Question

A certain particle of mass $m$ has momentum of magnitude $mc.$ What are $\left(a\right)\beta ,\left(b\right)\gamma ,$ and $\left(c\right)$ the ratio $K/E_0?$

Answer 1

(a) We set Eq. 37-41 equal to $mc$, as required by the problem, and solve for the speed.

Thus,

$\frac{mv}{\sqrt{1-v^2/c^2}}=mc$

leads to $\beta =1/\sqrt{2}=0.707$

(b) Substituting $\beta =1/\sqrt{2}$ into the definition of $\gamma$, we obtain

$\gamma =\frac{1}{\sqrt{1-v^2/c^2}}=\frac{1}{\sqrt{1-\left(1/2\right)}}=\sqrt{2}\approx 1.41$

(c) The kinetic energy is

$K=(\gamma -1)m{c}^2=(\sqrt{2}-1)m{c}^2=0.414m{c}^2=0.414E_0$

which implies $K/E_0=0.414$