Answer
Please see the work below.
Work Step by Step
(a) We know that
$\frac{m_{\circ}c^2}{\sqrt{1-\frac{v^2}{c^2}}}=m_{\circ}c^2$
This simplifies to:
$\sqrt{1-\frac{v^2}{c^2}}=\frac{1}{2}$
Squaring both sides, we obtain:
$1-\frac{v^2}{c^2}=\frac{1}{4}$
$\implies v=\frac{\sqrt {3}}{2}c$
$\implies v=0.866c$
(b) We know that the speed $v$ increases by less than a factor of two because of the relativistic increase in mass.
(c) We know that
$\frac{m_{\circ}c^2}{\sqrt{1-\frac{v^2}{c^2}}}=3m_{\circ}c^2$
This simplifies to:
$v^2=\frac{8}{9}c^2$
$\implies v=\frac{2\sqrt 2}{3}c$
$v=0.923c$
