants. The "missing" mass:

4

Compare burning petrol with solar fusion

Same equation, wildly different scale: work out the combustion mass defect and put it beside fusion.

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Worked Comparison, Fuel Combustion and $E = \Delta mc^2$
+5 XP

NESA syllabus (IQ4): mass converted to energy in "combustion of conventional fuel", alongside solar fusion and annihilation (ACSPH134)

We just saw that $E = \Delta mc^2$ applies to nuclear fission, fusion and annihilation. That raises a question: the syllabus explicitly lists a third example, "combustion of conventional fuel", so does burning petrol in a car engine also convert mass to energy? This card answers it → yes, in principle, but by a staggeringly smaller amount than any nuclear process.

Mass-energy equivalence is universal: every process that releases energy also loses a corresponding tiny amount of mass, whether that process is nuclear or chemical. Burning petrol releases energy through chemical bond rearrangement, not nuclear reactions, but $E = \Delta mc^2$ still applies, it must, since energy is being released. The question is only how large the mass defect turns out to be.

Problem

Burning 1.0 kg of petrol (octane) releases approximately 44 MJ of chemical energy. Calculate the mass defect $\Delta m$ associated with this energy release, and compare the fraction of mass converted to energy with nuclear fusion in the Sun (which converts about 0.7% of the fusing hydrogen's mass to energy).

Step 1, Mass defect from combustion

$\Delta m = \dfrac{E}{c^2} = \dfrac{44\times10^6\ \text{J}}{(3.00\times10^8)^2\ \text{m}^2/\text{s}^2} = \dfrac{4.4\times10^7}{9.0\times10^{16}} \approx \mathbf{4.9\times10^{-10}\ \text{kg}} \approx 0.49\ \mu\text{g}$

Burning an entire kilogram of petrol converts less than half a microgram of mass into energy, an amount no laboratory balance could ever detect directly.

Step 2, Fractional mass loss, combustion vs fusion

Fractional mass converted for combustion: $\dfrac{\Delta m}{m} = \dfrac{4.9\times10^{-10}\ \text{kg}}{1.0\ \text{kg}} \approx 4.9\times10^{-10}$ (about $0.00000005\%$).

For solar fusion, about 0.7% ($7\times10^{-3}$) of the fusing hydrogen's mass converts to energy. The ratio is $\dfrac{7\times10^{-3}}{4.9\times10^{-10}} \approx 1.4\times10^7$, nuclear fusion converts roughly 14 million times the fraction of mass to energy that chemical combustion does, for the same starting mass of fuel.

The pedagogical point

Chemical reactions, including everyday combustion, do technically obey $E = \Delta mc^2$ exactly like nuclear reactions do, mass and energy are equivalent in every process without exception. The difference is purely one of scale: chemical bond energies (electron volts) are minuscule compared with nuclear binding energies (millions of electron volts), so chemical mass defects are imperceptibly small (sub-microgram, per kilogram of fuel) next to the readily measurable mass defects of nuclear fission, fusion and annihilation.

Fraction of mass converted to energy (log scale, not to true linear scale) Combustion (petrol) ~4.9 × 10⁻¹⁰ (sub-microgram per kg) Nuclear fission (U-235) ~0.09% of mass Nuclear fusion (Sun) ~0.7% of mass Annihilation (e⁻e⁺) 100% of mass Combustion's bar is drawn at minimum visible width, it is roughly 10⁷–10⁹ times smaller than the nuclear bars

Figure 3, All four processes obey $E = \Delta mc^2$, but the fraction of mass converted to energy spans many orders of magnitude: combustion converts a sub-microgram per kilogram of fuel, while nuclear fusion converts about 0.7% and annihilation converts 100% of the interacting mass

Stop & Check

Natural gas (methane) releases about 55 MJ/kg when burned, slightly more energy per kilogram than petrol. Calculate the mass defect for burning 1.0 kg of methane. Is it still true that this mass loss is undetectably small compared with nuclear fission or fusion of the same starting mass? Justify your answer with a comparison.

Combustion of conventional fuel obeys $E = \Delta mc^2$ exactly like nuclear reactions, but the mass defect is vanishingly small: burning 1 kg of petrol (44 MJ) converts only about $4.9\times10^{-10}$ kg (sub-microgram) of mass to energy, roughly 14 million times smaller a fraction than solar fusion (0.7%). Chemical and nuclear energy release are the same physical phenomenon, mass-energy conversion, differing only in scale because chemical bond energies (eV) are far smaller than nuclear binding energies (MeV).

