Solutions to Week 11 Assignment

1. We determine the speed from the time dilation:

                  Dt = Dt₀/[1 - (v²/c^2<)]^1/2;

                 4.10 x 10^-8s = (2.60 x 10^-8s)/[1- (v²/c²)]^1/2, which gives v = 0.773c.

2. (a) To an observer on Earth, 76.0ly is the rest length, so the time will be

                  t_Earth = L₀/v = (75.0ly)/0.950c = 78.9yr.

    (b) We find the dilated time on the spacecraft from

                  Dt = Dt₀/[1 - (v²/c^2<)]^1/2;

                 78.9yr = Dt₀/[1 - (0.950)²]^1/2, which gives Dt₀ = 24.6yr.

    (c) To the spacecraft observer, the distance to the star is contracted:

                L = L₀[1 - (v²/c²)]^1/2 = (75.0ly)[1 - (0.950)²]^1/2 =  23.4ly.

    (d) To the spacecraft observer, the speed of the spacecraft is

               v = L/Dt = (23.4ly)/24.6yr = 0.950c, as expected.

3. We find the speed from

              m = m₀/[1 - (v²/c²)]^1/2;

             1.15m₀ = m₀/[1 - (v²/c²)]^1/2, which gives v = 0.494c.

4. (a) The kinetic energy is

             KE = (m - m₀)c² = m₀c²({1/[1 - (v²/c^2<)]^1/2} - 1)

                   = 37,000kg)(3.00 x 10⁸m/s)²({1 /[1- (0.21)²]^1/2}- 1) = 7.6 x 10¹⁹J.

    (b) When we use the classical expression, we get

             KE_c = 1/2mv² = 1/2(37,000kg)[(0.21)(3.00 x 10⁸m/s)]² = 7.34 x 10¹⁹J.

         The error is

             (7.43 - 7.6)/(7.6) = -0.04 = -4%.

5. (a) In the reference frame of the second spaceship, the Earth is moving at 0.5c, and the first spaceship is moving at 0.50c relative to the Earth. thus the speed of the first spacship relative to the second is

              u = (v + u')/(1 + vu'/c²) = (0.50c + 0.50c)[1 + (0.50)(0.50)] = 0.80c.

    (b) In the reference frame of the first spaceship,, the Earth moving at -0.50c, and the second spaceship is moving at -0.50c relative to Earth. Thus the speed of the second spaceship relative to the first is

             u = (v + u')/(1 + vu'/c²) = [-0.50c + (-0.50c)][1 + (-0.50)(-0.50)] = -0.80c, as expected.