1041_PartUniversity Physics Solution

1041_PartUniversity Physics Solution - 37-4 Chapter 37 v +...

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37-4 Chapter 37 EXECUTE: 22 0.950 0.650 0.300 0.784 . 1 / 1 (0.950 )( 0.650 )/ 1 0.6175 x x x vu c c c vc uv c c c c +− + == ++ The speed of the second particle, as measured in the laboratory, is 0.784 c . EVALUATE: The incorrect Galilean expression for the relative velocity gives that the speed of the second particle in the lab frame is 0.300 c . The correct relativistic calculation gives a result more than twice this. 37.20. IDENTIFY and SET UP: Let S be the laboratory frame and let S be the frame of one of the particles, as shown in Figure 37.20. Let the positive x direction for both frames be from particle 1 to particle 2. In the lab frame particle 1 is moving in the + x direction and particle 2 is moving in the x direction. Then 0.9520 uc = and 0.9520 =− . v is the velocity of particle 2 relative to particle 1. EXECUTE: 0.9520 0.9520 0.9988 1 / 1 (0.9520 )( 0.9520 )/ c c uv c c c c −− = . The speed of particle 2 relative to particle 1 is 0.9988 c . 0 v ′ < shows particle 2 is moving toward particle 1. Figure 37.20 37.21. IDENTIFY: The relativistic velocity addition formulas apply since the speeds are close to that of light. SET UP: The relativistic velocity addition formula is 2 1 x x x v uv c ′ = . EXECUTE: In the relativistic velocity addition formula for this case, x v is the relative speed of particle 1 with respect to particle 2, v is the speed of particle 2 measured in the laboratory, and u is the speed of particle 1 measured in the laboratory, u = – v . 2 () 2 1( ) 1 x vv v v vvc v c + . 2 2 20 x x v v c −+ = and 3 (0.890 ) 2 (0.890 ) 0 cv c −+ = . This is a quadratic equation with solution v = 0.611 c ( v must be less than c ). EVALUATE: The nonrelativistic result would be 0.445 c , which is considerably different from this result. 37.22. IDENTIFY and SET UP: Let the starfighter’s frame be S and let the enemy spaceship’s frame be S . Let the positive x direction for both frames be from the enemy spaceship toward the starfighter. Then 0.400 =+ . 0.700 ′=+ . v is the velocity of the missile relative to you. EXECUTE: (a) 2 0.700 0.400 0.859 1 / 1 (0.400)(0.700) c c uv c = (b) Use the distance it moves as measured in your frame and the speed it has in your frame to calculate the time it takes in your frame. 9 8 8.00 10 m 31.0 s (0.859)(3.00 10 m/s) t × × . 37.23. IDENTIFY and SET UP: The reference frames are shown in Figure 37.23. S = Arrakis frame S = spaceship frame The object is the rocket. Figure 37.23 u is the velocity of the spaceship relative to Arrakis. 0.360 ; 0.920 xx (In each frame the rocket is moving in the positive coordinate direction.)
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Relativity 37-5 Use the Lorentz velocity transformation equation, Eq.(37.22): 2 . 1/ x x x vu v uv c ′ = EXECUTE: 22 2 so and 1 xx x x x x x x x v u vv u v u u v v uv c c c ′′ ⎛⎞ =− = = ⎜⎟ ⎝⎠ 0.360 0.920 0.560 0.837 1 / 1 (0.360 )(0.920 )/ 0.6688 c c c uc vv c c c c −− == = = The speed of the spacecraft relative to Arrakis is 8 0.837 2.51 10 m/s.
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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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1041_PartUniversity Physics Solution - 37-4 Chapter 37 v +...

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