In Physics 2 you met $E = mc^2$ as the energy locked in mass. Here is where it comes from, and how energy and momentum change near the speed of light. The everyday formulas $p = mu$ and $K = \frac{1}{2}mu^2$ cannot be the whole story: they would let a steady push accelerate anything past $c$.
🔑 Key idea
For a particle of mass $m$ moving at speed $u$, with $\gamma = 1/\sqrt{1 - u^2/c^2}$: $$p = \gamma mu,\qquad E = \gamma mc^2,\qquad K = E - mc^2 = (\gamma - 1)mc^2,$$ and $$E^2 = (pc)^2 + (mc^2)^2.$$ The rest energy $mc^2$ is the energy at $u = 0$: 0.511 MeV for an electron, 938.3 MeV for a proton.
💡 Why it works
Momentum must be conserved in every inertial frame; checking collisions with the Lorentz transformation shows that only $p = \gamma mu$ works. Then kinetic energy is the work done by a net force starting from rest. Since $F = \frac{dp}{dt}$ and $u = \frac{dx}{dt}$, the work is $\int F\,dx = \int u\,dp$. Integrate by parts with $p = \gamma mu$: $$K = \int_0^u u'\,dp = pu - \int_0^u p\,du' = \gamma mu^2 - \int_0^u \frac{mu'\,du'}{\sqrt{1 - u'^2/c^2}}.$$ The last integral is $mc^2\left(1 - \frac{1}{\gamma}\right)$, and the algebra gives $K = \gamma mc^2 - mc^2$. At low speed, $\gamma \approx 1 + \frac{u^2}{2c^2}$ and $K \approx \frac{1}{2}mu^2$, as it should. As $u \to c$, $\gamma \to \infty$: reaching $c$ would take infinite energy.
Kinetic energy in units of the rest energy. The two curves agree at low speed; near $c$ the relativistic energy climbs without limit.
Worked examples
✏️ Worked example: An electron at 0.80c
Find $E$, $K$ and $p$ for an electron ($mc^2 = 0.511$ MeV) moving at $0.80c$.
The true value is 0.341 MeV, more than twice as much.
Answer: The classical formula is low by a factor of about 2
Particle accelerators live in this regime. Protons in the Large Hadron Collider at CERN reach 6.8 TeV of energy, so $\gamma = \frac{6800\ \text{GeV}}{0.938\ \text{GeV}} \approx 7250$ and their speed is about $0.99999999c$. Pushing harder adds a great deal of energy and momentum but almost no speed.
⚠️ Common mistake
Using $\frac{1}{2}mv^2$ near the speed of light, or taking $\gamma mc^2$ itself as the kinetic energy. $\gamma mc^2$ is the total energy; subtract the rest energy to get $K = (\gamma - 1)mc^2$. Check: at $u = 0$ the kinetic energy must be zero, and $(\gamma - 1)mc^2 = 0$ there, while $\gamma mc^2 = mc^2$ is not.
⚠️ Common mistake
Mixing units. In particle physics, energies are in MeV, momenta in MeV/$c$ and masses in MeV/$c^2$; then $pc$ is in MeV and the $c$'s take care of themselves.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.