1046_PartUniversity Physics Solution

# 1046_PartUniversity Physics Solution - Relativity 37-9 2 3...

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Relativity 37-9 EXECUTE: 22 3 (1 ) 4 mc mv γ −= . 13 1 4 1 x x . 2 1 14 x x ⎛⎞ =+ ⎜⎟ ⎝⎠ . After a little algebra this becomes 2 91 5 8 0 xx +− = . ( ) 2 1 15 (15) 4(9)(8) 18 x =− ± + . The positive root is 0.425 x = . / xvc = , so 0.652 vx c c == . EVALUATE: The fractional increase of the relativistic expression above the nonrelativistic one increases as v increases. 37.46. The fraction of the initial mass (a) that becomes energy is 3 (4.0015 u) 1 6.382 10 , 2(2.0136 u) × and so the energy released per kilogram is 38 2 1 4 (6.382 10 )(1.00 kg)(3.00 10 m s) 5.74 10 J. ×× = × (b) 19 4 14 1.0 10 J 1.7 10 kg. 5.74 10 J kg × × 37.47. (a) 2 6 8 2 9 , (3.8 10 J) (2.998 10 m s) 4.2 10 kg Em cmE c = × × = × . 1 kg is equivalent to 2.2 lbs, so 6 4.6 10 m tons (b) The current mass of the sun is 30 1.99 10 kg, × so it would take it 30 9 20 13 (1.99 10 kg) (4.2 10 kg s) 4.7 10 s 1.5 10 years = × = × to use up all its mass. 37.48. IDENTIFY: Since the final speed is close to the speed of light, there will be a considerable difference between the relativistic and nonrelativistic results. SET UP: The nonrelativistic work-energy theorem is 0 11 F x mv mv Δ= , and the relativistic formula for a constant force is 2 ) Fx m c . (a) Using the classical work-energy theorem and solving for x Δ , we obtain 9 8 2 0 6 ( ) (0.100 10 kg)[(0.900)(3.00 10 m s)] 3.65 m. ( 1 . 0 0 1 0 N ) mv v x F −× × = = × (b) Using the relativistic work-energy theorem for a constant force, we obtain 2 ) . mc x F For the given speed, 2 1 2.29, 10 . 9 0 0 thus 98 2 6 (2.29 1)(0.100 10 kg)(3.00 10 m s) 11.6 m. (1.00 10 N) x × = × EVALUATE: (c) The distance obtained from the relativistic treatment is greater. As we have seen, more energy is required to accelerate an object to speeds close to c , so that force must act over a greater distance. 37.49. (a) IDENTIFY and SET UP: 8 0 2.60 10 s t × is the proper time, measured in the pion’s frame. The time measured in the lab must satisfy , dc t where . uc Calculate t Δ and then use Eq.(37.6) to calculate u . EXECUTE: 3 6 8 1.20 10 m 4.003 10 s 2.998 10 m/s d t c × = = × × 0 1/ t t Δ so 1 / 2 0 / ) t t Δ Δ and 2 0 / ) t t Δ Δ Write ) Δ so that 2 (/) ( 1 ) 12 12 Δ=−Δ + Δ≈−Δ since Δ is small. Using this in the above gives 2 0 1( 12) t t Δ −−Δ= Δ 2 2 8 5 0 6 1 1 2.11 10 4 . 0 0 3 1 0 s t t Δ× = = × EVALUATE: An alternative calculation is to say that the length of the tube must contract relative to the moving pion so that the pion travels that length before decaying. The contracted length must be 88 0 (2.998 10 m/s)(2.60 10 s) 7.79 m. lc t =Δ = × × = 0 ll so 2 0 l l

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37-10 Chapter 37 Then (1 ) uc =− Δ gives 2 2 5 3 0 1 1 7.79 m 2.11 10 , 2 2 1.20 10 m l l ⎛⎞ Δ= = = × ⎜⎟ × ⎝⎠ which checks. (b) IDENTIFY and SET UP: 2 Em c γ = (Eq.(37.38).
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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1046_PartUniversity Physics Solution - Relativity 37-9 2 3...

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