Chapter 1
Relativity I. Home Work Solutions
1.1
Problem 1.2
A 2000-kg car moving with a speed of 20 m/s collides with and sticks to a 1500-kg car at
rest at a stop sign. Show that because momentum is conserved in the rest frame, momentum
is also conserved in a reference frame moving with a speed of 10 m/s in the direction of the
moving car.
Solution
In the rest frame, momentum is conserved. Since the collision is inelastic, kinetic energy is
not conserved. We then have:
m1 u1i + m2 u2i = (m1 + m2 )uf
2000 × 20 + 1500 × 0 = (2000 + 1500)uf
4 × 104 = 3.5 × 103 uf
4 × 104
uf =
3.5 × 103
(1.1)
where m1 , m2 , u1i , u2i are the masses and speeds of the cars before collision, uf is the speed
of the combined cars after collision.
Since the collision is taking place at non-relativistic speeds, we use the Galilean velocity
transformation to find the speeds in the moving frame.
2
CHAPTER 1. RELATIVITY I. HOME WORK SOLUTIONS
u01i = u1i − v = 20 − 10 = 10 m/s
u02i = u2i − v = 0 − 10 = −10 m/s
40
u0f = uf − v =
− 10
3.5
(1.2)
(1.3)
(1.4)
Let p0i and p0f be the momentum before and after collision in the moving frame:
p0i =
=
=
0
pf =
=
=
=
=
m1 u01i + m2 u02i
2000 × 10 − 1500 × 10
5000 kg · m/s
(m1 + m2 )uf
40
− 10
(2000 + 1500) ×
3.5
40
3500 ×
− 10
3.5
1000 × (40 − 35)
5000 kg · m/s
(1.5)
(1.6)
comparing Equations (1.5) and (1.6) we see that momentum is conserved in the moving
frame.
Physics 205:Modern Physics I, Chapter 1
Fall 2004
Ahmed H. Hussein
1.2. PROBLEM 1.8
1.2
3
Problem 1.8
A meter stick moving in a direction parallel to its length appears to be only 75 cm long to
an observer. What is the speed of the meter stick relative to the observer?
Solution
Length contraction gives:
L0
γ
1
=
γ
r
L =
L
L0
=
Physics 205:Modern Physics I, Chapter 1
L
L0
1−
v2
c2
2
v2
= 1− 2
c
s
2
L
v = c 1−
L0
s
2
75
= c 1−
100
= 0.661c
Fall 2004
Ahmed H. Hussein
4
CHAPTER 1. RELATIVITY I. HOME WORK SOLUTIONS
1.3
Problem 1.12
An astronaut at rest on earth has a heart beat rate of 70 beats/min. What will this rate be
when she is traveling in a spaceship at 0.90c as measured
(a) by an observer also in the space ship, and
(b) by an observer at rest on the Earth?
Solution
(a) On the ship, it is not possible to measure time dilation. So heart beat on the ship will
be observed as 70beats/min
(b) On the earth time dilation on the ship will be observed. The proper time is the time
measured at rest i.e. ∆tp = 1/70 min
∆t = γ∆tp
∆tp
= p
1 − v 2 /c2
1
p
=
70 1 − (0.9)2
1
=
70 × 0.4359
1
=
min
31.8
(1.7)
(1.8)
(1.9)
(1.10)
(1.11)
so the rate on the ship is:
rate =
Physics 205:Modern Physics I, Chapter 1
1
= 31.8 ≈ 32 beats/min
∆t
Fall 2004
Ahmed H. Hussein
1.4. PROBLEM 1.17
1.4
5
Problem 1.17
(a) How fast and in what direction must Galaxy A be moving if an absorption line found at
550 nm (green) for a stationary galaxy is shifted to 450 nm (blue) for A?
(b) How fast and in what direction is Galaxy B moving if it shows the same line shifted to
700 nm (red)?
Solution
(a) A blue shift (decreasing wavelength or increasing frequency) indicate that the Galaxy
and the observer are approaching each other. We can then calculate the speed of the
galaxy from:
λ2source − λ2observe
v
=
c
λ2source + λ2observe
(550)2 − (450)2
=
(550)2 + (450)2
= 0.198
v = 0.198c
(b) A red shift (increasing wavelength or decreasing frequency) indicate that the Galaxy and
the observer are receding away from each other. The speed of the galaxy is then:
v
λ2source − λ2observe
=
c
λ2source + λ2observe
(700)2 − (550)2
=
(700)2 + (550)2
= 0.237
v = 0.237c
Physics 205:Modern Physics I, Chapter 1
Fall 2004
Ahmed H. Hussein
6
CHAPTER 1. RELATIVITY I. HOME WORK SOLUTIONS
1.5
Problem 1.23
An observer in frame S sees lightning simultaneously strike two points 100 m apart. The
first strike occurs at x1 = y1 = z1 = t1 = 0 and the second at x2 = 100 m, y2 = z2 = t2 = 0.
(a) What are the coordinates of these two events in a frame S 0 moving in the standard
configuration at 0.70c relative to S.
(b) How far apart are the events in S 0 .
(c) Are the events simultaneous in S 0 ? If not, what is the difference in time between the
events, and which event occurs first?
Solution
(a) In the moving frame S 0 the first event have the following coordinates:
x01 =
=
y10 =
z10 =
t01
γ(x1 − vt1 )
0
y1
z1
h
v i
= γ t1 − 2 x1
c
= 0
where
1
γ = q
1−
= p
v2
c2
1
1 − (0.70)2
= 1.4
Physics 205:Modern Physics I, Chapter 1
Fall 2004
Ahmed H. Hussein
1.5. PROBLEM 1.23
7
Similarly the second event has the following coordinates:
x02 =
=
=
y20 =
z20 =
t02
γ(x2 − vt2 )
1.4 × (100 − 0)
140
y2
z2
h
v i
= γ t2 − 2 x2
c
0.7
= 1.4 × 0 −
× 100
3 × 108
= −0.33 µs
(b) Space separation between the events in the moving frame S 0 is:
∆x0 = x02 − x01
= 140 − 0
= 140 m
(c) The events are not simultaneous in the moving frame S 0 , the second event occurs 0.33 µs
earlier than the first event.
Physics 205:Modern Physics I, Chapter 1
Fall 2004
Ahmed H. Hussein