I don't understand this question.

Q. In the theory of relativity, the mass of a particle with velocity "v" is
m = mo/√(1-v2/c2) where mo is the mass of the particle at rest and "c" is the speed of light. What happends as v-->c-?

Thank You!

In the theory of relativity, the equation you provided describes how the mass of a particle changes as its velocity approaches the speed of light (c). The equation states that the mass (m) of a particle in motion is equal to the rest mass (mo) divided by the square root of (1 - v^2/c^2), where v is the velocity of the particle and c is the speed of light.

Now let's examine what happens as the velocity (v) of the particle approaches the speed of light (c-):

1. As v gets closer to c, the term v^2/c^2 in the denominator becomes significantly larger. This increases the value inside the square root, making it close to 1.

2. As the value inside the square root approaches 1, the square root itself approaches 1. Simplifying the equation, we have m = mo/√(1 - v^2/c^2) ≈ mo/√(1 - 1) = mo/0.

3. In the final step, we observe that we have a division by zero, which is undefined in mathematics. This implies that as v approaches c, the mass of the particle becomes infinitely large or, in other words, the mass approaches infinity.

Therefore, as the velocity of the particle approaches the speed of light, its mass becomes infinitely large or effectively infinite. This observation is a critical prediction of the theory of relativity and explains why objects with mass cannot reach or exceed the speed of light.