Physics · Units 3 & 4

Mass-Energy Equivalence

Understand mass-energy equivalence the easy way, with plain English intuition, a clear diagram, the rest energy and kinetic energy equations, worked examples and an auto marked practice test. VCE Physics Units 3 and 4.

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Albert Einstein discovered something that sounds impossible at first. Mass and energy are the same thing, just wearing different clothes. A lump of matter sitting perfectly still is secretly a huge store of energy, and the recipe that converts one into the other is the most famous equation in all of science: E0=mc2E_0 = mc^{2}.

Mass is frozen energy

Think of mass as energy that has been frozen solid. Even when an object is not moving at all, it still holds a vast amount of energy locked inside it. We call this its rest energy, and it depends only on the mass:

E0=mc2E_0 = mc^{2}

The reason the number is so enormous is the c2c^{2} part. The speed of light cc is 3.0×1083.0 \times 10^{8} m/s, and squaring it gives 9.0×10169.0 \times 10^{16}. So you multiply the mass by ninety thousand million million. A tiny mass becomes a colossal energy.

Moving objects carry even more

Once an object starts moving, it carries its rest energy plus extra energy from its motion. The total energy is the rest energy stretched by the Lorentz factor γ\gamma, a number that grows as the object speeds up:

Etot=γmc2E_{tot} = \gamma m c^{2}

The leftover, the part that is purely due to movement, is the kinetic energy:

Ek=(γ−1)mc2=Etot−E0E_k = (\gamma - 1)mc^{2} = E_{tot} - E_0

At everyday speeds γ\gamma is almost exactly 11, so EkE_k is tiny and matches the school formula. Near the speed of light γ\gamma shoots up, and the kinetic energy becomes huge.

When mass becomes pure energy

The cleanest demonstration of mass-energy equivalence is annihilation. When an electron meets its antimatter twin, a positron, the two particles vanish completely. All of their mass is converted into energy, carried away as two photons of light flying off in opposite directions.

e−electrone+positronphotonphotonmass becomes energy: E = mc²

The blue circle is the electron and the red circle is the positron. They drift together, touch, and disappear. In their place, two photons shoot off in opposite directions, carrying away every bit of energy that used to be mass. Nothing is destroyed, the mass is simply turned into energy.

How to actually solve one

The method is a short, repeatable recipe.

  1. For rest energy, square the speed of light first (c2=9.0×1016c^{2} = 9.0 \times 10^{16}), then multiply by the mass: E0=mc2E_0 = mc^{2}.
  2. For total energy of a moving object, multiply the rest energy by the Lorentz factor: Etot=γE0E_{tot} = \gamma E_0.
  3. For kinetic energy, take the rest energy back out: Ek=Etot−E0=(γ−1)E0E_k = E_{tot} - E_0 = (\gamma - 1)E_0.
  4. Keep your powers of ten lined up when you add or subtract energies.

Watch the trap. Total energy includes the rest energy, but kinetic energy does not. If a question asks only for the energy of motion, remember to subtract E0E_0.

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

Write the rest energy of an object of mass mm.
Why is the energy from a tiny mass so large?
Write the total energy of a moving object.
Write the kinetic energy of a fast particle two ways.
What is the difference between total and kinetic energy?
What happens in electron–positron annihilation?
Recall · Special Relativity
Write the Lorentz factor γ\gamma.
Recall · Atomic Spectra and Energy Levels
Where does the energy of an emitted photon come from in an atom?

See the recipe in action in the Worked Examples tab, then test yourself in Try It.

Worked examples

Worked Example 1The rest energy of an electron

Find the rest energy of an electron, which has a mass of 9.1×10−319.1 \times 10^{-31} kg. Take c=3.0×108c = 3.0 \times 10^{8} m/s.

  1. 1

    Rest energy is just the mass multiplied by the speed of light squared. Start by squaring cc.

    c2=(3.0×108)2=9.0×1016 m2/s2c^{2} = (3.0 \times 10^{8})^{2} = 9.0 \times 10^{16} \text{ m}^2/\text{s}^2
  2. 2

    Now multiply the mass by that number.

