1. An engineer in a train moving toward the station with a velocity v = 0.60c lights a signal flare as he reaches a marker 1.0 km from the station (according to a scale laid out on the ground). By how much time, on the stationmaster's clock, does the arrival of the optical signal precede the arrival of the train?
    Solution.
    The time required for the optical signal to reach the station, as measured by an observer at rest relative to the station is
    d/c = (1.0 km)/(3.00 x 105 km/s) = 3.33 µs.
    Measured by the same observer, the time needed for the train to arrive is 
    d/v = d/(0.60c) = (1.0 km)/(0.60 x 3.00 x 105 km/s) = 5.56 µs.
    The difference in the arrival times is
    5.56 µs - 3.33 µs = 2.2 µs.
  2. A body has a mass of 12.6 kg and a speed of 0.87c. (a) What is the magnitude of its momentum? (b) If a constant force of 16.4 x 108 N acts in the direction opposite to the body's motion, how long must the force act to bring the body to rest?
    Solution.
    (a) The magnitude of the momentum is calculated:
          p = γmv = [1 - (0.87c)2/c2]-1/2 x 12.6 kg x 0.87 x 3.00 x 105 km/s = 6.7 x 109 kg·m/s
    (b) To find the time needed to bring the body to rest, use the equation FnetΔt = Δp
         Δt = Δp/Fnet = (0 - 6.7 x 109 kg·m/s)/(-16.4 x 108 N) = 4.1 s.
  3. A muon with rest energy 106 MeV is created at an altitude of 4500 m and travels downward toward Earth's surface. An observer on Earth measures its speed as 0.980c. (a) What is the muon's total energy in the Earth's observer's frame of reference? (b) What is the muon's total energy in the muon's frame of reference?
    Solution.
    (a) The total energy is
          E = γmc2 = (1 - v2/c2)-1/2 x E0 = [1 - (0.980c)2/c2]-1/2 x 106 MeV = 530 MeV.
    (b) The muon is at rest in its own frame of reference. Its total energy is just its rest energy, 106 MeV.