Mass-Energy Equivalence and Relativistic Kinetic Energy Revision Notes | A Level AQA Physics

1. The Most Famous Equation in Physics — and What It Really Means

You have almost certainly seen $E = mc^2$ written on a T-shirt or a mug. But what does it actually say about the universe? At its core, it tells you that mass and energy are not two separate things — they are two faces of the same quantity, interchangeable under the right conditions.

Before Einstein, physicists treated mass as a fixed, inert property and energy as something entirely separate. Special relativity demolished that boundary. Even a stationary object — one with zero velocity and zero kinetic energy — possesses an enormous amount of energy locked inside its mass. That stored energy is called rest energy.

Notice the scale of $c^2$: it equals $9.0 \times 10^{16}$ m$^2$ s$^{-2}$. Even a mass of 1 kg stores $9.0 \times 10^{16}$ J of rest energy — roughly equivalent to the energy released by 20 megatons of TNT. This is why nuclear reactions, which convert tiny fractions of mass into energy, release such enormous amounts of it.

The equation also works in reverse: energy has an equivalent mass. When you heat an object, its mass increases — by an immeasurably small amount at everyday temperatures, but the principle holds exactly. Mass and energy are the same conserved quantity, expressed in different units.

2. Total Energy at Speed — Introducing the Lorentz Factor

The rest energy equation is only the starting point. When an object moves, its total energy increases beyond $m_0 c^2$. To describe this, we need the full relativistic energy equation, which introduces the Lorentz factor — the same factor you met in time dilation and length contraction.

The Lorentz factor $\gamma$ is defined as $\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$. When $v$ is small compared with $c$, $\gamma \approx 1$ and the object behaves classically. As $v$ approaches $c$, $\gamma$ grows without limit — and so does the total energy.

You can also write this compactly as $E = \gamma m_0 c^2$. The total energy is the rest energy multiplied by the Lorentz factor. At rest, $\gamma = 1$ and $E = m_0 c^2$, recovering the rest energy equation exactly — a reassuring consistency check.

The relativistic kinetic energy $E_k$ is then the difference between the total energy and the rest energy. Subtracting $m_0 c^2$ from both sides gives the kinetic contribution alone.

3. Worked Calculations — Applying the Equations

The equations look dense, but the method is always the same: calculate $\gamma$ first, then multiply by $m_0 c^2$. Let's walk through a full example so the process becomes automatic.

Worked example: Calculating relativistic kinetic energy at $0.8c$

An electron has rest mass $m_0 = 9.11 \times 10^{-31}$ kg and travels at $v = 0.8c$. Calculate its relativistic kinetic energy. ($c = 3.0 \times 10^8$ m/s) [4 marks]

1. $\frac{v^2}{c^2} = (0.8)^2 = 0.8 \times 0.8 = 0.64$ — Because $v = 0.8c$, the ratio $v/c = 0.8$ exactly. Squaring: $0.8 \times 0.8 = 0.64$. [M1]
2. $\gamma = \frac{1}{\sqrt{1 - 0.64}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.6} = 1.667$ — $1 - 0.64 = 0.36$; $\sqrt{0.36} = 0.6$; $\gamma = 1/0.6 = 1.667$. [M1]
3. $m_0 c^2 = 9.11 \times 10^{-31} \times (3.0 \times 10^8)^2$ $= 9.11 \times 10^{-31} \times 9.0 \times 10^{16}$ $= 8.20 \times 10^{-14}$ J — $(3.0 \times 10^8)^2 = 3.0 \times 10^8 \times 3.0 \times 10^8 = 9.0 \times 10^{16}$ m$^2$ s$^{-2}$. Then $9.11 \times 9.0 = 81.99 \approx 82.0$, giving $8.20 \times 10^{-14}$ J.
4. $E_k = (\gamma - 1)m_0 c^2 = (1.667 - 1) \times 8.20 \times 10^{-14}$ $= 0.667 \times 8.20 \times 10^{-14}$ $= 5.47 \times 10^{-14}$ J — $\gamma - 1 = 0.667$; $0.667 \times 8.20 = 5.469 \approx 5.47$. [A1] for correct kinetic energy with units.
5. $E_k^{\text{Newton}} = \frac{1}{2}m_0 v^2 = \frac{1}{2} \times 9.11 \times 10^{-31} \times (0.8 \times 3.0 \times 10^8)^2$ $= \frac{1}{2} \times 9.11 \times 10^{-31} \times (2.4 \times 10^8)^2$ $= \frac{1}{2} \times 9.11 \times 10^{-31} \times 5.76 \times 10^{16}$ $= 2.62 \times 10^{-14}$ J — $(2.4 \times 10^8)^2 = 2.4 \times 10^8 \times 2.4 \times 10^8 = 5.76 \times 10^{16}$. The Newtonian result ($2.62 \times 10^{-14}$ J) is roughly half the relativistic result ($5.47 \times 10^{-14}$ J) — confirming that at $0.8c$ the Newtonian formula significantly underestimates kinetic energy. [A1]

