Physics Solution Manual for 1100 and 2101

31c 065c v21v1e 1 1 2 c2 c

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 482 SPECIAL RELATIVITY SOLUTION Solving Equation 28.2 for v, we find that 2 2 L 0.500 m 8 v = c 1 − = (3.00 ×108 m/s) 1 − = 2.60 ×10 m/s L 1.00 m 0 ______________________________________________________________________________ 10. REASONING The distance between earth and the center of the galaxy is the proper length L0, because it is the distance measured by an observer who is at rest relative to the earth and the center of the galaxy. A person on board the spaceship is moving with respect to them and measures a contracted length L that is related to the proper length by Equation 28.2 as ( ) L = L0 1 − v 2 / c 2 . The contracted distance is also equal to the product of the spaceship’s speed v the time interval measured by a person on board the spaceship. This time interval is the proper time interval ∆t0 because the person on board the spaceship measures the beginning and ending events (the times when the trip starts and ends) at the same location relative to a coordinate system fixed to the spaceship. Thus, the contracted distance is also L = v∆t0. By setting the two expressions for L equal to each other, we can find the how long the trip will take according to a clock on board the spaceship. ( ) SOLUTION Setting L = L0 1 − v 2 / c 2 equal to L = v∆t0 and solving for the proper time interval ∆t0 gives ∆t0 = L0 v ( 1 − v2 / c2 ) 9.47 × 1015 m 1 ly ( 23 000 ly ) ( 0.9990c )2 1 yr = 1− = 1.0 × 103 yr 2 7 8 c 0.9990 3.00 × 10 m/s 3.16 × 10 s ______________________________________________________________________________ ( 11. ) SSM REASONING AND SOLUTION The length L0 that the person measures for the UFO when it lands is the proper length, since the UFO is at rest with respect to the person. Therefore, from Equation 28.2 we have L0 = L 2 = 230 m 2 = 530 m v (0.90 c) 1− 2 c c2 ______________________________________________________________________________ 1− Chapter 28 Problems 1483 12. REASONING The Martian measures the proper time interval ∆t0, because the Martian measures the beginning and ending events (the times when the trip starts and ends) at the same location relative to a coordinate system fixed to the spaceship. The given distance between Mars and Venus is the distance as measured by a person on earth. That person is at rest relative to the two planets and, hence, measures the proper length. The Martian, who is moving relative to the planets, does not measure the proper length, but measures a contracted length. According to the Martian, the time of the trip ∆t0 is equal to the contracted length that he measures divided by the speed v of the spaceship. SOLUTION a. The contracted length L measured by the Martian is related to the proper length L0 by Equation 28.2 as L = L0 1 − v2 c 2 ( = 1.20 × 10 m 11 ) ( 0.80c )2 1− c 2 = 7.2 × 1010 m b. The time of the trip as measured by the Martian is ∆t0 = L 7.2 × 1010 m 2 = = 3.0 × 10 s 8 v 0.80 3.00 × 10 m/s ( ) ______________________________________________________________________________ 13. REASONING Assume that traveler A moves at a speed of vA = 0.70c and traveler B moves at a speed of vB = 0.90c, both speeds being with respect to the earth. Each traveler is moving with respect to the earth and the distant star, so each measures a contracted length LA or LB for the distance traveled. However, an observer on earth is at rest with respect to the earth and the distant star (which is assumed to be stationary with respect to the earth), so he or she would measure the proper length L0. For each traveler the contracted length is given by the length-contraction equation as stated in Equation 28.2: LA = L0 1 − 2 vA c2 and LB = L0 1 − 2 vB c2 It is important to note that the proper length L0 is the same in each application of the lengthcontraction equation. Thus, we can combine the two equations and eliminate it. Then, since we are given values for LA, vA, and vB, we will be able to determine LB. 1484 SPECIAL RELATIVITY SOLUTION Dividing the expression for LB by the expression for LA and eliminating L0, we obtain LB = LA 1− LB = LA 1− L0 1 − L0 1 − 2 vB c2 = 2 vA c2 2 vB c 2 = 6.5 light-years ( ) 2 vA c2 1− 1− 2 vB c2 2 vA c2 ( ) 1 − 0.90c 2 c = 4.0 light-years ( 0.70c )2 1− c2 2 14. REASONING a. The two events are the creation of the particle and its subsequent disintegration. Relative to a stationary reference frame fixed to the laboratory, these two events occur at different locations, because the particle is moving relative to this reference frame. The proper −3 distance L0 is the distance (1.05 × 10 m) given in the statement of the problem, because this distance is measured by an observer in the laboratory who is at rest with respect to these locations. b. The distance measured by a hypothetical person traveling with the particle is a contracted distance, because it is measured by a person who is moving relative to the two locations. The contracted distance L is related to the proper distance L0 by the length-contraction formula, L = L0 1 − v 2 / c 2 (Equation 28.2). c. The proper lifetime ∆t0 of the particle is the lifetime as reg...
View Full Document

This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

Ask a homework question - tutors are online