Physics Solution Manual for 1100 and 2101

1 solution using a value of m 00280 kgmol we obtain

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: ( )2 = (1.50 km ) 1 − 0.609c = 1.19 km c2 c2 ______________________________________________________________________________ L = L0 1 − 2 vAB 41. REASONING The following relative velocities are pertinent to this problem. It is assumed that particle 1 is moving in the positive direction, and, therefore, particle 2 is moving in the negative direction. vP P = velocity of particle 1 (P1) relative to particle 2 (P2) 12 vP L = velocity of particle 1 (P1) relative to an observer in the Laboratory 1 vP L 2 = +2.10 × 108m/s = velocity of particle 2 (P2) relative to an observer in the Laboratory 8 = –2.10 × 10 m/s 1504 SPECIAL RELATIVITY The velocity vP P can be obtained from the velocity-addition formula, Equation 28.8: 12 vP L + vLP vP P = 12 1 1+ 2 vP L vLP 1 2 c2 The velocity vP L is given, but vLP , the velocity of the laboratory observer relative to 1 2 particle 2, is not. However, we know that vLP is the negative of vP L , so vLP = − vP L 2 = − ( −2.10 × 10 m/s ) = + 2.10 × 10 m/s. 8 2 2 2 8 SOLUTION a. According to the velocity-addition formula, the velocity of particle 1 relative to particle 2 is vP P = 12 vP L + vLP 1 1+ 2 vP L vLP 1 2 c 2 +2.10 × 108 m/s + 2.10 × 108 m/s = ( +2.10 × 10 8 1+ )( m/s +2.10 × 10 m/s 8 ) = +2.82 × 108 m/s c2 The speed of one particle as seen by the other particle is the magnitude of this result, or 2.82 × 108 m/s . b. The relativistic momentum is given by Equation 28.3, where the speed is that determined in part a. Therefore, p= mvP P 12 2 ( 2.16 × 10–25 kg ) ( +2.82 × 108 m/s ) = = 2 1.8 × 10 –16 kg ⋅ m/s vP P 2.82 × 108 m/s 1– 1– 1 2 8 c 3.00 × 10 m/s ______________________________________________________________________________ 42. REASONING AND SOLUTION The mass equivalent is given by E0 = KE = mc2 or m= KE 7.8 × 10 –13 J = c2 3.00 × 108 m/s ( ) 2 = 8.7 × 10 –30 kg (28.5) ______________________________________________________________________________ Chapter 28 Problems 43. 1505 SSM WWW REASONING AND SOLUTION The diameter D of the planet, as measured by a moving spacecraft, is given in terms of the proper diameter D0 by Equation 28.2. Taking the ratio of the diameter DA of the planet measured by spaceship A to the diameter DB measured by spaceship B, we find vA 2 (0.60 c) 2 DA c2 = c2 = = 1.3 DB vB2 (0.80 c) 2 1− D0 1 − 2 c2 c ______________________________________________________________________________ D0 1 − 1− 44. REASONING The magnitude of the relativistic momentum p of the proton is related to its relativistic total energy E by E 2 = p2c2 + m2c4 (Equation 28.7), where m = 1.67×10−27 kg is the mass of a proton (see the inside front cover of the text) and c is the speed of light in a vacuum. SOLUTION Solving Equation 28.7 for p, we obtain p 2 c 2 = E 2 − m2 c 4 or p2 = E2 − m2 c 2 2 c or p= E2 − m 2c 2 2 c Therefore, the magnitude of the relativistic momentum of the proton is ( 2.7 ×10−10 J ) − 1.67 ×10−27 kg 2 3.00 ×108 m/s 2 = 7.5×10−19 kg ⋅ m/s )( ) 2( ( 3.00 ×108 m/s) 2 p= 45. REASONING The total time for the trip is one year. This time is the proper time interval ∆t0, because it is measured by an observer (the astronaut) who is at rest relative to the beginning and ending events (the times when the trip started and ended) and who sees them at the same location in spacecraft. On the other hand, the astronaut measures the clocks on earth to run at the dilated time interval ∆t, which is the time interval of one hundred years. The relation between the two time intervals is given by Equation 28.1, which can be used to find the speed of the spacecraft. 1506 SPECIAL RELATIVITY SOLUTION The dilated time interval ∆t is related to the proper time interval ∆t0 by ( ) ∆t = ∆t0 / 1 − v 2 / c 2 . Solving this equation for the speed v of the spacecraft yields 2 2 1 yr ∆t (28.1) v = c 1− 0 = c 1− = 0.999 95c ∆t 100 yr ______________________________________________________________________________ 46. REASONING AND SOLUTION In all parts of this problem, the direction of the intergalactic cruiser, the ions, and the laser light is taken to be the positive direction. a. According to the second postulate of special relativity, all observers measure the speed of light to be c, regardless of their velocities relative to each other. Therefore, the aliens aboard the hostile spacecraft see the photons of the laser approach at the speed of light, c . b. To find the velocity of the ions relative to the aliens, we define the relative velocities as follows: vIS = velocity of the Ions relative to the alien Spacecraft vIC = velocity of the Ions relative to the intergalactic Cruiser = +0.950c vCS = velocity of the intergalactic Cruiser relative to the alien Spacecraft = +0.800c These velocities are related by the velocity-addition formula, Equation 28.8. The velocity of the ions relative to the alien spacecraft is: vIS = vIC + vCS +0.950c + 0.800c = = +0.994c vIC vCS ( +0.950c )( +0.800c ) 1+ 1+ 2 2 c c c. The aliens see the laser light (photons) moving with respect to the c...
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