( x vt) m (0.80)(3 10 m/s)( s) 1200 m m/s m/s m s 330 s c. 3.

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1 Solutions to HW 10 Problems and Exerises: 37.. Visualize: At t t t 0 s, the origins of the S, S, and S referene frames oinide. Solve: We have 1 ( v/ ) 1 (0.0) (a) Using the Lorentz transformations, 6 x ( xvt) m (0.0)(310 m/s)(.010 s) 100 m vx 6 (0.0)(310 m/s)(100 m) t t s.0 s 310 m/s (b) Using v, the above equations yield x 00 m and t.67 s assume that the axes of the two frames oinide at t=t = 0 se Solve: We have v In the earth s referene frame, the We Lorentz transformations yield x xvt ( ) m m/s m.3 10 m vx t t s 330 s m/s3.010 m m/s Model: S is the ground s frame and S is the roket s frame. S moves with veloity v 0.5 relative to S. Solve: (a) We have 1 v/ Applying the Lorentz transformations to the lightning strike at x 0 m and t 10 s, 5 x xvt (1.155) 0 m m/s0 s 173 m 1700 m vx 5 t t (1.155)(110 s 0 s) s 1 s For the lightning strike at x 30 km and t 10 s, 4 5 x m m/s 0 s 3.91 m 33 m 5 t s 46. s 46 s m/s3.010 m m/s (b) The events in the roket s frame are not simultaneous. The lightning is observed to strike the pole before the tree by s. Hene we ould infer, from diret appliation of Lorentz transformation (whih is less onfusing and more universal ), that simultaneity is relative.

2 Model: The roket and the earth are inertial frames. Let the earth be frame S and the roket be frame S. S moves with v 0. relative to S. The bullet s veloity in referene frame S is u 0.9. Solve: Using the Lorentz veloity transformation equation, u v u uv / / The bullet s speed is 0.36 along the x-diretion. Note that the veloity transformations use veloity, whih an be negative, and not speed. Hene it is absolutely important to define the positive diretion of motion. Is most ases the diretion of S w.r.t S is taken to be positive for bothe referene frames S and S Model: The proton and the earth (the Laboratory) are inertial frames. Let the earth be frame S and the proton be frame S. S moves with v 0.9. The eletron s veloity in the laboratory frame is 0.9. Notie that the diretion of motion of proton frame S w.r.t lab frame S is towards right. This diretion is taken to be positive for judging veloities in both S and S frames. Solve: Using the Lorentz veloity transformation equation, u v u uv / / The eletron s speed is Notie that under Galilean Transformation the veloity of the eletron w.r.t proton would exeed whih is impossible to observe Model: The earth and the other galaxy are inertial referene frames. Let the earth be frame S and the other galaxy be frame S. S moves with v 0.. The quasar s speed in frame S is u 0.. Solve: Using the Lorentz veloity transformation equation, u v u uv / / Assess: In Newtonian mehanis, the Galilean transformation of veloity would give u 0.6. Note: the diretion taken to be positive is away from the earth. Sine both move away from the earth hene we have substituted v 0. and u 0.. please keep a trak of the signs and while substituting in the formula take and given diretion to be positive, get your answer and depending on its sign reletive to your assumed positive diretion write down the motion (i.e, towards right/left, away/towards et.) Solve: (a) The relativisti momentum is kg m/s mu p 1 u / kg m/s In other words the relativisti momentum w.r.t a frame is the relativisti mass frame times the veloity of the body in that frame. Where m (b) The ratio of the relativisti momentum and the Newtonian momentum is prelativisti mu.4 p lassial 1 u / mu 1 u / 7 kg m0 m w.r.t that 1u is the rest mass.

3 Solve: The relativisti momentum is kg mu p 400,000 kg m/s 1u 1u 3 u kg u 3 u u 9 u 1 1 u ,000 kg m/s 4 16 Assess: In Newtonian mehanis, the momentum would be p mu ( kg)(0.0)( m/s) 40,000 kg m/s. Note: the partiles mass of 1 gram is atually the rest mass Model: The partile is highly relativisti sine u 0.. Solve: We have 5 5 p p 1 u K 1 E, where E 0 is The kineti energy is p 0 E 3 0 m kg m/s J J J K 3 The rest energy is E 0 m J. Here m is the rest mass. is the total relativisti energy whih is the sum of the rest mass energy E 0 and the kineti energy of p E 0 the partile. The total energy is E E 0 K J J J. u Solve: Ashown in the previous problem E p m E 0 K.E. For K.E E 0, the total energy of the moving partile in that frame is given by, E p m E 0 E 0 3m 1 u 1 p 3 1 u u / , This problemis similar to the previous problem, Solve: The total energy is E p m. For E E 0, E 0 p m p E 0 and p. Hene, 1 u 3 u 0 1 u / Model: Let the earth be referene frame S and let the spaeship be the referene frame S. S moves relative to S with speed v. Solve: For an observer in the earth s frame S, the length of the solar system is 10 lh. The time interval for the spaeship to ross is t 15 hours. The time interval measured in S is the proper time beause this an be measured with one lok at both positions (i.e., both edges of the solar system). The veloity v is 10 lh v lh/h 15 h 3 3

4 W.r.t the spae ship the two events (entering one side of the solar system and exiting from the other end ) happen at the same oordinate, i.e, fixed point. Hene the time for rossing the solar system as seen by the s[aeship is the proper time. Therefore the time witnessed by an observer from earth is the dilated time. Beause t, from Equation 36.9 for time dialationwe have t 1 15 h h Model: Let the earth be frame S and the train be frame S. S moves with veloity v 0.5 relative to S. The 30 m length is measured in the train s referene frame, frame S. Visualize: The light flashes at t t 0 s as the origins of S and S oinide. Solve: (a) For passengers on the train, light travels 15 m in both diretions at speed. The fat that the train is moving relative to the earth doesn t affet the speed of light. Thus the light flash arrives at both ends of the train simultaneously, ausing the bell and siren to be simultaneous. Sine the light flashed at t 0 s, the time of these two simultaneous events is t B t S (15 m)/(300 m/s) s. (b) The spaetime oordinates of the event bell rings are ( x B, t B ) (15 m, s). The oordinates of the event siren sounds are ( x S, t S ) ( 15 m, s). We an use the Lorentz time transformation to find the times of these events in frame S. To do so, we first need to alulate Consequently, the times are ( v / ) 1 (0.50) / ( / )( / ) t t vx t v x B B B B B s (0.50)((15 m)/(300 m/ s)) 0.07 s ts t S vx S/ t S ( v/ )( x S/ ) s (0.50)((15 m)/(300 m/ s)) 0.09 s Thus the siren sounds before the bell rings. The time interval between the two is t 0.07 s 0.09 s 0.05 s. This again illustrates the relativity of simultaneity Model: Let S be the galaxy s frame and S the alien spaeraft s frame. The spaetime interval s between the two events is invariant in all frames. Solve: (a) The light from Alpha s explosion will travel 10 ly in 10 years. Sine neither light nor any other signal from Alpha an travel 100 ly in 10 years to reah Beta, the explosion of Alpha ould not ause the explosion of Beta. (b) Beause the spaetime interval s between the two events is invariant,

5 1 ly 1 ly s t x t x t 10 y 100 ly 10 ly y y t 10 y 100 y 10 y t67.1 years This problem is an example of appliation of onservation of the spae-time interval. Eventhough a Lorentz transformation would be the more fundamental way of solving problems for moving frames but in this problem we do not have the neessary data given to us (for eg; the veloity of alien spaeraft w.r.t the galaxy) to arry out the transformation. Hene we are needed to use the spaetime interval onservation formula Model: The spaetime interval s between the two events is invariant in all frames. Solve: (a) Equating the two spaetime intervals, 300 m 300 m t x t x t 10 s 0 m 400 m s s 10 s s t1. s t (b) Note that t beause the event ours at the same point in spae. Hene, there is time dilation w.r.t the moving S frame. 10 s t 1. s v v/ 1 v / Model: The earth is frame S and the starship is frame S. S moves relative to S with a speed v. Solve: (a) The speed of the starship is 0 ly 0 y v y 5 y (b) The astronauts measure the proper time while they are traveling. This is v 1 t y 15 y Beause the explorers stay on the planet for one year, the time elapsed on their hronometer is 16 years Model: S is the eletron s frame and S is the ground s frame. S moves relative to S with a speed v Solve: For an experimenter in the S frame, the length of the aelerator tube is 3. km. This is the proper length L beause it is at rest and is always there for measurements. The eletron measures the tube to be length ontrated to L m0.7 m Model: Let the earth be frame S and the roket be frame S. S moves with speed v relative to S. Solve: (a) The round-trip distane is 60 ly. If the roket takes time t to make the round trip, as measured on earth, its speed (as a fration of ) is v 60 ly 60 yr t t

6 where we used 1 ly/yr (1 light year per year). The astronaut s elapsed time t is the proper time, so 0 yr. The time dilation equation is 0 yr t 1 (60 yr/ t) ( 0 yr/ t) 1 ( v / ) 1 (60 yr/ t) Solving for t gives t y, and thus v 60 y y v (b) The roket starts with rest energy E i m and aelerates to have energy E f p m. Thus the energy needed to aelerate the roket is E E f E 1 ( p 1)m This is just the kineti energy K gained by the roket. We know the roket s speed, so 1 E 1 ( ) 1 (0,000 kg)( m/s) J () The total energy used by the United States in 000 was J. To aelerate the roket would require roughly 760 times the total energy used by the United States Model: Let S be the earth s referene frame and S be the referene frame of one roket. S moves relative to S with v The speed of the seond roket in the frame S is u Visualize: Solve: Using the Lorentz veloity transformation equation, u v u uv Assess: In Newtonian mehanis, the Galilean transformation of veloity will give u 0.75 (0.75) This is not permissible aording to the theory of relativity. Note: we have hosen one diretion (right) to be positive and stik to it in both the frames Model: Let S be the sun s referene frame and S be the roket s referene frame. S moves with speed v 0. relative to S. The flare s speed in the frame S is u 0.9. Visualize:

7 Solve: Using the Lorentz veloity transformation equation, uv u 0.36 uv That is, the flare is approahing the roket at a speed of Note: we have hosen one diretion (right) to be positive and stik to it in both the frames Model: Use Equations and 37.4 for the momentum and total energy. Also, the quantity E 0 m is an invariant in all inertial referene frames. Solve: (a) The momentum and energy are 7 1 p mu mu kg m/s kg m/s 1 u E m m 1 u kg m/s 1.0 J 7 (b) From Equation 36.45, E p E m Thus, J p m/s 10 E 7 m kg m/s p kg m /s kg m/s Solve: The rest energy and the total energy are given by Equations and We have m p m e m p u m e 1 u / Model: Mass and energy are equivalent and given by Equation Solve: (a) The sun radiates energy for s per year. The amount of energy radiated per year is Sine E 0 m, the amount of mass lost is ( J/s)( s) J/y 34 E J m m/s kg kg (b) Sine the mass of the sun is kg, the sun loses % of its mass every year. () The lifetime of the sun an be estimated to be

8 kg kg/y T years The sun will not really last this long in its urrent state beause fusion only takes plae in the ore and it will beome a red giant when the ore hydrogen is all fused Model: Partiles an be reated from energy, and partiles an return to energy. When a partile and its antipartile meet, they annihilate eah other and reate two gamma ray photons. Solve: The energy of the eletron is E m kg m/s 31 eletron p e J The energy of the positron is the same, so the total energy is E total E eletron E positron J. The energy is onverted to two equal-energy photons. Thus, E total hf h. The wavelength is total 34 h J s 3 10 m/s 13 E J Assess: This wavelength is typial of -ray photons m 1 pm