How does a particle's relativistic kinetic energy $K$ compare to its Newtonian kinetic energy when the particle's speed $|\vec{v}|=\frac{4}{5}$ ? A) $K\frac{1}{2} m|\vec{v}|^{2}$ C) $K=\frac{1}{2} m|\vec{v}|^{2}$

Question

How does a particle's relativistic kinetic energy $K$ compare to its Newtonian kinetic energy when the particle's speed $|\vec{v}|=\frac{4}{5}$ ?
A) $K<\frac{1}{2} m|\vec{v}|^{2}$
B) $K>\frac{1}{2} m|\vec{v}|^{2}$
C) $K=\frac{1}{2} m|\vec{v}|^{2}$

Quizplus · Accepted Answer

The relativistic kinetic energy of a particle is given by $K = (\gamma - 1)mc^2$, where $\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$ and $v$ is the particle's speed. For speeds $v < c$, $\gamma > 1$, making $K$ larger than the Newtonian kinetic energy $\frac{1}{2} m|\vec{v}|^{2}$, especially as $v$ approaches $c$. At $|\vec{v}|=\frac{4}{5}c$, $\gamma$ is greater than 1, thus $K>\frac{1}{2} m|\vec{v}|^{2}$. 
Note that $K=m / \sqrt{1-|\vec{v}|^{2}}-m=m / \sqrt{\frac{9}{25}}-m=\frac{5}{3} m-m=\frac{2}{3} m$, but $\frac{1}{2} m|\vec{v}|^{2}=\frac{1}{2} m \frac{16}{25}=\frac{8}{25} m$.

How Does a Particle's Relativistic Kinetic Energy KKK Compare to Its Newtonian Kinetic Energy When the Particle's Speed

How Does a Particle's Relativistic Kinetic Energy $K$ Compare to Its Newtonian Kinetic Energy When the Particle's Speed