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T, the time between the arrival of signals on Earth is related to the proper time interval T0 (the time between signals in Al\u2019s frame of reference) through the time-dilation equation: ( )2201 cv TT \u2212= Substituting for T and simplifying yields: ( ) 0 22 Bert 600.1 1 T cv f \u2212= Substitute numerical values and evaluate Bertf : ( ) ( )( ) 1 2 Bert y 50y 0100.0600.1 600.01 \u2212=\u2212=f Chapter R 1068 (b) Express the number of signals N received by Bert in terms of the number of signals sent by Al: Al0 tfN \u394= Use the time dilation equation to express the elapsed time in Al\u2019s frame of reference in terms of the elapsed time in Bert\u2019s frame (the proper elapsed time): 2 BertAl 1\u394\u394 \u239f\u23a0 \u239e\u239c\u239d \u239b\u2212= c vtt Substitute for Al\u394t to obtain: 2 Bert0 1\u394 \u239f\u23a0 \u239e\u239c\u239d \u239b\u2212= c vtfN (1) Find Bert\u394t : y667.6 0.600 y00.4 \u394 Bert =\u22c5= c ct Substitute numerical values in equation (1) and evaluate N: ( )( ) 533 600.01y 667.6y100 2 1 = \u239f\u23a0 \u239e\u239c\u239d \u239b\u2212= \u2212 c cN (c) Express the number of signals N received by Bert in terms of the number of signals sent by Al before he returns: Al0 TfN \u394= Because 2 BertAl 1\u394\u394 \u239f\u23a0 \u239e\u239c\u239d \u239b\u2212= c vtt , the time in Al\u2019s frame for the round trip Al\u394T is given by: 2 BertAlAl 1\u3942\u3942\u394 \u239f\u23a0 \u239e\u239c\u239d \u239b\u2212== c vttT Substituting for Al\u394T in the expression for N yields: 2 Bert0 1\u3942 \u239f\u23a0 \u239e\u239c\u239d \u239b\u2212= c vtfN Substitute numerical values and evaluate N: ( )( ) 3 2 1 1007.1 600.01y667.6y1002 ×= \u239f\u23a0 \u239e\u239c\u239d \u239b\u2212= \u2212 c cN Special Relativity 1069 (d) Proceed as in Part (a) to find the rate at which Al receives signals as Bert is moving away from him: 1 Al y0.50 \u2212=f (e) Express the number of signals N received by Al: AlAl tfN \u394= Substitute for Alt\u394 to obtain: 2 BertAl 1\u394 \u239f\u23a0 \u239e\u239c\u239d \u239b\u2212= c vtfN Substitute numerical values and evaluate N: ( )( ) 267 600.01y667.6y0.50 2 1 = \u239f\u23a0 \u239e\u239c\u239d \u239b\u2212= \u2212 c cN (f) Express the total number of signals received by Al: return returnoutboundtot 267 N NNN += += The number, returnN , of signals received by Al on his return trip is given by: 2 Bertreturn Al, Alreturn Al,return 1\u394 \u394 \u239f\u23a0 \u239e\u239c\u239d \u239b\u2212= = c vtf tfN Substitute for returnN to obtain: 2 Bertreturn Al,tot 1\u394267 \u239f\u23a0 \u239e\u239c\u239d \u239b\u2212+= c vtfN Proceeding as in Part (a), find the rate at which signals are received by Al on the return trip: 1 return Al, y200 \u2212=f Substitute numerical values and evaluate totN : ( )( ) 321tot 1033.11334600.01y667.6y200267 ×==\u239f\u23a0\u239e\u239c\u239d\u239b\u2212+= \u2212 c cN (g) Their age difference is: AlBert ttt \u394\u2212\u394=\u394 Chapter R 1070 Substitute numerical values to obtain: y2.67 y 100 1067 y 100 1334 \u394 11 =\u2212= \u2212\u2212t and Bert.an younger thy 2.67 is Al Relativistic Energy and Momentum 39 \u2022 Find the ratio of the total energy to the rest energy of a particle of mass m moving with speed (a) 0.100c, (b) 0.500c, (c) 0.800c, and (d) 0.990c. Picture the Problem We can use Equation R-15 to find the ratio of the total energy to the rest energy for the given particle. From Equation R-15: 2 2 0 2 2 2 11 c v E c v mcE \u2212 = \u2212 = Solve for the ratio of E to E0 to obtain: 2 2 0 1 1 c vE E \u2212 = (a) Evaluate 0EE for v = 0.100c: ( ) 01.1100.01 1 2 2 100.00 = \u2212 =\u23a5\u23a6 \u23a4 = c cE E cv (b) Evaluate 0EE for v = 0.500c: ( ) 15.1500.01 1 2 2 500.00 = \u2212 =\u23a5\u23a6 \u23a4 = c cE E cv (c) Evaluate 0EE for v = 0.800c: ( ) 67.1800.01 1 2 2 800.00 = \u2212 =\u23a5\u23a6 \u23a4 = c cE E cv (d) Evaluate 0EE for v = 0.990c: ( ) 09.7990.01 1 2 2 999.00 = \u2212 =\u23a5\u23a6 \u23a4 = c cE E cv Special Relativity 1071 40 \u2022 A proton (rest energy 938 MeV) has a total energy of 1400 MeV. (a) What is its speed? (b) What is its momentum? Picture the Problem The rest energy E0 is equal to mc2. We are given E0 and