1041_PartUniversity Physics Solution

# 1041_PartUniversity Physics Solution - 37-4 Chapter 37 v u...

This preview shows pages 1–3. Sign up to view the full content.

37-4 Chapter 37 E XECUTE : 2 2 0.950 0.650 0.300 0.784 . 1 / 1 (0.950 )( 0.650 )/ 1 0.6175 x x x v u c c c v c uv c c c c ′ + + = = = = − + + The speed of the second particle, as measured in the laboratory, is 0.784 c . E VALUATE : 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. I DENTIFY and S ET U P : 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 u c = and 0.9520 v c = − . v is the velocity of particle 2 relative to particle 1. E XECUTE : 2 2 0.9520 0.9520 0.9988 1 / 1 (0.9520 )( 0.9520 )/ v u c c v 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. I DENTIFY : The relativistic velocity addition formulas apply since the speeds are close to that of light. S ET U P : The relativistic velocity addition formula is 2 1 x x x v u v uv c ′ = . E XECUTE : 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 2 ( ) 2 1 ( ) 1 x v v v v v v c v c − − ′ = = − − + . 2 2 2 0 x x v v v v c + = and 2 2 3 (0.890 ) 2 (0.890 ) 0 c v c v c + = . This is a quadratic equation with solution v = 0.611 c ( v must be less than c ). E VALUATE : The nonrelativistic result would be 0.445 c , which is considerably different from this result. 37.22. I DENTIFY and S ET U P : 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 u c = + . 0.700 v c ′ = + . v is the velocity of the missile relative to you. E XECUTE : (a) 2 0.700 0.400 0.859 1 / 1 (0.400)(0.700) v u c c v 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. I DENTIFY and S ET U P : 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 x x v c v c = + = + (In each frame the rocket is moving in the positive coordinate direction.)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Relativity 37-5 Use the Lorentz velocity transformation equation, Eq.(37.22): 2 . 1 / x x x v u v uv c ′ = E XECUTE : 2 2 2 so and 1 1 / x x x x x x x x x x x v u v v v v v v u v u u v v uv c c c = = = 2 2 0.360 0.920 0.560 0.837 1 / 1 (0.360 )(0.920 )/ 0.6688 x x x x v v c c c u c v v c c c c = = = − = − The speed of the spacecraft relative to Arrakis is 8 0.837 2.51 10 m/s. c = ×
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern