QUIZ 1 CH. 37---
discussion questions--1,3,7,8,14
     exercises--37.2: 1; 37.3: 2,4,6,8; 37.4: 9,10,11,12,13.
     problems---Note: They  tend to combine (i.e., synthesize)
the first 4 sections-49(a) only, 50, 51, 52, 54.
EVEN ANSWERS
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CHECK BACK FOR COMPLETED DISCUSSIONS ! Note: In this problem set we refer to the mirror clock. Whichever reference frame has the mirror clock experiences the proper time interval since the  light beam  returns to the same spot within the clock, like measuring the time interval at the same location in space. 
1. This problem expresses what I said in class:  you can re-do thought experiment of figure 37.10 by assuming the sources of lightening are hardwired to the moving train instead of the train station as assumed by   book. As we mentioned in that case, Mavis would experience  simultaneity and Stan  would not--- the 2  lightening signals would meet at Mavis and Mavis would be to the right of Stan.  Since the 2 light signals also started equidistant from Stan and they both approach him at the same universal  speed of light , he can only assume the signal coming from his right was launched  later than the signal coming from the left.
2.  (a) See example 1 and 2. In this case, the muon  has a proper time of 2.2 x 10 -6 s based on a "mirror-clock" (the clock in figure 37.6 attached to train)  located on the muon. So in a frame in which that  clock is moving,  the time interval will be  LONGER  by the factor  1/(1 - v2/c2)1/2  
(b) distance = 0.900*c*time. What value of time do you use? The time interval in the laboratory's rest frame  or the time interval  the muon's rest frame? Think about it.  
4. (a)  Where's the mirror-clock? It must be on the Mars, where the time measurement was made.
(b) Since the   clock is on Mars, a reference frame moving relative to Mars,   i.,e spaceship, would measure a  larger time interval  by a factor of  1/(1 - v2/c2)1/2   .  
6. (a) We assume the mirror-clock is hardwired to the race ship. Thus any time interval measured by SUV (Space Utility Vehicle)  will be  longer than  the time interval measured by an observer on  race ship. Another way of looking at it: If the SUV measures  a proper distance 1.20x10 8 m , the distance measured on the race ship will be shorter by the length contraction  formula.  The time on the race ship will be also shorter since time = contracted length/0.800c
(b) The distance = contracted length = 0.800c*time in part (a)
(c) The time interval read by the clock on SUV is longer than the value in part (a) by a factor  1/(1 - v2/c2)1/2   .
8. (a) We assume the mirror-clock is hardwired to the space craft since its time 0.0120 s < 0.190 s. The proper time is the smaller time interval.
(b)  Earth time interval = spacecraft time interval / (1 - v2/c2)1/2  . Square both sides and find v/c.  

9.  Once the spacecraft lands the scientist will measure a longer length, because there was contraction while the craft was moving. Once craft lands, the scientist would measure what a person on the moving craft would have measured.

 10.  You are on Earth.
Earth  measured length  = meter stick proper length * (1 - v2/c2)1/2  . Square both sides and find v/c and v.

11. See example 1 and conceptual example 37.6.
 2.20x 10 -6 s  is the proper lifetime time of the muon, where the mirror clock resides. Assume v = 0.999c. below.
General comments: Assume you are in the Earth reference frame: Over the lifetime of the muon, the muon travels distance = 0.999c*earth time, where earth measured lifetime is longer than the proper lifetime by factor
 1/(1 - v2/c2)1/2  . This provides context to part (b), at rest relative to Earth. Part(a) assumes you are in a reference frame at rest relative to muon.
(c) The muon measures a contracted length smaller than 10 km by factor  (1 - v2/c2)1/2  .  In the muon reference frame, the time to travel that length should be less than the proper lifetime 2.20x 10 -6 s  .

12. (a) Earth measured time  = distance /speed
(b) particle frame  measured length  =
Earth frame length * (1 - v2/c2)1/2  .
(c) The problem gives two hints~~in both the mirror clock is attached to the particle:
Method i:  muon measured time = 
                   Earth measured time*(1 - v2/c2)1/2   since clock on muon.
 Method ii: muon measured time  = contracted distance/speed.

13.
(a) Use length contraction formula to get a smaller value.
(b)  time = distance/speed
(c) time = contracted distance/speed.

49 (a)  v = contracted length/(proper lifetime). Note: contracted length = 1.20 km*(1 - v2/c2)1/2 .  
Thus: v = 1.20 km*(1 - v2/c2)1/2 /(2.6x10- 8 s) . Square both sides and solve for v or use the hint in the textbook to simplify the expression first before solving for v algebraically.
50. See figure 37. 12.  Only the dimension parallel to the motion will be contracted, so your new volume will contain a factor (1 - v2/c2)1/2  .   This problem explains why the charge density of a wire  is relative since the  volume changes for motion along the wire; charge density = charge/volume
51. You want a = b = 1.40*b*(1 - v2/c2)1/2  . Find v.
52. added time = contracted length/speed.  total age = 19yr + added time
54.  NY time = 4.00 h = plane time/(1 - v2/c2)1/2   or (4.00 h)*(1 - v2/c2)1/2 = plane time. Perform a binomial expansion and compute 4.00 h - plane time.
MORE DISCUSSIONS TO COME !! CHECK BACK LATER