E, where E0 = 938 MeV and the total energy E = 1400 MeV. (The total energy is the rest energy plus the kinetic energy). We can find the momentum p of the proton using 2 2 2 2 4E p c m c= + (Equation R-17), and once we have p we can solve for the speed v using v/c = pc/E (Equation R-16). (b) Use Equation R-17 to relate the momentum to the total energy and the rest energy: 42222 cmcpE += and 20 mcE = so 2 0 222 EcpE += \u21d2 c EE p 2 0 2 \u2212= Substitute numerical values and evaluate p: ( ) ( ) c c c p MeV/ 1040 MeV/ 1039 Mev 938Mev 1400 22 = = \u2212= (a) Use Equation R-16 to express the speed of the proton: E pc c v = \u21d2 c E pcv \u239f\u23a0 \u239e\u239c\u239d \u239b= Substitute numerical values and evaluate v: ( ) cccv 742.0 MeV 1400 MeV/ 1039 =\u239f\u23a0 \u239e\u239c\u239d \u239b= 41 \u2022 [SSM] How much energy would be required to accelerate a particle of mass m from rest to (a) 0.500c, (b) 0.900c, and (c) 0.990c? Express your answers as multiples of the rest energy, mc2. Picture the Problem We can use Equation R-14 to find the energy required to accelerate this particle from rest to the given speeds. From Equation R-14 we have: ( ) ( ) 2 2 2 2 2 1 1 1 1 mc cv mc cv mcK \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212 \u2212 = \u2212 \u2212 = Chapter R 1072 (a)Substitute numerical values and evaluate K(0.500c): ( ) ( ) 2 2 2 155.0 1 500.01 1500.0 mc mc cc cK = \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212 \u2212 = (b) Substitute numerical values and evaluate K(0.900c): ( ) ( ) 2 2 2 29.1 1 900.01 1900.0 mc mc cc cK = \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212 \u2212 = (c) Substitute numerical values and evaluate K(0.990c): ( ) ( ) 2 2 2 09.6 1 990.01 1990.0 mc mc cc cK = \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212 \u2212 = 42 \u2022 If the kinetic energy of a particle equals its rest energy, what percentage error is made by using p = mv for its momentum? Is the non- relativistic expression always low or high compared to the relativistically correct expression for momentum? Picture the Problem We can use Equations R-10 and R-14 to express the error made in using p = mv for the momentum of the particle when K = E0. The error in using p = mv for the momentum of the particle is given by: relrel rel 1 p p p pp \u2212=\u2212 (1) From Equation R-14 we have: ( ) ( ) 2 2 2 2 2 1 1 1 1 mc cv mc cv mcK \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212 \u2212 = \u2212 \u2212 = For K = E0: ( ) KcvK \u239f \u239f \u23a0 \u239e \u239c\u239c\u239d \u239b \u2212 \u2212 = 1 1 1 2 or ( ) 21 1 2 = \u2212 cv Special Relativity 1073 From Equation R-10, the relativistic momentum of the particle is: ( ) mvcv mvp 2 1 2 rel =\u2212 = Substitute in equation (1) and simplify to obtain: %50 2 11 rel =\u2212=\u2212 mv mv p p The ratio of the non-relativistic momentum of a particle to its relativistic momentum is given by: 22 22 rel 1 1 cv cv p p p p \u2212= \u2212 = Because 1 1 22 <\u2212 cv , the non-relativistic expression is always low compared to the relativistically correct expression for momentum. 43 \u2022 What is the total energy of a proton whose momentum is 3mc? Picture the Problem We can use Equation R-17 to find the total energy of any proton whose momentum is given. See Problem 42 for the rest energy of a proton. The total energy, momentum, and rest energy of the proton are related by Equation R-17: ( )22222 mccpE += Substitute for the momentum of the proton: ( ) ( ) 424242 22222 109 3 cmcmcm mccmcE =+= += Solving for E yields: 210mcE = Substitute for m and evaluate E: ( ) GeV97.2MeV/93810 22 == ccE 44 \u2022\u2022 Using a spreadsheet program or graphing calculator, make a graph of the kinetic energy of a particle with rest energy of 100 MeV for speeds between 0 and c. On the same graph, plot 221 mv by way of comparison. Using the graph, estimate at about what speed the non-relativistic expression is no longer a good approximation to the kinetic energy. As a suggestion, plot the energy in units of MeV and the speed in the dimensionless form v/c. Picture the Problem We can create a spreadsheet program to plot both the classical and relativistic kinetic energy of the particle. Chapter R 1074