PHYS 1112
Chapter 26 Problems

Selected Solutions

  1. a) The observer on the spaceship is in the exact same reference frame as the astronaut. Therefore, the observer experiences time exactly like the astronaut, meaning that the heartbeat measurement will be 70 beats/minutes

    b) An observer on the Earth sees time on the spaceship as being dilated with respect to their reference frame. In the astronaut's reference frame, the heart beats at a rate of

    60 sec/70 beats = .86 sec/beat

    This means that their is .86 sec between beats. In the observer's reference frame, this amount of time is dilated to

    Dt = Dt/g = (.86 sec)/(1 - (.9c)2/c2)1/2 = (.86 sec)/(1 - .81)1/2 = 1.97 sec

    Therefore, the astronaut's heart beats 1.97 sec/beat, which means that it beats at a rate of 30 beats per minute.

  2. The momentum of an object is given by p = gmov. For an electron (mo= 9.1 x 10-31 kg) moving at v = .5c, we get

    p = (9.1 x 10-31 kg)(.5)(3.0 x 108m/s)/(1 - (.5c)2/c2)1/2 = (1.4 x 10-22 kg m/s)/(1 - .25)1/2

    p = 1.6 x 10-22 kg m/s

  3. From the equation for the radius of a charged object moving in a magnetic field, we can see that the radius will increase due to the speed of the object increasing. However, we must also notice that the mass of the object will increase with the increase in velocity, thereby increasing the radius as well. Therefore, to solve this problem, we cannot just set r1/v1 = r2/v2 as we would if the speeds involved were non-relativistic. Instead, since q and B are not changing in this problem, we must solve the equations 1/qB = r1/m1v1 = r2/m2v2 and m2 = g m1 . Combining these two, we get

    r2 = r1m2v2 /m1v1 = r1g m1v2 /m1v1 = r1g v2 /v1

    r2 = (10 cm)(.96)(3.0 x 108 m/s)/((1.0 x 105 m/s)(1 - (.96c)2/c2)1/2)

    r2 = (2.9 x 104 cm)/(1 - .92)1/2 = 1.0 x 105 cm

  4. From the book, we know that the wavelength of an object changes as

    l' = ((1 - v/c)/(1 + v/c))1/2 l

    where l' is the observed wavelength and l is the emitted wavelength. Using the values that are given in the problem and substituting bc for v (v = bc), we get

    ((1 - b)/(1 + b))1/2 = 525 nm/650 nm = .808

    (1 - b)/(1 + b) = .652

    1.65 b = .348

    b = .211

    This means that the object must be moving at v = .211 c.

If you find an error in these, or just have a comment, send an e-mail message to jpratte@kennesaw.edu