Solve using the lorentz velocity transformation

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Solve: Using the Lorentz velocity transformation equation, ( )( ) 2 2 0.8 0.2 0.71 1 / 1 0.8 0.2 / u v c c u c uv c c c c ′ = = = Assess: In Newtonian mechanics, the Galilean transformation of velocity would give u = 0.6 c .
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37.34. Solve: (a) The relativistic momentum is ( ) ( ) ( ) ( ) 27 8 17 2 2 2 1.67 10 kg 0.999 3.0 10 m/s 1.12 10 kg m/s 1 / 1 0.999 mu p u c × × = = = × (b) The ratio of the relativistic momentum and the Newtonian momentum is relativistic 2 2 2 2 classical 1 1 22.4 1 / 1 / p mu p mu u c u c = = =
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37.35. Solve: The relativistic momentum is ( ) 3 2 2 2 2 2 3 2 2 2 2 2 1.0 10 kg 400,000 kg m/s 1 1 1.0 10 kg 3 9 1 1 0.80 400,000 kg m/s 4 16 u mu p u c u c u u u u u c u c c c c c c × = = × = = = = Assess: In Newtonian mechanics, the momentum would be p = mu = (1.0 × 10 3 kg)(0.80)(3.0 × 10 8 m/s) = 240,000 kg m/s.
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37.36. Solve: The Newtonian momentum is p Newton = mu . We have 2 2 2 2 1 3 2 1 / 0.866 4 2 1 / mu p mu u c u c c u c = = = = =
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37.37. Solve: We have 2 2 2 2 2 2 2 1 1 0.707 2 1 mu u u c p mc u c u c c c u c = = = = = = Assess: The particle’s momentum being equal to mc does not mean that the particle is moving with the speed of light. We must use the relativistic formula for the momentum as the particle speeds become high.
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37.38. Model: The particle is highly relativistic since u = 0.8 c . Solve: We have ( ) p p 2 2 2 1 1 5 5 2 1 1 3 3 3 1 1 0.80 u c γ γ = = = = = The kinetic energy is ( ) p 0 1 K E γ = , where E 0 is ( )( ) 2 2 3 8 13 0 1.0 10 kg 3.0 10 m/s 9.0 10 J E mc = = × × = × ( ) 13 13 2 9.0 10 J 6.0 10 J 3 K = × = × The rest energy is E 0 = mc 2 = 9.0 × 10 13 J. The total energy is E = E 0 + K = 9.0 × 10 13 J + 6.0 × 10 13 J = 1.5 × 10 14 J.
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37.39. Model: The hamburger is a classical particle whose rest energy is E 0 = mc 2 . Solve: (a) We have ( )( ) 2 2 3 8 16 0 200 10 kg 3.0 10 m/s 1.8 10 J E mc = = × × = × (b) The ratio of the energy equivalent to the food energy is 16 9 6 1.8 10 J 9.0 10 2 10 J × = × ×
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37.40. Solve: The rest energy and the total energy are given by Equations 37.43 and 37.42. We have m p c 2 = γ m e c 2 p 2 2 e 1 1833 1 / m m u c γ = = = u = 0.99999985 c
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37.41. Solve: Equation 37.42 is E = γ p mc 2 = E 0 + K . For K = 2 E 0 , γ p mc 2 = E 0 + 2 E 0 = 3 mc 2 2 p 2 2 2 1 1 8 3 1 0.943 9 3 1 / u u c c c u c γ = = = = =
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37.42. Solve: The total energy is E = γ p mc 2 . For E = 2 E 0 , 2 E 0 = γ p mc 2 = γ p E 0 and γ p = 2. Hence, 2 2 2 2 1 3 2 0.866 4 1 / u u c c u c = = =
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37.43. Model: Let S be the laboratory frame and S be the reference frame of the 100 g ball. S moves to the right with a speed of v = 2.0 m/s relative to frame S. The 50 g ball’s speed in frame S is u 50 = 4.0 m/s. Because these speeds are much smaller than the speed of light, we can use the Galilean transformations of velocity. Visualize: Solve: Transform the collision from frame S into frame S , where 100 initial 0 m/s u = . Using the Galilean velocity transformation, 50 initial 50 initial 4.0 m/s 2.0 m/s 2.0 m/s u u v = = = Using Equation 10.43, ( ) 50 final 50 initial 50 g 100 g 1 2.0 2.0 m/s m/s 50 g 100 g 3 3 u u = = − = − + ( ) ( ) 100 final 50 initial 2 50 g 2 4.0 2.0 m/s m/s 50 g 100 g 3 3 u u = = = + +
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