Ask Question
6 November, 08:19

A rocket moves forward with velocity v = 3c/4, with respect to the earth. Inside this rocket, a second, smaller rocket is launched. In the inertial frame of reference where the larger rocket is at rest, the smaller rocket has velocity v = 3c/4. What is the velocity of the smaller rocket with respect to the earth? You may express the answer as a fraction times c, if you wish (but simplify as much as possible).

+3
Answers (1)
  1. 6 November, 08:25
    0
    We shall apply the formula for relative velocity in relativistic mechanics

    Vr = v₁ - v₂ / (1 - v₁v₂/c²), v₁ is velocity of smaller rocket with respect to earth, v₂ is velocity of large rocket with respect to earth.

    Given Vr = 3c/4, v₁?, v₂ = 3c/4

    3c/4 = (v₁ - 3c/4) / (1 - v₁ 3c/4c²)

    3c/4 - (9v₁ / 16) = v₁ - 3c/4

    v₁ + 9v₁/16 = 3c/2

    = 25v₁ / 16 = 3c / 2

    v₁ = 16 / 25 x 3c / 2

    = 24/25 x c
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A rocket moves forward with velocity v = 3c/4, with respect to the earth. Inside this rocket, a second, smaller rocket is launched. In the ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers