Question

Relativity problem.

Two particle beams, A and B, are travelling collinearly, beam A with a velocity corresponding to $\gamma$ = 10,000, while the other, beam B, has a velocity corresponding to $\gamma$ =1,000.

Answer 1

Velocity corresponding to r_1 = 10,000 implies r = 1/ squareroot 1-x^2/c^2 x^2/c^2 = 1-1/r^2 implies v_1 = 0.999999995 c (a) v_BA = r_A(v_B-v_A) = (10,000) [-4.95 times 10^-7 c] v_BA = -4.95 times 10^-3 c (b) v_BA = r_B (v_A-v_B) = (1,000) [4.95 times 10^-7 c] v_BA = 4.95 times 10^-4 c (c) When 2 particles with same speeds start moving towards each other then time of collision in lab frame 't' t = [10-(-10)]km/v_1 + v_1] = 20 times 10^3/1.999999495 times c t = 3.33 times 10^-5 seconds the collision will take place at distance V_1 t from x_A = (0.999999995 c)(20 times 10^3)/1.999999495 times c = 10.0000025 km Hence collision will take place at x = 0.25cm (d) from the prospective of particle bunch rightarrow A, it did not travelled any distance because he is at rest in its own frame. Only B moved 20km all.