Problem

At what speed would the mass of a proton triple, given its rest mass?

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Data

• Relativistic mass: $m = 3m_0$
• Rest mass of proton: $m_0 = 1.673 \times 10^{-27} \, \mathrm{kg}$
• Speed of light: $c = 3 \times 10^8 \, \mathrm{ms^{-1}}$

To Find:

• Speed ($v$) at which the proton’s mass triples.

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Prerequisite Concepts

1. Relativistic mass formula:

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

2. Solve for speed $v$:

$$\sqrt{1 - \frac{v^2}{c^2}} = \frac{m_0}{m}$$ $$\frac{v^2}{c^2} = 1 - \left(\frac{m_0}{m}\right)^2$$ $$v = c \sqrt{1 - \left(\frac{m_0}{m}\right)^2}$$

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Solution

Step 1: Substituting the given values

The relativistic mass is triple the rest mass:

$$m = 3m_0$$

Using the formula:

$$\sqrt{1 - \frac{v^2}{c^2}} = \frac{m_0}{3m_0}$$ $$\sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{3}$$

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Step 2: Solving for $v^2/c^2$

Squaring both sides:

$$1 - \frac{v^2}{c^2} = \frac{1}{9}$$

Rearranging:

$$\frac{v^2}{c^2} = 1 - \frac{1}{9}$$ $$\frac{v^2}{c^2} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9}$$

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Step 3: Solving for $v$

Taking the square root:

$$v = c \sqrt{\frac{8}{9}}$$ $$v = c \times \frac{\sqrt{8}}{3}$$

Approximating:

$$v = 0.9428c$$

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Answer

The speed at which the mass of the proton triples is:

$$v = 0.9428c$$