Phys 181b Chapter 37 Solutions

# Phys 181b Chapter 37 Solutions - 1 From the time dilation...

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1. From the time dilation equation t = γ t 0 (where t 0 is the proper time interval, γβ =− 11 2 /, and β = v / c ), we obtain F H G I K J 1 0 2 t t . The proper time interval is measured by a clock at rest relative to the muon. Specifically, t 0 = 2.2000 µ s. We are also told that Earth observers (measuring the decays of moving muons) find t = 16.000 s. Therefore, 2 2.2000 s 1 0.99050. 16.000 s §· = ¨¸ ©¹

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10. Only the “component” of the length in the x direction contracts, so its y component stays == ° = AAA yy sin . 30 05000m while its x component becomes =− = ° = AA A xx 1 30 1 0 90 0 3775 22 β cos . . m. Therefore, using the Pythagorean theorem, the length measured from S' is () () 2 2 0.63m. xy ′′ =+ = A
11. The length L of the rod, as measured in a frame in which it is moving with speed v parallel to its length, is related to its rest length L 0 by L = L 0 / γ , where γβ =− 11 2 / and β = v / c . Since must be greater than 1, L is less than L 0 . For this problem, L 0 = 1.70 m and = 0.630, so L = 170 1 0 630 132 2 .. . mm . b g b g

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12. (a) We solve Eq. 37-13 for v and then plug in: 2 2 0 1 1 1 0.866. 2 L L β §· =− = = ¨¸ ©¹ (b) The Lorentz factor in this case is () 2 1 2.00 1/ vc γ == .
15. (a) Let d = 23000 ly = 23000 c y, which would give the distance in meters if we included a conversion factor for years seconds. With t 0 = 30 y and t = d / v (see Eq. 37-10), we wish to solve for v from Eq. 37-7. Our first step is as follows: 0 22 23000 y 30 y , 11 t d t v β ββ ∆= = ¡ = −− at which point we can cancel the unit year and manipulate the equation to solve for the speed parameter . This yields () 2 1 0.99999915. 1 30/ 23000 == + (b) The Lorentz factor is 2 1/ 1 766.6680752 γβ =− = . Thus, the length of the galaxy measured in the traveler’s frame is 0 23000 ly 29.99999 ly 30 ly.

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Phys 181b Chapter 37 Solutions - 1 From the time dilation...

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