Conceptual Physics (12th Edition)

Published by Addison-Wesley
ISBN 10: 0321909100
ISBN 13: 978-0-32190-910-7

Chapter 35 - Think and Explain - Page 684-685: 87


This is a writing exercise for the student, and answers will vary.

Work Step by Step

Sample 1: An alien in a 77-meter-diameter flying saucer whizzes by at v = 0.866c. A clock on the flying saucer ticks off 100 seconds. What is the time interval that you measure? a. 50 seconds b. 100 seconds c. 150 seconds d. 200 seconds The answer is d, 200 seconds. Find $\gamma$ at v = 0.866c. It is $\frac{1}{\sqrt{1-v^{2}/c^{2}}}$, or $\frac{1}{\sqrt{1-0.866^{2}}}$, which is 2. This is the time dilation factor - you measure a longer time interval that is twice the proper time interval. Sample 2: An alien in a 77-meter-diameter flying saucer whizzes by at v = 0.99c. What is the diameter of the flying saucer that you measure? a. 11 meters b. 77 meters c. 836 meters d. 76.2 meters The answer is a. The 77-meter craft is measured to be about 11 meters in diameter when it is moving at 0.99c. Let $L_{o}$ represent the original, rest length of the craft, and L the measured length of the moving craft. $$L = L_{o} \sqrt{1-\frac{ v^{2} } { c^{2} }} $$ $$L = (77 meters) \sqrt{1-0.99^{2}} = 10.9 meters $$ This is discussed on page 676. Alternatively, find $\gamma$ at v = 0.99c. It is $\frac{1}{\sqrt{1-v^{2}/c^{2}}}$, or $\frac{1}{\sqrt{1-0.99^{2}}}$, which is 7.09. The factor $\gamma$ is the length contraction factor. Divide the diameter by $\gamma$ to find the contracted length. Sample 3: An alien in a 33-meter-diameter flying saucer whizzes by at v = 0.995c. Its momentum is NOT mv. Which expression below represents its true momentum? a. 2 mv b. mv/10 c. 10 mv d. 0.995 mv The answer is c, 10 mv. The relativistic formula for momentum, $p = \gamma mv$, is greater than the classical formula $p = mv$. Find $\gamma$ at v = 0.995c. It is $\frac{1}{\sqrt{1-v^{2}/c^{2}}}$, or $\frac{1}{\sqrt{1-0.995^{2}}}$, which is 10. Sample 4: A 1-gram fuel pellet is completely transformed into energy in a futuristic reactor. How much energy is released? a. 0 b. $9.0 \times 10^{13} J$ c. $3.0 \times 10^{5} J$ d. $9.0 \times 10^{16} J$ The answer is b, $9.0 \times 10^{13} J$. The mass that is transformed to energy is 0.001 kg. The energy released by this transformation is $E = mc^{2} = (0.001 kg)(3.00 \times 10^{8} m/s)^{2} = 9.0 \times 10^{13} J$.
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