Pré-visualização10 páginas

second wad of putty of mass m2 which is initially at rest. Do you expect that after the collision the combined putty mass will be (a) greater than, (b) less than, (c) the same as m1 + m2. Explain your answer. Determine the Concept Because the collision is inelastic, kinetic energy is converted into mass energy. ( )a is correct. 11 \u2022\u2022 [SSM] Many nuclei of atoms are unstable; for example, 14C, an isotope of carbon, has a half-life of 5700 years. (By definition, the half-life is the time it takes for any given number of unstable particles to decay to half that number of particles.) This fact is used extensively for archeological and biological dating of old artifacts. Such unstable nuclei decay into several decay products, each with significant kinetic energy. Which of the following is true? (a) The mass of the unstable nucleus is larger than the sum of the masses of the decay products. (b) The mass of the unstable nucleus is smaller than the sum of the masses of the decay products. (c) The mass of the unstable nucleus is the same as the sum of the masses of the decay products. Explain your choice. Determine the Concept Because mass is converted into the kinetic energy of the fragments, the mass of the unstable nucleus is larger than the sum of the masses of the decay products. ( )a is correct. 12 \u2022\u2022 Positron emission tomography (PET) scans are common in modern medicine. In this procedure, positrons (a positron has the same mass, but the opposite charge as an electron) are emitted by radioactive nuclei that have been introduced into the body. Assume that an emitted positron, traveling slowly (with Special Relativity 1045 negligible kinetic energy) collides with an electron traveling at the same slow speed in the opposite direction. They undergo annihilation and two quanta of light (photons) are formed. You are in charge of designing detectors to receive these photons and measure their energy. (a) Explain why you would expect these two photons to come off in exactly opposite directions. (b) In terms of the electron mass me , how much energy would each photon have? (1) less than mc2, (2) greater mc2, (3) exactly mc2. Explain your choice. Determine the Concept (a) The linear momentum of the electron-positron pair was zero just before their annihilation and the emission of the two photons and therefore the linear momentum of the two photons must be zero. (b) ( )3 is correct. Each photon has the rest mass energy, mec2, equivalent of an electron. Because there are two photons and they have the same momentum, they share equally, each getting half the available initial rest energy of 2mec2. Estimation and Approximation 13 \u2022\u2022 [SSM] In 1975, an airplane carrying an atomic clock flew back and forth at low altitude for 15 hours at an average speed of 140 m/s as part of a time- dilation experiment. The time on the clock was compared to the time on an atomic clock kept on the ground. What is the time difference between the atomic clock on the airplane and the atomic clock on the ground? (Ignore any effects that accelerations of the airplane have on the atomic clock which is on the airplane. Also assume that the airplane travels at constant speed.) Picture the Problem We can use the time dilation equation to relate the elapsed time in the frame of reference of the airborne clock to the elapsed time in the frame of reference of the clock kept on the ground. Use the time dilation equation to relate the elapsed time \u394t according to the clock on the ground to the elapsed time \u394t0 according to the airborne atomic clock: 2 0 1 \u394 \u394 \u239f\u23a0 \u239e\u239c\u239d \u239b\u2212 = c v tt (1) Because v << c, we can use the approximation x x 2 11 1 1 +\u2248\u2212 to obtain: 2 2 2 11 1 1 \u239f\u23a0 \u239e\u239c\u239d \u239b+\u2248 \u239f\u23a0 \u239e\u239c\u239d \u239b\u2212 c v c v Chapter R 1046 Substitute in equation (1) to obtain: 0 2 0 0 2 2 1 2 11 t c vt t c vt \u394\u239f\u23a0 \u239e\u239c\u239d \u239b+\u394= \u394\u23a5\u23a5\u23a6 \u23a4 \u23a2\u23a2\u23a3 \u23a1 \u239f\u23a0 \u239e\u239c\u239d \u239b+=\u394 (2) where the second term represents the additional time measured by the clock on the ground. Evaluate the proper elapsed time according to the