Answer: $E_k = 5.47 \times 10^{-14}$ J (relativistic); Newtonian formula gives only $2.62 \times 10^{-14}$ J — a significant underestimate at $0.8c$.

That comparison in Step 5 is worth pausing on. At $0.8c$, the relativistic kinetic energy is more than twice the Newtonian prediction. Particle accelerators operate in exactly this regime, which is why their designers must use relativistic equations — Newtonian mechanics would give completely wrong answers for the energies involved.

4. Graphs of Mass and Kinetic Energy Against Speed

The AQA specification requires you to describe and interpret graphs showing how relativistic mass and kinetic energy vary with speed. These graphs carry a powerful physical message: approaching the speed of light demands infinite energy, which is why no massive object can ever reach it.

Some treatments use the concept of relativistic mass $m = \gamma m_0$, which increases with speed. Although modern physics prefers to keep mass as the invariant rest mass $m_0$ and attribute the increase to momentum and energy, the AQA specification uses the relativistic mass picture for graph interpretation, so we will work with it here.

The shape of this graph is the key exam point. From $v = 0$ up to about $0.3c$, the curve is nearly flat — relativistic mass barely differs from rest mass. Beyond $0.5c$ the curve rises steeply, and as $v \to c$ the relativistic mass tends towards infinity. The curve never reaches $v = c$ on the horizontal axis because that would require infinite mass.

The kinetic energy graph shows two curves that start together at low speeds and diverge dramatically above $0.5c$. The Newtonian curve ($\frac{1}{2}m_0 v^2$) grows as a simple parabola and would predict a finite kinetic energy at $v = c$. The relativistic curve shoots upward without limit as $v \to c$, demonstrating that infinite energy would be required to reach the speed of light.

When describing these graphs in an exam, use the language of asymptotes: the relativistic mass and kinetic energy curves are asymptotic to the vertical line $v = c$. The curves approach but never reach $v = c$, confirming that $c$ is an absolute speed limit for any object with rest mass.

5. Why a Massive Object Can Never Reach the Speed of Light

The graphs make the argument visually, but it is worth stating the reasoning explicitly because AQA exam questions often ask you to explain — not just describe — the speed limit.

As an object's speed increases towards $c$, the denominator $\sqrt{1 - v^2/c^2}$ approaches zero. This makes $\gamma$ grow without bound, and therefore both the total energy $E = \gamma m_0 c^2$ and the kinetic energy $E_k = (\gamma - 1)m_0 c^2$ tend towards infinity. No finite energy source can supply an infinite amount of energy, so $v = c$ is unattainable for any object with $m_0 > 0$.

