Ch.39 | 6 | 11 | 17 | 15 | 22 | 23 | 25 | 43 | 48 | 50 |
6. See lecture notes from the day that I derived the time relationship using a space ship and mirror system with flash light. That system was moving past a person at rest in the space station and we got the relationship between the times in the ship and in the station.... | ||||||||||
11. This is easy once you
think about it for about 30 minutes (on the average.......) The quantity you have is the time measured on earth. v = length/time ...where you have the time. The only thing you need now is the length, which you must write in terms of the speed v since the length observed by the earth observer is contracted: length = 300·(1 -v2/c2)½ . You will have to set v = 300·(1 -v2/c2)½ /time and square both sides to find v.... |
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17. (a) 20 m...why? (b) 19 m ....why??? (c) 19 = 20·(1 -v2/c2)½ |
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15. 10 m is measured
by us sitting in the laboratory as the particle whizzes by us at a great
speed!!!! time - 26ns (nanoseconds) 10 m = v·time/(1 -v2/c2)½ |
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22. 0.950c = ux = expression on the right hand side of equation 39.18. Solve for ux' . You can also use equation 39.16 and get the same answer for ux' . Try them both!!!! |
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23. Observers on the earth see two jets. One (jet 1) moves to the left at
speed 0.75c relative to the earth and the other (jet 2) moves to the right at speed 0.75c
relative to the earth. Let S' represent the reference frame of the earth. Let S
represent the reference frame of jet 1. Then the velocity of jet 2 as measured by jet 1
is: Here, ux ' = +0.75c = the velocity of jet 2 measured by an observer on earth. v is the velocity component of S' (Earth) relative to S (jet1). Note: v = +0.75c |
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43. This is a tough problem! From
conservation of energy, we have: Substituting the momentum equation into the energy equation, we get: Now, we simplify the notation by using these definitions: |
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48. The momentum of the system is zero. Thus, the Fe nucleus has an equal
in magnitude but oppositely directed momentum compared to that of the photon. Since the
photon has energy 14-KeV, its momentum is 14 KeV/c. Thus, the magnitude of the momentum of
the nucleus is also: . Now, from eqn. 39.26: Here p = the momentum of the nucleus and m is the mass. Substituting into this equation the expression for the total energy in terms of the kinetic energy K we get: Now, on both sides of the equation, factor out mc2 Take the square root of both sides: Apply the binomial theorem to the right hand side. See page B.6 in appendix B of the text on series expansions. We assume that: Solve for K. Note that pc = 14-KeV |
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50. 2me c2 =1.02 MeV. This gives the lower limit on
the energy of the gamma ray. |