Question

1) A particle has a 4-momentum of [2266, 1813, -1133, 453] kg-m-s with respect to the Home frame a) What are the components of the particle's velocity, [vx, Vy, vz], in m/s? b) What is the particle's mass, in kg? c) W J? d) Calculate the particle's 4-momentum in an Other frame traveling with a relative velocity of u 0.9c hat are the particle's total energy and kinetic energy, in

Answer 1

Solution:

Given that

Given that we Recognize that The y-momentum P = (E?C, P_x, P_y, P_z) [2266, 1813, -1133, 453] kg middot ms^-1 therefore E/C = 2266 From relativity we have E = gamma mc^2 Here gamma = 1/squareroot 1 - v^2/c^2 c = 3 times 10^8 m/s E/C = gamma mc therefore gamma m = 2266/C \r\n here P_x = gamma mv_x Thus v_x = P_x/rm 1813/2266 times e 0.8 e v_y = P_y/rm -1133/2266 times c -0.5 c v_z = 453/2266 e 0.2 c v_x = 0.8 c m/s, v_y = 0.5 e m/s, v_z = 0.2 c Here total v = squareroot v_x^2 + v_y^2 + v_z^2 \r\n Thus v = 0.9643 c so gamma = 1/squareroot 1 - v^2/c^2 3.77989 gamma_m = 2266/c therefore m = 2266/gamma c = 2266/3.7787 times c m = 599.67/3 times 10^8 199.89 times 10^-8 kg Thus m = 1.9989 times 106-6 kg almostequalto 2 times 10^-6 kg \r\n particle total energy E = 2266 c 2266 times 3 times 10^8 Jule