An unstable particle at rest breaks into two fragments
of unequal mass. The mass of the first fragment is 2.50
× 10–28 kg, and that of the other is 1.67 × 10–27 kg. If the
lighter fragment has a speed of 0.893c after the
breakup, what is the speed of the heavier fragment?
An unstable particle at rest breaks into two fragments of unequal
mass. The mass of the first fragment is 2.50 × 10–28 kg, and that of
the other is 1.67 × 10–27 kg. If the lighter fragment has a speed of
0.893c after the breakup, what is the speed of the heavier
fragment?
Relativistic momentum of the system of fragments must be conserved. For total
momentum to be zero after as it was before, we must have, with subscript 2
referring to the heavier fragment, and subscript 1 to the lighter,
p2  p1
 2m 2u2   1m 1u1 


1.67  1027 kg u2
1  u2 c
2
2.50  1028 kg
1  0.893

2

  0.893c
 4.960  1028 kg c
An unstable particle at rest breaks into two fragments of unequal
mass. The mass of the first fragment is 2.50 × 10–28 kg, and that of
the other is 1.67 × 10–27 kg. If the lighter fragment has a speed of
0.893c after the breakup, what is the speed of the heavier
fragment?
27
2
 1.67  10
u2 
u22
 1 2


28
c
c
 4.960  10
12.3
u22
2
c
1
u2  0.285c
An unstable particle with a mass of 3.34 × 10–27 kg is
initially at rest. The particle decays into two fragments that
fly off along the x axis with velocity components 0.987c and
–0.868c. Find the masses of the fragments. (Suggestion:
Conserve both energy and momentum.)
An unstable particle with a mass of 3.34 × 10–27 kg is
initially at rest. The particle decays into two fragments that
fly off along the x axis with velocity components 0.987c and
–0.868c. Find the masses of the fragments. (Suggestion:
Conserve both energy and momentum.)
1 
2 
1
1  0.868
2
1
1  0.987
2
 2.01
 6.22
E1  E2  Etotal
 1m 1c2   2m 2c2  m totalc2
2.01m 1  6.22m 2  3.34  1027 kg
An unstable particle with a mass of 3.34 × 10–27 kg is
initially at rest. The particle decays into two fragments that
fly off along the x axis with velocity components 0.987c and
–0.868c. Find the masses of the fragments. (Suggestion:
Conserve both energy and momentum.)
2.01m 1  6.22m 2  3.34  1027 kg
This reduces to:
m 1  3.09m 2  1.66  1027 kg
p1  p2
 1m 1u1   2m 2u2
 2.01 0.868c m 1   6.22 0.987c m 2
m 1  3.52m 2
m 1  8.84  1028 kg
m 2  2.51 1028 kg
Suppose a heat engine is connected to two energy reservoirs, one
a pool of molten aluminum (660°C) and the other a block of solid
mercury (–38.9°C). The engine runs by freezing 1.00 g of
aluminum and melting 15.0 g of mercury during each cycle. The
heat of fusion of aluminum is 3.97  105 J/kg; the heat of fusion
of mercury is 1.18  104 J/kg. What is the efficiency of this
engine?
Suppose a heat engine is connected to two energy reservoirs, one a pool of
molten aluminum (660°C) and the other a block of solid mercury (–38.9°C).
The engine runs by freezing 1.00 g of aluminum and melting 15.0 g of mercury
during each cycle. The heat of fusion of aluminum is 3.97  105 J/kg; the heat
of fusion of mercury is 1.18  104 J/kg. What is the efficiency of this engine?
The heat to melt 15.0 g of Hg is:



Q c  m L f  15  103 kg 1.18  104 J kg  177 J
The energy absorbed to freeze 1.00 g of aluminum is:



Q h  m L f  103 kg 3.97  105 J/kg  397 J
and the work output is:
e
W eng
Qh

220 J
 0.554
397 J
W eng  Q h  Q c  220 J
or
55.4%
2. (a) Find the rest energy of a proton in electron volts; (b)
If the total energy of a proton is three times its rest
energy, what is the speed of the proton? (c) Determine
the kinetic energy of the proton in electron volts.
2. (a) Find the rest energy of a proton in electron volts; (b) If the total energy of a
proton is three times its rest energy, what is the speed of the proton? (c)
Determine the kinetic energy of the proton in electron volts.
(a)Rest energy:
E0=mpc2 = (1.67 x 10-27kg) x (3 x 108m/s)2 = 1.5 x 10-10J x (1eV / 1.6 x 10-19J)
= 938 MeV
(b) The total energy
Etot  3E0  3m p c 2 
3
1
u2
1 2
c
mpc2
u2
1 2
c
 u2  1
1  2  
 c  9
u2 8