Record the combustion mass-defect calculation and the "same physics, different scale" comparison in your book.

Burning 1.0 kg of petrol releases 44 MJ of energy. Using $E = \Delta mc^2$, the mass converted to energy is closest to:

5

Work a particle-accelerator problem end to end

Voltage to kinetic energy to $\gamma$ to speed to momentum, then check it against the classical prediction.

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Worked Example, Particle Accelerator Energy
+5 XP

From voltage to velocity

We just saw the real-world manifestations of mass-energy equivalence, fission, fusion and annihilation. That raises a question: if a proton is accelerated through a large voltage, how do you calculate its $\gamma$, speed and momentum in sequence? This card answers it → a five-step worked example at 500 MV.

Problem

A proton is accelerated from rest through a potential difference of 500 MV in a particle accelerator. ($m_p = 1.67\times10^{-27}$ kg, $e = 1.60\times10^{-19}$ C)

  1. Calculate the proton's kinetic energy in joules and in MeV.
  2. Calculate the proton's Lorentz factor $\gamma$.
  3. Calculate the proton's speed as a fraction of $c$.
  4. Calculate the proton's momentum.
  5. Compare the relativistic kinetic energy to the classical prediction.
Step 1, Kinetic energy

$E_k = qV = (1.60\times10^{-19})(500\times10^6) = 8.00\times10^{-11}$ J

$E_k = 500$ MeV (since 1 eV per volt per elementary charge)

Step 2, Lorentz factor

$E_k = (\gamma - 1)mc^2$

$mc^2 = (1.67\times10^{-27})(3.00\times10^8)^2 = 1.50\times10^{-10}$ J = 938 MeV

$\gamma - 1 = 500/938 = 0.533$

$\gamma = $ 1.533

Step 3, Speed

$\gamma = 1/\sqrt{1-v^2/c^2}$

$v/c = \sqrt{1 - 1/\gamma^2} = \sqrt{1 - 1/1.533^2} = \sqrt{1 - 0.426} = \sqrt{0.574} = $ 0.758

Step 4, Momentum

$p = \gamma mv = 1.533 \times (1.67\times10^{-27}) \times (0.758 \times 3.00\times10^8)$

$p = 5.82\times10^{-19}$ kg·m/s

Alternatively, using $E^2 = (pc)^2 + (mc^2)^2$:

$E = E_k + mc^2 = 500 + 938 = 1438$ MeV

$pc = \sqrt{E^2 - (mc^2)^2} = \sqrt{1438^2 - 938^2} = 1090$ MeV/$c$

Step 5, Classical comparison

Classical: $E_k = \frac{1}{2}mv^2 = \frac{1}{2}(1.67\times10^{-27})(0.758c)^2 = 4.31\times10^{-11}$ J = 269 MeV

Relativistic: 500 MeV. Classical underestimates by 46% at this speed.

Stop & Check

An electron is accelerated to $\gamma = 50$. Calculate its kinetic energy in MeV, its speed as a fraction of $c$, and its momentum in MeV/$c$. ($m_e c^2 = 0.511$ MeV)

Worked strategy (proton at 500 MV): step 1, $E_k = qV = 500$ MeV. Step 2, $\gamma = 1 + E_k/mc^2 = 1.533$. Step 3, $v/c = \sqrt{1-1/\gamma^2} = 0.758$. Step 4, $pc = \sqrt{E^2 - (mc^2)^2} = 1090$ MeV/$c$. Classical would give 269 MeV: underestimate by 46%.

Write the four-step energy-to-speed chain and note when classical fails.

As $v \to c$, the kinetic energy of a massive particle:

6

Clear the three misconceptions

What $E = mc^2$ is not, when classical kinetic energy stops working, and why "relativistic mass" is deprecated.

5
Common Misconceptions and HSC Strategies
+5 XP

Avoiding the traps that cost marks

We just saw the step-by-step strategy for particle accelerator energy problems. That raises a question: what conceptual confusions most often trip students up on relativistic energy in the HSC? This card answers it → three misconceptions and a five-step exam strategy.

Misconception 1: $E = mc^2$ gives kinetic energy.
Wrong. $E = mc^2$ is the rest energy, the energy of a stationary particle. Total energy is $E = \gamma mc^2$. Kinetic energy is $E_k = (\gamma - 1)mc^2 = E_{total} - mc^2$. A common exam trap is to confuse these three quantities.

