Problem

At what fraction of the speed of light must a particle move so that its kinetic energy is one and a half times its rest energy?

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Data

• Kinetic Energy ($KE$) = $1.5 E_0$
• Rest Energy ($E_0$) = $m_0 c^2$

To Find:

• Fraction of the speed of light ($v/c$).

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Prerequisite Concepts

1. Total energy ($E$) of a particle:

$$E = E_0 + KE$$

Substituting $KE = 1.5 E_0$:

$$E = E_0 + 1.5 E_0 = 2.5 E_0$$

2. Relativistic total energy formula:

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

3. Rearrange for $v$:

$$\sqrt{1 - \frac{v^2}{c^2}} = \frac{E_0}{E}$$

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Solution

Step 1: Express total energy

Given $KE = 1.5 E_0$, total energy is:

$$E = E_0 + 1.5 E_0 = 2.5 E_0$$

Step 2: Apply relativistic energy formula

Using $E = \frac{m_0 c^2}{\sqrt{1 - \frac{v^2}{c^2}}}$:

$$\sqrt{1 - \frac{v^2}{c^2}} = \frac{E_0}{E} = \frac{E_0}{2.5 E_0} = \frac{1}{2.5}$$ $$\sqrt{1 - \frac{v^2}{c^2}} = 0.4$$

Step 3: Solve for $v^2/c^2$

Squaring both sides:

$$1 - \frac{v^2}{c^2} = 0.16$$

Rearranging:

$$\frac{v^2}{c^2} = 1 - 0.16 = 0.84$$

Step 4: Solve for $v/c$

Taking the square root:

$$v/c = \sqrt{0.84} \approx 0.9165$$

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Answer

The particle must move at approximately $0.917c$ for its kinetic energy to be one and a half times its rest energy.