2
c
9
8
u
c  0.943c  2.83108 m / s
3
(c) Kinetic energy:
K = Etot – E0 = 3mpc2 – mpc2 = 2mpc2 = 2 x (938 MeV) = 1876 MeV
A proton in high-energy accelerator moves with a
speed c/2. Use the work kinetic energy theorem to
find the work required to increase the speed to (a)
0.750c and (b) 0.995c.
An experimenter arranges to trigger two flashbulbs
simultaneously, a blue flush located at the origin of his
reference frame and a red flush at x = 30.4 km. A second
observer, moving at speed 0.247c in the direction of
increasing x, also views the flushes.
(a) What time interval between them does she find?
(b) Which flush does she say occurs first?
An experimenter arranges to trigger two flashbulbs
simultaneously, a blue flush located at the origin of his reference
frame and a red flush at x = 30.4 km. A second observer, moving
at speed 0.247c in the direction of increasing x, also views the
flushes.
(a) What time interval between them does she find?
(b) Which flush does she say occurs first?
Solution:
We use Lorentz transformation equation in the “interval” form:
ux 

t    t  2 
c 

'
First, evaluate γ:
 
1
u
1  
c
2

1
1  0.2472
 1.032
An experimenter arranges to trigger two flashbulbs
simultaneously, a blue flush located at the origin of his reference
frame and a red flush at x = 30.4 km. A second observer, moving
at speed 0.247c in the direction of increasing x, also views the
flushes.
(a) What time interval between them does she find?
(b) Which flush does she say occurs first?
We also have:
x  xr  xb  30.4  0  30.4km
t  tr  tb  0
so,
tr  tb
An experimenter arranges to trigger two flashbulbs
simultaneously, a blue flush located at the origin of his reference
frame and a red flush at x = 30.4 km. A second observer, moving
at speed 0.247c in the direction of increasing x, also views the
flushes.
(a) What time interval between them does she find?
(b) Which flush does she say occurs first?


ux 
(0.247)(30.4  103 m) 
5
t    t  2   (1.032) 0 


2
.
58

10
s  25.8s

8
c 
3  10 m / s



'
We got
t  tr  tb
and therefore t '  tr'  tb'
since Δt’< 0, we have tb > tr. Hence, the red flush is seen
first.
For electron :
me  9.11  10  31 kg
me c 2  (9.11  10  31 kg )(3  108 m / s ) 2  8.2  10 14 J
1eV  1.602  10 19 J
Ee, rest  mc 2  (8.2  10 14 J ) /(1.602  10 19 J / eV ) 
 0.511  106 eV  0.511MeV
For proton:
m p  1.67310 27 kg
m p c 2  (1.67310 27 kg )(3 108 m / s) 2  1.511010 J
E p ,rest  (1.511010 J ) /(1.6021019 J / eV )  0.938.6 109 eV
 938.6MeV
SUMMARY
1. Einstein’s Postulates:
• Postulate 1: Absolute uniform motion can not
be detected.
• Postulate 2: The speed of light is independent
of the motion of the source.
SUMMARY
• 2. The Lorentz Transformation:
x'  γ x  ut 
y' y
z'  z
xu 

t'  γ  t  2 
c 

SUMMARY
6. The Velocity Transformation:
'
vx  u
vx 
'
uv x
1 2
c
v

u
v'x  x
uv
1 x
2
c
SUMMARY
• 3. Time Dilation:
• 4. Length Contraction:
• 5. The Relativistic
Doppler Effect:
To
T 
u
1  
 c
L' 
2
 T0
L0

cu
f '
f0
cu
approaching
cu
f '
f0
cu
receding
SUMMARY
p
7. Relativistic Momentum:
8. Relativistic Energy:
mu
u2
1 2
c
E  m c2  K  m c2 
m c2
1
9. Rest Energy:
10. Kinetic Energy
u2
c
2
E0  mc2
K
m c2
1
u2
c
2
 m c2  m c2 (  1)
SUMMARY
10. Useful Formulas for Speed, Energy, and Momentum:
 
2 2
E  p c  mc
2
E  pc
2 2
for
E  m c
u pc

c E
2
• Specific Heat:
• Latent Heat:
Q
c
mT
Q
L
m
• Work done by engine:
• Thermal Efficiency:


Q  mcT
Q   Lm
Wengine  Qnet  Qh  Qc
Qh  Qc
Qc
e

 1
Qh
Qh
Qh
Weng