1 Chapter Homework Solutions
1. All else being equal, relativistic phenomena would be less conspicuous if the speed of light were larger. We would have to obtain an even larger speed than we do now for relativistic effects to manifest themselves.
2.
t = t0
To the observer, the watch in the rest frame of the runner records a 10% smaller time so t0=0.9t and =1/0.9=1.11 so and v=(1-.81)1/2c=0.44c v2 1- 2 = 0.9 c 3. Assume the car on earth is moving slowly compared to the speed of light so the speed of the spacecraft relative to the car is .6C In eq. 1.3 t = 50 mins is the time measured by observer in the moving spacecraft. While t0 is the time the driver notices to elapse during the course of the trip so, t0= t* sqrt(1-v2/c2) = 40 minutes. 4. The proper astronaut's length (height) is 5 ft, which is what an observer in the spaceship would measure. The external observer measures a shorter length (from Eq. 1.9) L = L0 / L =(1 - L0 5. v 2 1/ 2 ) = - .642 )1/ 2 = ft 5(1 3.8 2 c
6.
7.
8.
2 E = mc=
1 = v2 1- 2 c
1 = 1.15 (0.5c) 2 1- c2
so E = 1.15x0.938 GeV=1.08 GeV
KE =- 1)mc 2 = ( .15(.938) = GeV = (mc 2 ) 2 + p 2 c 2 0.14 E2 p 2 c 2 =2 - (mc 2 ) 2 = 2 - .9382 = GeV 2 E 1.15 .44 p = 0.67GeV / c 9.
KE = ( - 1)mc 2 = KE / m = ( - 1)c 2 = 9 x1016 ((1/ 1 - .7 2 ) - 1) = 3.6 x1016 J / kg
10.
KE = - 1)mc 2 = 2 ((1/ 1 - 2 ) - 1) ( mc KE.2 / KE.1 (1/ 1 - .22 ) - 1) /(1/ 1 - .12 ) -= .0206 / .00504 4.1 = 1) = KE.4 / KE.1 (1/ 1 - .42 ) - 1) /(1/ 1 - .12 ) -= .0901/ .00504 18 = 1) =