Observers in system S\' measure the speed of a 1.60 times 10^-29 kg particle tra
ID: 1538334 • Letter: O
Question
Observers in system S' measure the speed of a 1.60 times 10^-29 kg particle travelling parallel to their X'-axis to be 0.6c. If the relative speed between S and S' is 0.8c. what do observers in S measure for the momentum of the particle? Solution: We know m_0 = 1.60 times 10^-29 kg, nu_s' = nu' = 0.6c rightarrow Gamma' = 5/4, and u = 0.8c rightarrow gamma = 5/3 and need to find p. From Equation 4.66 we have p_A = gamma(P_a' + e'u/c^2). which is expressible in terms of the given information as P_x = gamma (Gamma' m_2 nu' + E' u/c^2). = gamma (Gamma' m_0 nu + Gamma' m_0 u) = gamma Gamma' m_0 (nu' + u) Now, direct substitution yields P_x = (5/3) (5/4) m_0 (0.6c + 0.8c) =(5/3) (5/4) (16.0 times 10^-30 kg) (14/10) c = (5/3) (7/4) (16.0 times 106-30 kg) (3 times 1068 m/s) = (5)(7) (4 times 10^-22 kg middot m/s) = 1.4 times 10^-20 kg middot m/s Actually, this problem could have been solved more directly by using Equation 4.66. since we had already calculated its momentum and energy for an inertial system in Problems 4.3 and 4.5. respectively. In Problem 4.19. what do observers in system S measure for the particle's total energy?Explanation / Answer
The total relativistic energy will be given by:
E = sqrt (p^2c^2 + mo^2c^4) = c sqrt (p^2 + mo^2c^2)
E = 3 x 10^8 x sqrt [(1.4 x 10^-20)^2 + (1.6 x 10^-29)^2(3 x 10^8)^2] = 4.44 x 10^-12 J
Hence, E = 4.44 x 10^-12 J
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