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Unformatted text preview: Phys 263/Amath 261 Assignment 8 SOLUTIONS 1. (a) The speed of the second ship is defined on the first ship by u = x t (1) where x is the measured length of the 2nd ship. This is length contracted by x = 1 L (2) where L is the proper length of 100m. Thus u = 1 100 2 . 5 10 6 = radicalbigg 1- u 2 c 2 100 2 . 5 10 6 (3) Solving for u gives 3 . 965 10 7 m/sec. (b) In this case, we are solving for u = x t (4) where x is now the length of the first ship in its own reference frame, i.e. L . So t = 100 3 . 965 10 7 (5) giving 2 . 522 6 s. 2. This question is the relativistic doppler effect for receding bodies. The effect for two bodies in motion towards each other was derived in class. (a) When the source and receiver are receding from each other, the derivation is similar, with one important exception. Now the length of the wavetrain measured in S , or the distance between the beginning and end of the waves is length = c t + v t (6) This can be put into the derivation as done in class, to get...
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This note was uploaded on 09/20/2010 for the course AMATH 261 taught by Professor Rogermelko during the Spring '10 term at Waterloo.
- Spring '10