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# solutions5 - Homework Set 5 Due Thursday 07/28 1 Time...

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Homework Set 5 Due Thursday, 07/28 1. Time dilation and length contraction Problem 1: A spaceship travels from Earth to a star 100 light-years away. The ship is moving at a constant speed of 0.85 c . (a) How long does the trip take according to an observer on Earth? (b) How long does the trip take according to an observer on board the spaceship? (c) What is the distance from Earth to the star, according to the observer on the spaceship? Solution: (a) According to the observer on Earth, the ship moves at a speed of 0.85 c = 0.85 light- years per year. The time for the trip is therefore 100ly / 0.85 ly/yr = 118 yrs. (b) The time according to the observer on Earth is larger than the proper time measured by an observer on the spaceship due to time dilation: The time for an observer on the spaceship is thus only 62 years. (c) The observer on the spaceship moves at a speed of 0.85 c relative to the star and to Earth, and takes 62 years to get there. Therefore, according to him, the distance from Earth to the star is (62 yr) x (0.85 ly / yr) = 53 light-years. One could use length contraction to get the same result. Problem 2: Unstable particles are produced in a nuclear reaction in a laboratory. The particles have an average lifetime of 1.0 x 10 -10 seconds when at rest. The particles travel an average distance of 5 cm from the point of production before decaying, according to an observer at rest in the laboratory. What is the speed of the particles relative to the stationary observer? Solution: The lifetime of the particles is longer in the frame where they are moving. The range is therefore given by

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Solve for the speed v : Problem 3: Suppose that at t = 0, Bob synchronizes his watch with Alice, and gets on a high-speed train that travels around and around the Earth at a speed of 500 km/h . If the train never stops, how long would Bob need to stay on it before his watch was off from Alice's by one second, due to time dilation? Express the answer in years. Solution: Alice sees Bob's clock run slower; the amount of time that passes for Alice is bigger by a factor of Thus Alice's watch accumulates an extra 10 -13 seconds or so for every second Bob is on the train. Thus it would take 10 13 seconds for the watches to be off by a second. In years, this is Thus the high-speed train is not a very practical time machine. It would take over 300,000 years to generate a time difference of 1 second between someone on board the train and an inertial observer. 2. Velocity addition Problem 4: Alice observes a particle moving with a speed of 0.8 c in the x direction, while Bob observes the same particle moving with a speed of 0.5 c in the x direction. How fast is Bob moving relative to Alice? Solution: Let v be the speed of Bob relative to Alice, v B be the speed of the particle relative to Bob, and v A be the speed of the particle relative to Alice. Then, by the velocity addition equation,
We know v A and v B , and need to solve for v : Bob is thus moving at a speed of 0.5 c relative to Alice.

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solutions5 - Homework Set 5 Due Thursday 07/28 1 Time...

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