Question:
The velocity, \( v \), at which the mass of a particle is double its rest mass is:
The velocity, \( v \), at which the mass of a particle is double its rest mass is:
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In relativity, the velocity at which the particle's mass doubles is given by \( v = \sqrt{3} c \), which is less than the speed of light.
Updated On: Apr 28, 2026
- \( v = c \)
- \( v = \sqrt{3} c \)
- \( v = \sqrt{2} c \)
- \( v = 2c \)
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The Correct Option is B
Solution and Explanation
Step 1: Relativistic mass relation
The relativistic mass is given by:
\( m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \)
Step 2: Substitute given condition
Given \( m = 2m_0 \), so:
\[ 2 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \]
Step 3: Solve the equation
\[ \sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{2} \] \[ 1 - \frac{v^2}{c^2} = \frac{1}{4} \] \[ \frac{v^2}{c^2} = \frac{3}{4} \] \[ v = \frac{\sqrt{3}}{2}c \]
Final Answer:
\( v = \frac{\sqrt{3}}{2}c \)
The relativistic mass is given by:
\( m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \)
Step 2: Substitute given condition
Given \( m = 2m_0 \), so:
\[ 2 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \]
Step 3: Solve the equation
\[ \sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{2} \] \[ 1 - \frac{v^2}{c^2} = \frac{1}{4} \] \[ \frac{v^2}{c^2} = \frac{3}{4} \] \[ v = \frac{\sqrt{3}}{2}c \]
Final Answer:
\( v = \frac{\sqrt{3}}{2}c \)
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