Answer to Question #216609 in Mechanics | Relativity for Peael

Question

Determine the speed of an object whose Kinect energy equals it's rest energy(2.6×10^8m/s)

Answer 1

The Kinetic Energy of the object is

$K=\text{total energy - rest mass energy}\\\implies K= \dfrac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}-m_0c^2$

Since Kinetic Energy of the particle is equal to its rest energy. So,

$\implies \dfrac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}-m_0c^2=m_0c^2\\\ \\\implies \dfrac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}=2m_0c^2\\\ \\\implies \dfrac{1}{\sqrt{1-\frac{v^2}{c^2}}}=2\\\ \\\implies\dfrac{1}{4}=1-\dfrac{v^2}{c^2}\\\ \\\implies \dfrac{v^2}{c^2}=1-\dfrac{1}{4}=\dfrac{3}{4}\\\ \\\implies v=\dfrac{\sqrt 3}{2}c$

$\because c=2.6\times 10^8\ \ m/s$

So, speed of object (v) = $\dfrac{\sqrt 3}{2}\times 2.6\times 10^8 =2.25\times 10^8\ \ m/s$