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UH PHYS 1302 - Ch29

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Chapter 29: RelativityAnswers to Even-Numbered Conceptual Questions 2. The second postulate of relativity specifically refers to the speed of light in a vacuum. The speed of light inother mediums will always be less than the speed of light in a vacuum.4. If the speed of light were only 35 mi/h, we would experience relativistic effects everyday. For example, acommuter would age more slowly than a person who works at home; a moving car would be noticeablyshorter and distorted; you wouldn’t be able to drive faster than 35 mi/h, no matter how powerful your carwas and no matter how long you held the “pedal to the metal.”6. No, in both cases. The theory of relativity imposes no limits on the energy or momentum an object can have.Solutions to Problems and Conceptual Exercises1. Picture the Problem: You are in a spaceship, traveling directly away from the Moon with a speed of 0.9c. A light signalis sent in your direction from the surface of the Moon.Strategy: Use Einstein’s second postulate to answer the conceptual question.Solution: 1. (a) A light signal will always pass your ship with a speed of c, regardless of the speed of the ship. Therefore, you measure the speed to be greater than 0.1c. 2. (b) The best explanation is I. The speed you measure will be greater than 0.1c; in fact, it will be c, since all observers in inertial frames measure the same speed of light. Statements II and III are each false.Insight: Statement III is the most correct answer from the standpoint of classical mechanics.2. Picture the Problem: Two ships approach each other at a relative speed of 0.90c. Light emitted from one of the ships travels to the second ship.Strategy: Use the second postulate of relativity to determine the speed of the light beam in all reference frames.Solution: Because the speed of light in vacuum is c, Isaac measures the speed of Albert’s light beam as c .Insight: The speed of light in a vacuum is the same to all observers in any inertial frame of reference.3. Picture the Problem: A street performer tosses a ball straight up into the air (event 1) and then catches it in his mouth (event 2).Strategy: Recall that the proper time is the time between two events that occur at the same location, as seen by a given observer. Solution: 1. (a) The street performer sees the two events occur at the same location (his body) and therefore observes proper time.2. (b) A stationary observer on the other side of the street sees the two events occur at the same location (the street performer’s body) and therefore observes proper time.3. (c) A person sitting at home watching the performance on TV sees the two events occur at the same location (the television set) and therefore observes proper time.4. (d) A person observing the performance from a moving car sees the two events occur at different locations relative to his car and therefore observes dilated time.Insight: What if the TV camera in part (c) were moving at high speed relative to the street performer? The viewer would still see the proper time interval because of issues involved in the transmission of the information from the TV Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.29 – 1Chapter 29: Relativity James S. Walker, Physics, 4th Editioncamera at the finite speed of light. The Special Theory of Relativity cannot be thwarted by a television broadcast!Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.29 – 2Chapter 29: Relativity James S. Walker, Physics, 4th Edition4. Picture the Problem: An observer determines that a clock in a moving (relative to the observer) rocket is running slow.Strategy: Use the principle of time dilation to answer the conceptual question.Solution: 1. (a) If the rocket reverses direction the clock will be observed to run slow. The direction of motion does not matter in time dilation, only the relative speed between the observer and the clock.2. (b) The best explanation is I. The clock will run slow, just as before. The rate of the clock depends only on relative speed, not on direction of motion. Statements II and III are each false.Insight: Each observer in separate frames of reference will observe the other person’s clock to be running slow, independent of the direction of motion.5. Picture the Problem: As a salesman you travel on a spaceship that reaches speeds near the speed of light, and you are paid by the hour. Strategy: Use the principle of time dilation to answer the conceptual question.Solution: 1. (a) You would definitely want to be paid according to the clock at Spacely Sprockets universal headquarters on Earth. The clock in the spaceship will run slow compared to the headquarters clock, and hence it would give you a much smaller paycheck.2. (b) The best explanation is I. You want to be paid according to the clock on Earth, because the clock on the spaceshipruns slow when it approaches the speed of light. Statements II and III are each false.Insight: Statement II is true as long as you are moving relative to the Earth. As soon as you slow down and land on Earth, you re-enter Earth’s frame of reference and discover that the spaceship clock has been running slow. The dispute over which clock is running slow is unsolvable as long as both observers determine that their frame of reference is at rest. As soon as one observer experiences the inertial forces due to slowing down or speeding up, the symmetry is broken and the two observers will agree on who is moving and who is not.6. Picture the Problem: An astronaut travels at a high rate of speed while observing the periodic flashes of a neon sign.Strategy: The two events to be considered in this case are two successive flashes of the neon sign. From the point of view of the astronaut, these events occur at different places, so the astronaut observes the dilated time interval t and theobserver on Earth measures the proper time interval t0 = 4.1s. Use equation 29-2 to calculate t.Solution: Find the time interval observed by the astronaut: Δt =Δt01−v2c2=4.1 s1− 0.84( )2= 7.6 sInsight: Time


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UH PHYS 1302 - Ch29

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