Ch37Word - Chapter 37 p.1 CHAPTER 37 Special Theory of...

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Chapter 37 p.1 CHAPTER 37 – Special Theory of Relativity 1. ( a ) [1 – ( v / c ) 2 ] 1/2 = {1 – [(20,000 m/s)/(3.00 × 10 8 m/s)] 2 } 1/2 = 1.00 . ( b ) [1 – ( v / c ) 2 ] 1/2 = [1 – (0.0100) 2 ] 1/2 = 0.99995 . ( c ) [1 – ( v / c ) 2 ] 1/2 = [1 – (0.100) 2 ] 1/2 = 0.995 . ( d ) [1 – ( v / c ) 2 ] 1/2 = [1 – (0.900) 2 ] 1/2 = 0.436 . ( e ) [1 – ( v / c ) 2 ] 1/2 = [1 – (0.990) 2 ] 1/2 = 0.141 . ( f ) [1 – ( v / c ) 2 ] 1/2 = [1 – (0.999) 2 ] 1/2 = 0.0447 . 2. You measure the contracted length. We find the rest length from L = L 0 [1 – ( v / c ) 2 ] 1/2 ; 28.2 m = L 0 [1 – (0.750) 2 ] 1/2 , which gives L 0 = 42.6 m . 3. We find the lifetime at rest from ? t = ? t 0 /[1 – ( v 2 / c 2 )] 1/2 ; 4.76 × 10 –6 s = ? t 0 /{1 – [(2.70 × 10 8 m/s)/(3.00 × 10 8 m/s)] 2 } 1/2 , which gives ? t 0 = 2.07 × 10 –6 s . 4. You measure the contracted length: L = L 0 [1 – ( v / c ) 2 ] 1/2 = (100 ly){1 – [(2.50 × 10 8 m/s)/(3.00 × 10 8 m/s)] 2 } 1/2 = 55.3 ly . 5. We determine the speed from the time dilation: ? t = ? t 0 /[1 – ( v 2 / c 2 )] 1/2 ; 4.10 × 10 –8 s = (2.60 × 10 –8 s)/[1 – ( v / c ) 2 ] 1/2 , which gives v = 0.773 c . 6. We determine the speed from the length contraction: L = L 0 [1 – ( v / c ) 2 ] 1/2 ; 25 ly = (75 ly)[1 – ( v / c ) 2 ] 1/2 , which gives v = 0.94 c . 7. For a 1.00 per cent change, the factor in the expressions for time dilation and length contraction must equal 1 – 0.0100 = 0.9900: [1 – ( v / c ) 2 ] 1/2 = 0.9900, which gives v = 0.141 c . 8. In the Earth frame, the clock on the Enterprise will run slower. ( a ) We find the elapsed time on the ship from ? t = ? t 0 /[1 – ( v 2 / c 2 )] 1/2 ; 5.0 yr = ? t 0 /[1 – (0.84) 2 ] 1/2 , which gives ? t 0 = 2.7 yr . ( b ) We find the elapsed time on the Earth from ? t = ? t 0 /[1 – ( v 2 / c 2 )] 1/2 = (5.0 yr)/[1 – (0.84) 2 ] 1/2 = 9.2 yr . 9. ( a ) To an observer on Earth, 95.0 ly is the rest length, so the time will be t Earth = L 0 / v = (95.0 ly)/0.960 c = 99.0 yr . ( b ) We find the dilated time on the spacecraft from ? t = ? t 0 /[1 – ( v 2 / c 2 )] 1/2 ; 99.0 yr = ? t 0 /[1 – (0.960) 2 ] 1/2 , which gives ? t 0 = 27.7 yr . ( c ) To the spacecraft observer, the distance to the star is contracted: L = L 0 [1 – ( v / c ) 2 ] 1/2 = (95.0 ly)[1 – (0.960) 2 ] 1/2 = 26.6 ly . ( d ) To the spacecraft observer, the speed of the spacecraft is v = L /? t = (26.6 ly)/27.7 yr = 0.960 c , as expected.
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Chapter 37 p.2 10. ( a ) You measure the contracted length. We find the rest length from L = L 0 [1 – ( v / c ) 2 ] 1/2 ; 4.80 m = L 0 [1 – (0.660) 2 ] 1/2 , which gives L 0 = 6.39 m . Distances perpendicular to the motion do not change, so the rest height is 1.25 m . ( b ) We find the dilated time in the sports vehicle from ? t = ? t 0 /[1 – ( v 2 / c 2 )] 1/2 ; 20.0 s = ? t 0 /[1 – (0.660) 2 ] 1/2 , which gives ? t 0 = 15.0 s . ( c ) To your friend, you moved at the same relative speed: 0.660 c . ( d ) She would measure the same time dilation: 15.0 s . 11. In the Earth frame, the average lifetime of the pion will be dilated: ? t = ? t 0 /[1 – ( v 2 / c 2 )] 1/2 . The speed as a fraction of the speed of light is v / c = d / c ? t = d [1 – ( v 2 / c 2 )] 1/2 / c ? t 0 ; v / c = (15 m)[1 – ( v 2 / c 2 )] 1/2 /(3.00 × 10 8 m/s)(2.6 × 10 –8 s), which gives v = 0.89 c = 2.7 × 10 8 m/s . 12. With the standard orientation of the reference frames, the Galilean transformation is x = x + vt , y = y , z = z . ( a ) For the given data we have x = 25 m + (30 m/s)(2.5 s) = 100 m; y = 20 m; z = 0. Thus the coordinates of the person are (100 m, 20 m, 0). ( b ) For the given data we have x = 25 m + (30 m/s)(10.0 s) = 325 m; y = 20 m; z = 0.
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Ch37Word - Chapter 37 p.1 CHAPTER 37 Special Theory of...

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