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( )2 2 2 1 cv mcmcK \u2212 =+ Solving for ( )21 cv\u2212 yields: ( ) K mc K mc cv 2 2 2 1 1 + =\u2212 (2) Evaluate K mc2 to obtain: ( )( ) 11048.5 GeV 1035.3 MeV/amu 932amu 197 3 4 2 <<×= ×= \u2212 K mc Because mc2 << K, equation (2) can be approximated by: ( ) K mccv 2 21 \u2248\u2212 (3) Solve for v/c to obtain: 22 1 \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b\u2212= K mc c v Chapter R 1086 Expanding the radical binomially yields: sorder termhigher 2 111 2222 2 1 + \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b\u2212=\u23a5\u23a5\u23a6 \u23a4 \u23a2\u23a2\u23a3 \u23a1 \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b\u2212 K mc K mc Because mc2 << K: 2222 2 111 2 1 \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b\u2212\u2248\u23a5\u23a5\u23a6 \u23a4 \u23a2\u23a2\u23a3 \u23a1 \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b\u2212 K mc K mc and \u23a5\u23a5\u23a6 \u23a4 \u23a2\u23a2\u23a3 \u23a1 \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b\u2212= 22 2 11 K mccv Substituting for v in equation (1) yields: c K mc K mcccv 22 22 2 1 2 11\u394 \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b= \u23a5\u23a5\u23a6 \u23a4 \u23a2\u23a2\u23a3 \u23a1 \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b\u2212\u2212= Substitute numerical values and evaluate \u394v: ( ) km/s .504m/s 10998.2 GeV 1035.3 MeV/amu 932amu 197 2 1 \u394 8 2 4 =×\u239f\u23a0 \u239e\u239c\u239d \u239b × ×=v (b) Because the nuclei are, to three significant figures, traveling at the speed of light: s 334.0 m/s102.998 m 100\u394 \u394 8 \u3bc=×=\u2248 c xt 56 \u2022\u2022\u2022 Consider the flight of a beam of neutrons produced in a nuclear reactor. These neutrons have kinetic energies of up to 1.00 MeV. The rest energy of a neutron is 939 MeV. (a) What is the speed of 1.00 MeV neutrons? Express your answer in terms of v/c. (b) If the average lifetime of such a neutron is 15.0 min, what is the maximum length of a beam of such neutrons (in a vacuum, in the absence of any interactions between the neutrons and other material)? Estimate this maximum range by calculating the length corresponding to five mean lifetimes. After five mean lifetimes only e\u22125 or 0.007 (0.7%) of the neutrons remain. (c) Compare this to the range of so-called \u2033thermally moderated\u2033 neutrons, whose kinetic energies are around 0.025 eV. Express your answer as a percentage. That is, what percent of the 1.00 MeV-neutron range is the thermally moderated-neutron range? (Note that our assumption of a vacuum continues, however in reality neutrons of this energy interact readily with matter, such as air or water, and \u2033real\u2033 ranges are much shorter.) Special Relativity 1087 Picture the Problem We can apply ( )222 1 cvmcmcK \u2212=+ , the expression for the total relativistic energy of a particle, to find the speed of 1.00 MeV neutrons. We can use the equation relating distance, rate, and time to estimate their maximum beam length. Finally, we can use the non-relativistic expression for their kinetic energy to find the range of \u2033thermally moderated\u2033 neutrons. (a) The total relativistic energy of a neutron is given by Equation R-15: ( )2 2 2 1 cv mcmcK \u2212 =+ Solving for v/c yields: 2 21 11 \u239f\u239f \u239f\u239f \u23a0 \u239e \u239c\u239c \u239c\u239c \u239d \u239b + \u2212= mc Kc v Substitute numerical values and evaluate v/c: %61.4 MeV 939 MeV 00.11 11 2 = \u239f\u239f \u239f\u239f \u23a0 \u239e \u239c\u239c \u239c\u239c \u239d \u239b + \u2212= c v (b) Express the maximum expected beam length of the neutrons as a function of the number n of lifetimes: tc c vnx \u394\u394 \u239f\u23a0 \u239e\u239c\u239d \u239b= (1) Evaluate this expression for n = 5: ( )( )( ) m 106.22 min s 60min 0.15m/s 102.9980461.05\u394 108 ×=\u239f\u23a0 \u239e\u239c\u239d \u239b ××=x (c) Express the ratio r of the range of \u2033thermally moderated\u2033 neutrons to the range of 1.00 MeV neutrons: MeV 1.00 moderated thermally \u394 \u394 x x r = (2) Because the energy of the \u2033thermally moderated\u2033 neutrons is 