Question

A spacecraft is launched from the surface of the Earth with a velocity of 0.60c at an angle of 50.0 degree above the horizontal positive axis. Another spacecraft is moving past with a velocity of 0.70c in the negative x direction. Determine the magnitude and direction of the velocity of the first spacecraft as measured by the pilot of the second spacecraft.

Answer 1

The velocity of the first spacecraft as measured by the pilot of the second spacecraft is approximately 0.1724 times the speed of light in the negative x direction.

Step-by-step explanation:

In order to determine the magnitude and direction of the velocity of the first spacecraft as measured by the pilot of the second spacecraft, we can use the relativistic velocity addition formula.

Let's assume the velocity of the first spacecraft as measured by the pilot of the second spacecraft is v'.

The formula for relativistic velocity addition is:

v' = (v1 + v2) / (1 + (v1 * v2) / c^2)

where v1 is the velocity of the first spacecraft and v2 is the velocity of the second spacecraft.

Plugging in the values, we have:

v' = (0.60c + (-0.70c)) / (1 + (0.60c * -0.70c) / c^2)

v' = (-0.10c) / (1 - 0.42)

v' = -0.10c / 0.58

v' = -0.1724c

Therefore, the magnitude of the velocity of the first spacecraft as measured by the pilot of the second spacecraft is approximately 0.1724 times the speed of light, and the direction is in the negative x direction.