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Homework # 2
Relativity
Chapter 37
Part 2
37.16 Space pilot Mavis zips past Stanley at a constant speed relative to him of 0.80c. Mavis and Stanley
start timers at zero when the front of Mavis's ship is directly above Stanley. When Mavis reads 5.0
sec on her timer, she turns on a bright light under the front of her spaceship.
(a) Use the Lorentz coordinate transformation to calculate x as measured by Stanley for the event of
turning on the light.
(b) Use the Lorentz coordinate transformation to calculate t as measured by Stanley for the event of
turning on the light.
(c) Use the time dilation formula to calculate the time interval between the two events (the front of the
spaceship passing overhead and turning on the light) as measured by Stanley.
(d) Multiply the time interval by Mavis's speed, both as measured by Stanley, to calculate the distance
she has traveled as measured by him when the light turns on.
(a) u = 0.8c, γ = [1 − (u/c)2]-½ = 5/3
x = γ[x’ + ut’] = (5/3) [0 + 0.8c (5)] = 20c/3 = 2 x 109 m
(b)
t = γ[t’ + x’u/c2] = (5/3) [5 + 0] = 25/3 = 8.3 sec
(c)
Mavis reads 5 s on her watch which is the proper time.
Stanley measured the events at a time interval longer than ∆to by γ,
such that ∆t = γ ∆to = (5/3)(5) = 25/3 = 8.3 sec
which is the same as part (b).
(d) According to Stanley,
dist = u ∆t = 0.8c (8.3) = 2 x 109 m
which is the same as in part (a)
37.30 Calculate the magnitude of the force required to give a 0.145-kg
baseball an acceleration 1 m/s2 in the direction of the baseball's
initial velocity when this velocity has a magnitude of
(a) 10 m/s.
(b) 0.9c
(c) 0.99c
(d) 10 m/s with the force and acceleration perpendicular to v.
(e) 0.9c with the force and acceleration perpendicular to v.
(f) 0.99c with the force and acceleration perpendicular to v.
(a) F = γ3moa = (1)(0.145)(1) = 0.145 N
(b) γ = [1 – 0.92]-½ = 2.29, F = γ3moa = (2.29)3(0.145)(1) = 1.74 N
(c) γ = [1 – 0.992]-½ = 7.09, F = γ3moa = (7.09)3(0.145)(1) = 51.7 N
(d) F = γmoa = (1)(0.145)(1) = 0.145 N
(e) γ = [1 – 0.92]-½ = 2.29, F = γmoa = (2.29)(0.145)(1) = 0.333 N
(f) γ = [1 – 0.992]-½ = 7.09, F = γmoa = (7.09)(0.145)(1) = 1.03 N
37.31 (a) What is the speed of a particle whose kinetic energy is
equal to its rest energy? (b) What is the speed of a particle
whose kinetic energy is five times its rest energy?
(a)
KE = ( γ − 1) m o c 2 = m o c 2
  u 2 
γ = 2 = 1 −   
  c  
u = c 1−
−1 / 2
1
3
=
= 0.866c
c
2
γ
4
(b)
KE = ( γ − 1)m o c 2 = 5m o c 2
  u 2 
γ = 6 = 1 −   
  c  
−1 / 2
1
35
u = c 1− 2 = c
= 0.986c
36
γ
37.33 A proton (rest mass 1.67 x 10-27 kg ) has total energy that is
4 times its rest energy.
(a) What is the kinetic energy of the proton?
(b) What is the magnitude of the momentum of the proton?
(c) What is the speed of the proton?
1.67 x 10-27 kg = 1.67 x 10-27 kg (3 x 108 m/s)2 (1 eV/1.6 x 10-19 J) = 939 MeV
(a) E = γm o c 2 = 4m o c 2
γ=4
KE = ( γ − 1) m o c 2 = 3m o c 2 = 2818 MeV = 4.51 x 10-10 J
(b) E 2 = (pc )2 + (m o c 2 )2
(4m c ) = (pc) + (m c )
pc = 15 (m c )
2 2
2 2
2
o
o
2
o
p = 3638 MeV/c = [3638 MeV/(3 x 108 m/s)] [1.6 x 10-19 J/eV] = 1.94 x 10-18 kg.m/s
  u 2 
(c) γ = 4 = 1 −   
  c  
u = c 1−
−1/ 2
1
15
=
c
= 0.968c = 2.9 x 108 m/s
2
γ
16
note: p = γmov = γ(moc2/c2)(0.968c) = (4)(939)(0.968) MeV/c = 3636 MeV/c, same as (b)
37.47 The sun produces energy by nuclear fusion reactions, in which matter is
converted into energy. By measuring the amount of energy we receive from
the sun, we know that it is producing energy at a rate of 3.8 x 1026 W.
(a) How many kilograms of matter does the sun lose each second?
(b) Approximately how many tons of matter is this? (1 ton = 907.2 kg)
(c) At this rate, how long would it take the sun to use up all its mass?
1 kg = 1 kg (3 x 108 m/s)2 = 9 x 1016 J
(a) 3.8 x 1026 J/s [ 1 kg / 9 x 1016 J ] = 4.22 x 109 kg/s
(b) 4.22 x 109 kg [ 1 ton / 907.2 kg ] = 4.6 x 106 tons
(c) Mass of sun ~ 2 x 1030 kg
2 x 1030 kg / (4.22 x 109 kg/s)
= (4.74 x 1020 s)(1 hr/3600 s)(1 day/24 hr)(1yr/365 days) = 1.5 x 1013 yrs
37.55 Physicists and engineers from around the world have come together to build
the largest accelerator in the world, the Large Hadron Collider (LHC) at the
CERN Laboratory in Geneva, Switzerland. The machine will accelerate protons
to kinetic energies of 7 TeV in an underground ring 27 km in circumference. (For
the latest news and more information on the LHC, visit www.cern.ch.)
(a) What speed v will protons reach in the LHC? [Since v is very close to c, write v =
(1 – ∆)c and give your answer in terms of ∆.]
(b)
Find the relativistic mass of the accelerated protons in terms of their rest mass.
−1 / 2
(a)
  u 2 
2
γ = 1 −   
= 1 − (1 − ∆ )
  c  
KE = ( γ − 1)m o c 2
[
7 ⋅1012 eV = ( γ − 1)939MeV
1
γ = 7455 =
2∆
∆ = 9 ⋅10 −9
(b) m = γmo = 7455 mo → 7 TeV
]
−1 / 2
≈ [1 − (1 − 2∆)]
−1 / 2
=
1
2∆
37.56 A nuclear bomb containing 8.50 kg of plutonium explodes. The
sum of the rest masses of the products of the explosion is less than
the original rest mass by one part in 104.
(a) How much energy is released in the explosion?
(b) If the explosion takes place in 4.40 µs , what is the average power
developed by the bomb?
(c) What mass of water could the released energy lift to a height of
1.00 km?
1 kg = 1 kg (3 x 108 m/s)2 (1 eV/1.6 x 10-19 J) = 5.625 x 1035 eV
(a)
E = (∆m)c2 = (8.5 kg / 104)(3 x 108 m/s)2 = 7.65 x 1013 J
or ∆m = 8.5 x 10-4 kg (5.625 x 1035 eV/kg) = 4.78 x 1032 eV
(b)
P = E/t = 7.65 x 1013 J/4.4 µs = 1.74 x 1019 W
(c)
E = mgh
7.65 x 1013 J = m(9.8)(1000)
m = 7.8 x 109 kg
37.68 A baseball coach uses a radar device to measure the speed of an
approaching pitched baseball. This device sends out electromagnetic
waves with frequency fo and then measures the shift in frequency ∆f of
the waves reflected from the moving baseball. If the fractional
frequency shift produced by a baseball is ∆f/fo =2.73×10−7, what is the
baseball's speed? (Hint: Are the waves Doppler-shifted a second time
when reflected off the ball?)
f = fo
1+
c+u
1+ u / c
= fo
c−u
1− u / c
u
u
≈ 1+
c
2c
 u
= 1 − 
u  c
1−
c
1
2
−1 / 2
≈ 1+
f 
u 
u
≈ 1 +  = 1 +
f o  2c 
c
∆f f − f o f
u
=
= −1 =
fo
fo
fo
c
u
2c
 ∆f