Misconception 2: Classical $E_k = \frac{1}{2}mv^2$ is a good approximation at any speed.
Wrong. When $\gamma > 1.1$ (roughly $v > 0.4c$), the classical formula is already off by more than 5%. At $v = 0.758c$ (the worked example), classical underestimates by 46%. Always check whether the problem is relativistic before applying classical formulas.

Misconception 3: Mass increases with speed.
The older "relativistic mass" concept ($m_{rel} = \gamma m_0$) is deprecated. It is clearer to use invariant mass $m$ and say that the particle's energy and momentum increase with speed; the relation between force and acceleration becomes direction-dependent, while $m$ remains constant.

v E c (barrier) E = γmc² Infinite energy required to reach exactly c Photon (m = 0): E = pc naturally travels at c

Figure 3, Why $c$ is a cosmic speed limit: total energy $E = \gamma mc^2$ diverges to infinity as $v \to c$ for any massive particle. Massless photons ($m = 0$) naturally travel at $c$ with $E = pc$.

HSC Exam Strategy, Energy Problems

Step-by-step for relativistic energy problems:

  1. Identify what is given: voltage → $E_k = qV$; $\gamma$ → $E_k = (\gamma-1)mc^2$; speed → calculate $\gamma$ first
  2. Calculate $\gamma = 1/\sqrt{1-v^2/c^2}$ if speed is given
  3. Choose the right formula: total energy ($E = \gamma mc^2$) vs kinetic energy ($E_k = (\gamma-1)mc^2$)
  4. Use $E^2 = (pc)^2 + (mc^2)^2$ for momentum without knowing speed
  5. Check: $E_{total} = E_k + mc^2$? If $E_{total} < mc^2$, something is wrong
Stop & Check

An electron ($m_ec^2 = 0.511$ MeV) has total energy 1.022 MeV. (a) Calculate its kinetic energy. (b) Find its Lorentz factor $\gamma$. (c) Calculate its speed as a fraction of $c$.

Three traps: (1) $E = mc^2$ is rest energy, NOT kinetic, kinetic energy is $E_k = (\gamma-1)mc^2 = E_{total} - mc^2$. (2) Classical $\frac{1}{2}mv^2$ fails when $\gamma > 1.1$ ($v > 0.4c$). (3) "Relativistic mass" is deprecated, rest mass $m_0$ is constant; it is inertia and energy that increase. HSC strategy: total energy check: $E_{total} = E_k + mc^2$.

Write the three misconception corrections and the total-energy identity.

7

Apply and consolidate

Run your own relativistic energy calculations, then connect $E = mc^2$ to real nuclear reactions.

Activities

Activity 1, Relativistic Energy Calculations
ApplyBand 5

Apply $E = \gamma mc^2$ and $E_k = (\gamma-1)mc^2$ across a range of scenarios

  1. Set $v/c = 0.6$. Calculate $\gamma$, then find $E_{total}/mc^2$ and $E_k/mc^2$. Verify: $E_{total} = E_k + mc^2$?
  2. Find the speed (as a fraction of $c$) at which kinetic energy equals rest energy ($E_k = mc^2$). What is $\gamma$ at this speed?
  3. An electron in a hypothetical high-energy electron accelerator has $E_{total} = 7.0$ TeV ($m_e c^2 = 511$ keV). Calculate $\gamma$ and express the electron's speed as $(1 - \epsilon)c$ where $\epsilon$ is very small. The LHC itself accelerates hadrons, not electrons.
  4. The Sun emits $3.8\times10^{26}$ W. Calculate the mass lost per second. If the Sun's mass is $2.0\times10^{30}$ kg, what fraction is lost per year?
Activity 2, Mass-Energy in Nuclear Physics
AnalyseBand 6

Connect $E = mc^2$ to real nuclear reactions

  1. Calculate the energy released when 1.0 kg of matter is completely converted to energy. Express in joules and compare to the annual electricity output of Australia (~200 TWh).
  2. Uranium-235 fission releases ~200 MeV per reaction. Calculate the mass equivalent of this energy. If 1 kg of U-235 undergoes complete fission, how much mass is converted to energy?
  3. In electron-positron annihilation ($e^- + e^+ \to \gamma + \gamma$), each photon carries energy $m_e c^2 = 0.511$ MeV. (a) Show that momentum is conserved if the photons travel in opposite directions. (b) Why must there be two photons, not one?
  4. The energy-momentum relation gives $E^2 = (pc)^2 + (mc^2)^2$. Show that for a photon ($m = 0$), this reduces to $E = pc$, and calculate the momentum of a 511 keV photon in kg·m/s.