    E0=mc2=9.1×10−31×9.0×1016E_0 = mc^{2} = 9.1 \times 10^{-31} \times 9.0 \times 10^{16}
  3. 3

    Multiply the front numbers, then add the powers of ten.

    E0=8.2×10−14 JE_0 = 8.2 \times 10^{-14} \text{ J}
Answer
E0=8.2×10−14 JE_0 = 8.2 \times 10^{-14} \text{ J}
Worked Example 2Total and kinetic energy of a fast particle

A particle has a rest energy of 8.0×10−148.0 \times 10^{-14} J and moves at a speed where the Lorentz factor is γ=2.0\gamma = 2.0. Find its total energy and its kinetic energy.

  1. 1

    Total energy is the rest energy stretched by the Lorentz factor. Just multiply the two together.

    Etot=γE0=2.0×8.0×10−14=1.6×10−13 JE_{tot} = \gamma E_0 = 2.0 \times 8.0 \times 10^{-14} = 1.6 \times 10^{-13} \text{ J}
  2. 2

    Kinetic energy is whatever total energy is left over once you take the rest energy back out.

    Ek=Etot−E0=1.6×10−13−8.0×10−14E_k = E_{tot} - E_0 = 1.6 \times 10^{-13} - 8.0 \times 10^{-14}
  3. 3

    Line up the powers of ten and subtract.

    Ek=1.6×10−13−0.80×10−13=8.0×10−14 JE_k = 1.6 \times 10^{-13} - 0.80 \times 10^{-13} = 8.0 \times 10^{-14} \text{ J}
Answer
Etot=1.6×10−13 J,Ek=8.0×10−14 JE_{tot} = 1.6 \times 10^{-13} \text{ J}, \quad E_k = 8.0 \times 10^{-14} \text{ J}

Practice questions

Practice test

Try it yourself

5 questions, 7 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.The rest energy of an object of mass mm is given by:

1mark
Need a hint?
This is the most famous equation in physics. The speed of light appears squared.

Q2.A small mass of 2.0×10−32.0 \times 10^{-3} kg is completely converted to energy. Using c=3.0×108c = 3.0 \times 10^{8} m/s, the energy released is:

1mark
Need a hint?
Use E0=mc2E_0 = mc^{2} with c2=9.0×1016c^{2} = 9.0 \times 10^{16}.

Q3.A particle has rest energy E0E_0 and moves at a speed where γ=3.0\gamma = 3.0. Its total energy is:

1mark
Need a hint?
Total energy is Etot=γmc2=γE0E_{tot} = \gamma m c^{2} = \gamma E_0.

Q4.A particle has a rest energy of 5.0×10−145.0 \times 10^{-14} J. When moving, its total energy is 9.0×10−149.0 \times 10^{-14} J. Its kinetic energy is:

1mark
Need a hint?
Kinetic energy is total energy minus rest energy.

Q5.A proton has a mass of 1.7×10−271.7 \times 10^{-27} kg. Taking c=3.0×108c = 3.0 \times 10^{8} m/s, calculate its rest energy. Show your working.

3marks

Work this on paper. The worked solution appears once you submit.

Frequently asked questions

What does mass-energy equivalence actually mean?
It means mass and energy are two forms of the same thing. A given amount of mass is equivalent to a fixed amount of energy, set by E equals mc squared, and one can be converted into the other.
Why is the energy from a tiny mass so large?
Because the mass is multiplied by the speed of light squared, and the speed of light is about 300 million metres per second. Squaring it gives a factor of 9 followed by 16 zeros, so even a gram of mass holds an enormous amount of energy.
What is the difference between total energy and kinetic energy?
Total energy is everything the object has, the rest energy plus the kinetic energy. Kinetic energy is only the extra energy due to motion, which is the total energy with the rest energy taken back out.