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The equation given is:

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

where:
- \( m_0 \) is the rest mass of the particle,
- \( v \) is the velocity of the particle,
- \( c \) is the speed of light.

The problem asks what happens as \( v \to c \).

### Analysis:
As \( v \) approaches \( c \), the term inside the square root, \( 1 - \frac{v^2}{c^2} \), approaches zero because \( \frac{v^2}{c^2} \) approaches 1. This makes the denominator, \( \sqrt{1 - \frac{v^2}{c^2}} \), approach zero.

Thus, as \( v \to c \), the expression for the mass becomes:

\[
m = \frac{m_0}{\sqrt{0}} \to \infty
\]

This implies that the mass of the particle increases without bound (approaches infinity) as its velocity approaches the speed of light.

### Answer:
- The correct choice is: **\( m \to \infty \)**.

Would you like more details or have any questions?

### Related Questions:
1. What is the physical significance of mass approaching infinity at the speed of light?
2. How does this equation relate to Einstein's theory of special relativity?
3. Why can no object with mass reach the speed of light based on this formula?
4. What would happen to time as an object approaches the speed of light?
5. How does relativistic momentum change as \( v \to c \)?

### Tip:
Relativistic effects become significant as an object's velocity approaches the speed of light, causing increases in mass and time dilation effects.

In the theory of relativity, the mass m of a particle with velocity v is m = m_0 / sqrt(1 - v^2/c^2), where m_0 is the mass of the particle at rest and c is the speed of light. What happens as v → c?

Explore how the mass of a particle increases as its velocity approaches the speed of light, based on Einstein's theory of relativity. This problem explains the relativistic mass equation and the significance of the Lorentz factor. Suitable for university-level physics students.

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