clock on the airplane: ( ) s1040.5 h s3600h15 40 ×=\u239f\u23a0 \u239e\u239c\u239d \u239b=\u394t Substitute numerical values and evaluate the second term in equation (2): ( ) ns9.5s1089.5 s1040.5 m/s10998.2 m/s140 2 1 \u394 9 4 2 8 \u2248×= ×\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b ×= \u2212 t' 14 \u2022\u2022 (a) By making any necessary assumptions and finding certain stellar distances, estimate the speed at which a spaceship would have to travel for its passengers to make a trip to the nearest star (not the Sun!) and back to Earth in 1.0 Earth years, as measured by an observer on the ship. Assume the passengers make the outgoing and return trips at constant speed and ignore any effects due to the spaceship stopping and starting. (b) How much time would elapse on Earth during their roundtrip? Include 2.0 Earth years for a low-speed exploration of the planets in the vicinity of this star. Picture the Problem We can use the length-contraction equation ( )20 1 cvLL \u2212= and vLt =\u394 to estimate the speed of the spaceship. Note that there is no time dilation effect during the two-year exploration period. Take the distance to the nearest sun to be 4.0 c\u22c5y. (a) According to the travelers, the elapsed time for the trip is: ( ) v cvL v Lt 2 0 1\u394 \u2212== Multiplying both sides of the equation by c yields: ( ) ( ) cv cvL v cvcL tc 2 0 2 0 11\u394 \u2212=\u2212= Because c\u394t = 1.0 c\u22c5y and L0 = 4.0 c\u22c5y: ( )cv cv 210.4 0.1 \u2212= Special Relativity 1047 Solve for v to obtain: ccv 97.0 17 16 == (b) Express the elapsed time as the sum of the travel and exploration times: nexploratio wayl1 nexploratiotravel \u394\u3942 \u394\u394\u394 tt ttt += += (1) From the Earth frame-of reference: y 12.40.97 y 0.4 \u394 0 way1 =\u22c5== c c v Lt Substitute numerical values in equation (1) and evaluate \u394t: ( ) y 2.10y 0.2y 12.42\u394 =+=t 15 \u2022\u2022 (a) Compare the kinetic energy of a moving car to its rest energy. (b) Compare the total energy of a moving car to its rest energy. (c) Estimate the error made in computing the kinetic energy of a moving car using non-relativistic expressions compared to the relativistically correct expressions. Hint: Use of the binomial expansion may help. Picture the Problem We can compare these energies by expressing their ratios. Assume that the speed of the car is 30 m/s (\u2248 67 mi/h). (a) Express the ratio of the kinetic energy of a moving car to its rest energy: 2 2 2 2 1 2 2 1 \u239f\u23a0 \u239e\u239c\u239d \u239b== c v mc mv mc K For a car moving at 30 m/s: %105 m/s 10998.2 m/s 30 2 1 13 2 82 \u2212×\u2248 \u239f\u23a0 \u239e\u239c\u239d \u239b ×=mc K (b) Express the ratio of the total energy of a moving car to its rest energy: 500 000 000 000 000.1 122 2 2 \u2248 +=+= mc K mc mcK mc E (c) The error made in computing the kinetic energy of a moving car using the non-relativistic expression compared to the relativistically correct expression is given by: 1 \u394 rel rel-non rel relrel-non rel \u2212= \u2212= K K K KK K K (1) From Equation R-14, the relativistic kinetic energy is given by: 2 2 2 rel 11 2 1 mc c vK \u23a5\u23a5\u23a6 \u23a4 \u23a2\u23a2\u23a3 \u23a1 \u2212\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212= \u2212 (2) Chapter R 1048 Expand 2 1 2 2 1 \u2212 \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212 c v binomially to obtain: 4 4 2 2 4 4 2 2 2 2 8 3 2 11... 8 3 2 111 2 1 c v c v c v c v c v ++\u2248+++=\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212 \u2212 Substituting in equation (2) and simplifying yields: \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b += \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b+= \u23a5\u23a6 \u23a4\u23a2\u23a3 \u23a1 += 2 2 rel-non 2 2 22 2 4 4 2 2 rel 4 31 8 3 2 1 8 3 2 1 c vK c vmvmv mc c v c vK Substitute for Krel in equation (1) and simplify to obtain: 1 4 31 1 4 31 \u394 1 2 2 2 2 rel-non rel-non rel \u2212\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b += \u2212 \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b + = \u2212 c v c vK K K K Expanding 1 2 2 4 31 \u2212 \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b + c v binomially yields: ( ) ( )( ) \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b\u2212\u2248+\u239f\u239f\u23a0