 fo

u
 = 2
c
 total
c
u = (2.73 ⋅10 −7 ) = 41 m/s
2
1 mph = 0.45 m/s
41 m/s = 91 mph
37.73 Using a method analogous to the one in the text to find the Lorentz
transformation formula for velocity, we can find the Lorentz transformation for
acceleration. Let frame S’ have a constant x-component of velocity u relative
to frame S. An object moves relative to frame S along the x-axis with
instantaneous velocity vx and instantaneous acceleration ax.
(a) Find its instantaneous acceleration in frame S’.
xu 
udx 
udx 



 uv x 
t' = γ t – 2  ⇒ dt' = γ dt – 2  = γdt 1 – 2  = γdt 1 – 2 
c 
c 
c 


 c dt 

2
(
)
−
−
v
u
dv
v
u
u
/
c
dv x
x
⇒ dv 'x =
+ x
v 'x = x
2
vx u
vxu
v
u


x
1− 2
1− 2
−
1

2 
c
c
c 

(
)
 v u


u2 
2
x
1 − 2 + (v x − u ) u / c 
 1− 2 
dv x
c
c
 = dv x 
=
dv 'x = dv x 
2
2


  vx u 2 
vxu 
 vxu 
2
1 − 2 


 1 − 2   γ  1 − 2 
c 
c  
c 





dv 'x
dv x
ax
'
=
=
ax =
2
3
dt'
uv
v
u
uv






γ 3dt 1 – 2x 1 − x2 
γ 3 1 − 2x 
c 
c 
c 


(
)
37.73 (b) Show that the acceleration in frame S can be expressed as
 u
a x = a 'x 1 − 2
 c
2



3/ 2
 uv
1 +
c

'
x
2



−3
where v’x = dx’/dt’ is the velocity of the object in frame S’.
dv 'x
a =
=
dt'
'
x
ax
 uv 
γ 3 1 − 2x 
c 

3
Interchange S → S’, u → −u, γ stays the same
ax =
a 'x
 uv
γ 1 +
c

3
'
x
2



3