# At what speed of motion is the relativistic mass of a particle twice as large as the rest mass of this particle?

Problem data: m (relativistic mass of the taken particle) = 2m0 (rest mass).

Constants: C (speed of light movement) = 3 * 10 ^ 8 m / s.

To determine the required speed of the taken particle, we use the formula: m = m0 / √ (1 – V ^ 2 / C ^ 2); m / m0 = 2 = 1 / √ (1 – V ^ 2 / C ^ 2).

√ (1 – V ^ 2 / C ^ 2) = 1/2.

(1 – V ^ 2 / C ^ 2) = 1/4.

V ^ 2 / C ^ 2 = 1 – 1/4.

V ^ 2 = (1 – 1/4) * C ^ 2.

V = √ ((1 – 1/4) * C ^ 2).

Let’s calculate: V = √ ((1 – 1/4) * (3 * 10 ^ 8) ^ 2) ≈ 2.6 * 10 ^ 8 m / s.

Answer: The taken particle must have a velocity of 2.6 * 10 ^ 8 m / s.

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