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Interactive Tutorial, Chapter 29 Problem 43 Fundamentals of College Physics, 3rd. Ed. Dr. Peter J. Nolan "Fundamentals of College Physics" Third Edition Dr. Peter J. Nolan, SUNY Farmingdale Chapter 29 Special Relativity Computer Assisted Instruction Interactive Tutorial 43. Length Contraction. The length of a rod at rest is found to be Lo = 2.55 m. Find the length L of the rod when observed by an observer in motion at a speed v = 0.25 c. Initial Conditions Lo = 2.55 m c = 3.00E+08 m/s v = 2.50E-01 c = 7.50E+07 m/s For speeds that are not given in terms of the speed of light c use the following converter to find the equivalent speed in terms of the speed of light c. Then place the equivalent speed into the yellow cell for v above. v= 1610 km/hr = 447.58 m/s = 1.49E-06 c v= 2E+06 m/s = 5.37E-03 c The length contraction is given by equation 29.60 as L = Lo sqrt[1 - (v2)/(c2)] L=( 2.55 ) x sqrt[1 -( 7.50E+07 m/s)2 / ( 3.00E+08 m/s)2] L = 2.47E+00 m Page 1 Problem 44 "Fundamentals of College Physics" Third Edition Dr. Peter J. Nolan, SUNY Farmingdale Chapter 29 Special Relativity Computer Assisted Instruction Interactive Tutorial 44. Time Dilation. A clock in a moving rocket ship reads a time duration Dt0 = 15.5 hr. What time elapses, Dt, on earth if the rocket ship is moving at a speed v = 0.355 c? Initial Conditions Dto = 15.5 hr c = 3.00E+08 m/s v = 3.55E-01 c = 1.07E+08 m/s For speeds that are not given in terms of the speed of light c use the following converter to find the equivalent speed in terms of the speed of light c. Then place the equivalent speed into the yellow cell for v above. v= 1610 km/hr = 447.58 m/s = 1.49E-06 c v = 1.61E+06 m/s = 5.37E-03 c The time dilation is given by equation 29.64 as Dt = Dto / sqrt[1 - (v2)/(c2)] Dt = ( 2 2 15.5 hr) /sqrt[1- ( 1.07E+08 m/s) / ( 3.00E+08 m/s) ] Dt = 16.57991 hr Page 2 Problem 45 "Fundamentals of College Physics" Third Edition Dr. Peter J. Nolan, SUNY Farmingdale Chapter 29 Special Relativity Computer Assisted Instruction Interactive Tutorial 45. Relative velocities. Two spaceships are approaching a space station as in figure 29.15. Spaceship one has a velocity of 0.55 c to the left and spaceship two has a velocity of 0.75 c to the right. Find the velocity of rocket ship one as observed by rocket ship two. Assume the space station to be at rest. Spaceship one then has the velocity Vx = - 0.55c. Spaceship two is considered to be a moving coordinate system approaching with the speed v = 0.75 c. Initial Conditions Vx = -4.85E-01 c = -1.45E+08 m/s c = 3.00E+08 m/s v = 6.55E-01 c = 1.96E+08 m/s The relative velocity of approach of spaceship one as observed by the S' observer on rocket ship two is found from equation 29.75 as Vx' = [Vx - v] / [1 - (v/c2) Vx] Vx' = [( -4.85E-01 c) - ( 6.55E-01 c)] / [(1 - {( 1.96E+08 m/s) / ( 3.00E+08 m/s)2} x ( -1.45E+08 m/s)] Vx' = -0.86516 c = -2.59E+08 m/s Page 3 Problem 46 "Fundamentals of College Physics" Third Edition Dr. Peter J. Nolan, SUNY Farmingdale Chapter 29 Special Relativity Computer Assisted Instruction Interactive Tutorial 46. Relativistic mass. A mass at rest has a value mo = 2.55 kg. Find the relativistic mass m when the object is moving at a speed v = 0.355 c. Initial Conditions mo = 2.55 kg c= 3.00E+08 m/s v = 3.55E-01 c = 1.07E+08 m/s For speeds that are not given in terms of the speed of light c use the following converter to find the equivalent speed in terms of the speed of light c. Then place the equivalent speed into the yellow cell for v above. v= 1610 km/hr = 447.58 m/s = 1.49E-06 c v = 1.61E+06 m/s = 5.37E-03 c The relativistic mass is given by equation 29.86 as m = mo / sqrt[1 - (v2)/(c2)] m=( 2.55 kg)/sqrt[1 - ( 1.07E+08 m/s)2 / ( 3.00E+08 m/s)2] m = 2.72766277 kg Page 4 Problem 47 "Fundamentals of College Physics" Third Edition Dr. Peter J. Nolan, SUNY Farmingdale Chapter 29 Special Relativity Computer Assisted Instruction Interactive Tutorial 47. Plot of length contraction and mass change versus speed The length of a rod at rest is Lo = 1.00 m and its mass is mo = 1.00 kg. Find the length L and mass m of the rod as its speed v in the axial direction increases from 0.00 c to 0.90 c, where c is the speed of light (c = 3.00 x 108 m/s). Plot the results. Initial Conditions Lo = 1m c= 3.00E+08 m/s v= 0.1 c = 3.00E+07 m/s mo = 1 kg The length contraction is given by equation 29.60 as L = Lo sqrt[1 - (v2)/(c2)] L=( 1 ) x sqrt[1- ( 3.00E+07 m/s)2 / ( 3.00E+08 m/s)2] L = 0.994987 m The relativistic mass is given by equation 29.86 as m = mo / sqrt[1 - (v2)/(c2)] m=( 1 kg)/sqrt[1-( 3.00E+07 m/s)2 / ( 3.00E+08 m/s)2] m = 1.005038 kg The contraction in length and increase in mass as the speed increases is calculated by the above equations and is shown in the table below. v (%c) v (m) L (m) m (kg) 0 0.00E+00 1.000 1.000 0.1 3.00E+07 0.995 1.005 0.2 6.00E+07 0.980 1.021 0.3 9.00E+07 0.954 1.048 0.4 1.20E+08 0.917 1.091 0.5 1.50E+08 0.866 1.155 0.6 1.80E+08 0.800 1.250 0.7 2.10E+08 0.714 1.400 0.8 2.40E+08 0.600 1.667 0.9 2.70E+08 0.436 2.294 The contraction of the length of the rod L versus the speed v is shown in the graph Page 5 Problem 47 below. Length Contraction 1.200 1.000 0.800 Length L 0.600 0.400 0.200 0.000 0 0.2 0.4 0.6 0.8 1 Speed v The increase in the mass m of the rod versus the speed v is shown in the graph below. Increase in mass with speed 2.500 2.000 Mass m 1.500 1.000 0.500 0.000 0 0.2 0.4 0.6 0.8 1 Speed v Page 6 Problem 48 "Fundamentals of College Physics" Third Edition Dr. Peter J. Nolan, SUNY Farmingdale Chapter 29 Special Relativity Computer Assisted Instruction Interactive Tutorial 48. An accelerated charged particle. An electron is accelerated from rest through a potential difference V = 4.55 x 105 V. Find (a) the kinetic energy of the electron, (b) the rest mass energy of the electron, (c) the total relativistic energy of the electron, (d) the speed of the electron, (e) the relativistic mass of the electron, and (f) the momentum of the electron. Initial Conditions mo = 9.11E-31 kg c = 3.00E+08 m/s q = 1.60E-19 C V = 4.55E+05 V a. The kinetic energy of the electron is found from equation 19.36 as KE = work done = q V KE = ( 1.60E-19 C) x ( 4.55E+05 V) KE = 7.29E-14 J = 4.56E-01 Mev b. The rest mass energy of the electron is found from equation 29.101 as Eo = mo c2 2 Eo = ( 9.11E-31 kg) x ( 3.00E+08 m/s) Eo = 8.19E-14 J = 5.12E-01 Mev c. The total relativistic energy is found from equation 29.100 as E = KE + Eo E = ( 4.56E-01 Mev) + ( 5.12E-01 Mev) E= 9.67E-01 Mev d. The speed of the electron is found by solving equation 29.96 for the speed v. That is, KE = {(mo c2) / sqrt[1 - (v2)/(c2)]} - mo c2 KE = {(Eo) / sqrt[1 - (v2)/(c2)]} - Eo and the speed v is found as v = c sqrt[1 - [1 / {(KE/Eo) + 1}]2] v = c sqrt[ 1 - [1 /{ ( 4.56E-01 Mev) / ( 5.12E-01 ) + 1} ]2 ] Page 7 Problem 48 2 v = c sqrt[ 1 - [1 /{ ( 8.90E-01 ) + 1} ] ] v= 0.8486 c = 2.54E+08 m/s e. The relativistic mass is given by equation 29.86 as m = mo / sqrt[1 - (v2)/(c2)] m = ( 9.11E-31 kg)/sqrt[1- ( 2.54E+08 m/s)2 / ( 3.00E+08 m/s)2] m = 1.722E-30 kg f. The momentum of the electron is found from equation 29.90 as p=mv p = ( 1.72E-30 kg) x ( 2.54E+08 m/s) p = 4.381E-22 kg m/s Page 8