Question

A particle of rest mass m0 has a speed v= 0.8c. Find its relativistic momentum, its kinetic energy and total energy

Answer 1

Answer on Question # 74787, Physics - Mechanics - Relativity:

Question: A particle of rest mass $m_0$ has a speed $v = 0.8c$. Find its relativistic momentum, its kinetic energy and total energy.

Solution: Rest mass of the particle = $m_0$

Speed of the particle $v = 0.8c$, where $c =$ speed of light.

Now consider a constant $\beta = \frac{v}{c} = 0.8$

And $\sqrt{1 - \beta^2} = 0.6$

We know relativistic momentum $p = \frac{m_0 \times v}{\sqrt{1 - \beta^2}} = \frac{m_0 \times 0.8c}{0.6} = 1.33 \, \text{m}_0 \text{c}$

Again relativistic kinetic energy $T = \frac{m_0 \times c^2}{\sqrt{1 - \beta^2}} - m_0 c^2 = m_0 c^2 \left( \frac{1}{\sqrt{1 - \beta^2}} - 1 \right)$

$$= m_0 c^2 \left( \frac{1}{0.6} - 1 \right) = 0.67 \, \text{m}_0 \text{c}^2$$

Now, total relativistic energy $E = \sqrt{(p^2 c^2 + m_0^2 c^4)} = \sqrt{(1.7689 \, m_0^2 c^4 + m_0^2 c^4)}$

$$= 1.664 \, \text{m}_0 \text{c}^2.$$

Answer: So, relativistic momentum is $1.33 \, \text{m}_0 \text{c}$, kinetic energy $0.67 \, \text{m}_0 \text{c}^2$ and total energy $1.664 \, \text{m}_0 \text{c}^2$.

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