Time Dilation and Length Contraction Show that the experiment depicted in Figure

Question

Time Dilation and Length Contraction Show that the experiment depicted in Figure 2.11 and discussed in the text leads directly to the derivation of length contraction. A rocket ship carrying passengers blasts off to go from New York to Los Angeles, a distance of about 5000 km. (a) How fast must the rocket ship go to have its own length shortened by 1%? (b) Ignore effects of general relativity and determine how much time the rocket ship's clock and the ground-based clocks differ when the rocket ship arrives in Los Angeles. Astronomers discover a planet orbiting around a star similar to our sun that is 20 lightyears away. How fast must a rocket ship go if the round trip is to take no longer than 40 years in time for the astronauts aboard? How much time will the trip take as measured on Earth? Particle physicists use particle track detectors to determine the lifetime of short-lived particles. A muon has a mean lifetime of 2.2 mu s and makes a track 9.5 cm long before decaying into an electron and two neutrinos. What was the speed of the muon?

Explanation / Answer

19)d = 5000 km

a)intial length = L0 = 5000 km

l = 1% of 5000 km = 50 km

Observed length = L = Lo - l = 5000 - 50 = 4950 km

We know that the formula of length contaction is:

L = L0 x sqrt(1 - v^2/c^2)

L/Lo = sqrt(1 - v^2/c^2)

4950/5000 = sqrt (1 - v^2/c^2)

0.9801 = (1 - v^2/c^2)

v^2/c^2 = 1 - 0.9801 = 0.0199

v = sqrt (0.0199 c^2) = 0.141 c

Hence, v = 0.141 c

b)s = d/t => t = d/s

t = 5000 x 10^3 m/0.141 x 3 x 10^8 m/s = 0.118 s

from the formula of time dilatation,

T = T0/sqrt (1 - v^2/c^2)

T = 0.118/sqrt (1 - 0.141^2c^2/c^2) = 0.119 s = 120 ms

delta T = 1.2 ms

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