Relativistic Mass and Momentum (HSC SSCE Physics): Flashcards

$m_0$

$m$

$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$

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

$p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}$

Mass increases, acceleration decreases toward zero

Circular accelerator using electric & magnetic fields

Decreases: $a = \frac{F_R}{\gamma m}$

Higher energy needs larger radius due to relativistic mass

$3.0 \times 10^8$ m s$^{-1}$