Phy 121-Assignment 14
A. 1. As discussed in class, relativistic effects start to affect 3 digit numbers
somewhere around 10% of c. As also discussed in class, KE = ½mv2 is only
valid at non-relativistic speeds. So, the ones that come out above (.1)c = 3 x
107 m/s are incorrect:
a – c: incorrect.
d. correct.
2. a. It says both ships have the same rest length. (“identical when docked”)
He’s at rest relative to his ship, and sees it as 20 m. So, L0 = 20 m for your ship
too.
b. You see him fly past you at the same speed that he sees you fly past him. So,
each of you sees the same amount of contraction for the other. He sees L = 19
m for you, so that’s what you see for him.
c.
B. 1. KE = Etot – E0. Look up an electron’s rest energy on the formula sheet.
1.17 MeV - .51 MeV = .66 MeV.
2.
C. 1. Since mass is equivalent to energy, giving a body kinetic energy
increases its inertia. (Gravitational effects depend on just the rest mass, but in
the sense of how hard it would hit you at a given speed, the body gets
“heavier.”) As an object approaches the speed of light, m approaches infinity.
This infinite inertia makes further acceleration impossible.
2. a) Mass difference: (9.12099 - 9.12041) x 10-26 = .00058 x 10-26 kg
To convert form mass units to energy units, E = mc2 =
(.00058 x 10-26 kg)(2.998 x 108)2 = 5.21 x 10-13 J
Then convert J to MeV:
(5.21 x 10-13 J)(
) = 3.26 x 106 eV = 3.26 MeV
b) KE = Etot – E0 = 3.26 MeV - .51 MeV =
2.57 MeV
.51 MeV is the electron’s rest energy; that is, the energy equivalent of its rest
mass. See “electron rest mass” on formula sheet.
D. 1. (a) To someone on Earth, the astronaut is shortened along the direction of
motion, and has a slowed pulse. (b) The astronaut sees no change in herself.
(but sees the observer on Earth as shortened and slowed down.)
2. A projectile’s x velocity is constant. Since the rivet’s vx was 18 m/s when it
separated from the train, it will still be 18 m/s on reaching the ground.
The y velocity increases downward at 9.8 m/s2:
vfy = v0y + ayt = 0 + (9.8)(1.25) = 12.25 m/s (Using down for positive)

Then, find v' s magnitude from its components:
√
=
21.8 m/s
E. 1. The speed of light in a vacuum, and their relative speed.
2.
F. a. Adiabatic: P1V1 = P2V2
(This  is the ratio CP/CV, not the  used in special relativity. From the table in
the formula sheet,  = 1.4 for both oxygen and nitrogen, the main things in air.)
(100 kPa)V11.4 = P2(V1/3)1.4
Divide both sides by V11.4: 100 kPa = P2(1/3)1.4
(100)(31.4) = P2
P2 = 466 kPa
b. Isothermal: P1V1 = P2V2
(100 kPa)V1 = P2(V1/3)
Divide both sides by V1: 100 kPa = P2(1/3)
P2 = (100)(3) = 300 kPa