Photons, by contrast, have zero rest mass ($m_0 = 0$). The formula $E = \gamma m_0 c^2$ would give $0/0$ for a photon, which is why photons are described by a different relation ($E = hf$). They travel at $c$ precisely because they have no rest mass to resist acceleration.

6. Comparing Newtonian and Relativistic Kinetic Energy

A common exam task is to compare the two expressions for kinetic energy and evaluate when each is appropriate. The table below sets out the key differences clearly.

The critical point for evaluation questions is that the Newtonian formula is not wrong — it is an approximation that works extremely well at low speeds. At $v = 0.1c$, the two formulas agree to within about 0.75%. At $v = 0.5c$, the relativistic value is already 24% larger than the Newtonian prediction. By $v = 0.9c$, the Newtonian formula underestimates kinetic energy by a factor of approximately 3.2.

7. Exam Technique and Mark-Winning Guidance

Mass-energy equivalence and relativistic kinetic energy appear in the Turning Points topic, which is assessed at A-level only. Questions tend to combine calculation with conceptual explanation, so you need both the arithmetic and the reasoning.

For extended-response questions asking you to evaluate the two kinetic energy models, structure your answer as: (1) state that both agree at low speeds; (2) explain the divergence at high speeds with reference to $\gamma$; (3) state the physical consequence — infinite energy barrier at $v = c$; (4) give a conclusion about which model applies in a given context. This four-point structure reliably covers the marking points.

Worked example: Exam-style: Calculate total energy and kinetic energy of a proton at $0.6c$

A proton has rest mass $m_0 = 1.67 \times 10^{-27}$ kg and travels at $v = 0.6c$. Calculate (a) its total relativistic energy and (b) its relativistic kinetic energy. ($c = 3.0 \times 10^8$ m/s) [5 marks]

1. $\frac{v^2}{c^2} = (0.6)^2 = 0.6 \times 0.6 = 0.36$ — $v = 0.6c$ so $v/c = 0.6$; squaring: $0.6 \times 0.6 = 0.36$. [M1]
2. $\gamma = \frac{1}{\sqrt{1 - 0.36}} = \frac{1}{\sqrt{0.64}} = \frac{1}{0.8} = 1.25$ — $1 - 0.36 = 0.64$; $\sqrt{0.64} = 0.8$; $\gamma = 1/0.8 = 1.25$. [M1]
3. $m_0 c^2 = 1.67 \times 10^{-27} \times (3.0 \times 10^8)^2$ $= 1.67 \times 10^{-27} \times 9.0 \times 10^{16}$ $= 1.503 \times 10^{-10}$ J — $(3.0 \times 10^8)^2 = 3.0 \times 10^8 \times 3.0 \times 10^8 = 9.0 \times 10^{16}$; $1.67 \times 9.0 = 15.03$, so $1.503 \times 10^{-10}$ J.
4. $E = \gamma m_0 c^2 = 1.25 \times 1.503 \times 10^{-10}$ $= 1.879 \times 10^{-10}$ J — $1.25 \times 1.503 = 1.87875 \approx 1.88$. Total energy $E = 1.88 \times 10^{-10}$ J. [A1]
5. $E_k = E - m_0 c^2 = 1.879 \times 10^{-10} - 1.503 \times 10^{-10}$ $= 0.376 \times 10^{-10}$ $= 3.76 \times 10^{-11}$ J — Alternatively: $E_k = (\gamma - 1)m_0 c^2 = 0.25 \times 1.503 \times 10^{-10} = 3.76 \times 10^{-11}$ J. Both routes give the same answer. [A1]

Answer: (a) Total energy $E = 1.88 \times 10^{-10}$ J; (b) Relativistic kinetic energy $E_k = 3.76 \times 10^{-11}$ J

These two worked examples cover the full range of what AQA can ask on this topic: a single-step kinetic energy calculation, a two-part total-and-kinetic energy question, and a comparison with the Newtonian prediction. Practise each method until the $\gamma$-first approach is automatic.