Question

A. A rocket is travelling in space with a speed of 0.6c away from the origin along the negative x-axis. A second rocket passes by the first. The second rocket has speed 0.6c, heading in a direction in the x-y plane at an angle of 40° above the positive x-axis. Find the magnitude and direction of the velocity of the second rocket as measured in the rest frame of the first rocket.

Answer 1

Here u�_x = 0.6c cos 90 = 0.46c u�_y = 0.6c sin 90 = 0.38c u_x = -0.6c, u_y = 0 v_x = ? u_y = ? now u� = u - v/1 - uv implies u� - u�uv/c^2 = u - v u� - u = u� uv/c^2 - v u� - u = -v(1 - u�u/c^2) u - u�/1 - u�u/c^2 = v so, v_x = u_x - u�x/1 - u�_x u_x/c^2 = -0.6c - 0.46c 1 + 0.6c 0.45c/c^2 = -1.06/2 v_x = -0.53c v_y = u_y - u�_y/1 - u�_y u_y/c^2 = 0 - 0.38 c/1 v_y = -0.38c so, v = squareroot v_x^2 + v_y^2 = 0.65c theta = tan^-1(0.38/0.53) = 35 theta� = 180 + 35 theta� = 215 from + x-axis