0.025 eV, we can use the non-relativistic equation for their kinetic energy: 2 2 2 12 2 1 c c vmmvK \u239f\u23a0 \u239e\u239c\u239d \u239b== \u21d2 22mc K c v = Substitute for v/c in equation (1) to obtain: tcmc Knx \u3942\u394 2moderated thermally = Chapter R 1088 Substitute numerical values and evaluate moderated thermally \u394x : ( ) ( )( ) m 1084.9min 0.15m/s 102.998 MeV 939 eV 025.025\u394 68moderated thermally ×=×=x Substitute numerical values in equation (2) and evaluate r: %016.0m1022.6 m 1084.9 10 6 =× ×=r 57 \u2022\u2022\u2022 You and Ernie are trying to fit a 15-ft-long ladder into a 10-ft-long shed with doors at each end. You suggest to Ernie that you open the front door to the shed and have him run toward it with the ladder at a speed such that the length contraction of the ladder shortens it enough so that it fits in the shed. As soon as the back end of the ladder passes through the door, you will slam it shut. (a) What is the minimum speed at which Ernie must run to fit the ladder into the shed? Express it as a fraction of the speed of light. (b) As Ernie runs toward the shed at a speed of 0.866c, he realizes that in the reference frame of the ladder, it is the shed which is shorter, not the ladder. How long is the shed in the rest frame of the ladder? (c) In the reference frame of the ladder is there any instant that both ends of the ladder are simultaneously inside the shed? Examine this from the point of view of relativistic simultaneity. Picture the Problem Let the letter \u2033L\u2033 denote the ladder and the letter \u2033S\u2033 the shed. We can apply the length contraction equation to the determination of the minimum speed at which Ernie must run to fit the ladder into the shed as well as the length of the shed in rest frame of the ladder. (a) Express the length LL of the shed in Ernie\u2019s frame of reference in terms of its proper length LL,0: 2 0L,L 1 \u239f\u23a0 \u239e\u239c\u239d \u239b\u2212= c vLL \u21d2 2 L,0 2 L1 L Lcv \u2212= Substitute numerical values and evaluate v: ccv 75.0 ft15 ft101 2 =\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b\u2212= (b) Express the length LS of the shed in the rest frame of the ladder in terms of its proper length LL,0: 2 2 0S,S 1 c vLL \u2212= Substitute numerical values and evaluate LS: ( ) ft0.5866.01ft10 2S =\u239f\u23a0 \u239e\u239c\u239d \u239b\u2212= c cL (c) No. In your rest frame, the back end of the ladder will clear the door before the front end hits the wall of the shed, while in Ernie\u2019s rest frame, the front end will Special Relativity 1089 hit the wall of the shed while the back end has yet to clear the door. Let the ladder be traveling from left to right. To \u2033explain\u2033 the simultaneity issue we first describe the situation in the reference frame of the shed. In this frame the ladder has length LL = 7.5 m and the shed has length LS,0 = 10 m. We (mentally) put a clock at each end of the shed. Let both clocks read zero at the instant the left end of the ladder enters the shed. At this instant the right end of the ladder is a distance LS, 0 \u2212 LL = 2.5 m from the right end of the shed. At the instant the right end of the ladder exits the shed both clocks read \u394t, where ( ) ns63.9\u394 LS,0 =\u2212= vLLt . There are two space-time coincidences to consider: the left end of the ladder enters the shed and the clock at the left end of the shed reads zero, and the right end of the ladder exits the shed and the clock on the right end of the shed reads ( ) ns63.9LS,0 =\u2212 vLL . In the reference frame of the ladder the two clocks are moving to the left at speed v = 0.866c. In this frame the clock on the right (the trailing clock) is ahead of the clock on the left by ns, 9.282S,0 =cvL so when the clock on the right reads 9.63 ns the one on the left reads \u201319.3 ns. This means the left end of the ladder is yet to enter the shed when the right end of the ladder is exiting the shed. This is consistent with the assertion that in the rest frame of the ladder, the ladder is longer than the shed, so the entire ladder is never entirely inside the shed. Chapter R 1090