Problem
Set 26: 3
 6  8  12  2
 4  11  13  14
 17  45  47  18
 21  23  26  28
 31  Go up

Scott Parsons, 2003
3.
A beam of a certain type of elementary particle travels at a speed of
2.70 x 10^8 m/s. At this speed, the
average lifetime is measured to be 4.76 x 10^6 s.
What is the particle’s lifetime at rest?
Use
the following equation:
Dt
= gDt_{0
}
Where
g = 1/Ö1  (v^{2}/c^{2})
Put
these together and you have
Dt
= Dt_{0
}/Ö1  (v^{2}/c^{2})
Now replace
the variables in the equation and solve for Dt
V = 2.70 x
10^8 m/s
Dt_{0
}= 4.76 x 10^6 s
c = 3.00 x 10^8 m/s
Dt
= (4.76 x 10^6 s) /Ö1 – ((2.70 x 10^8 m/s)^2/(c^{2})) = 2.07 x 10^6 s
6.
What is the speed of a beam of pions if their average lifetime is
measured to be 4.10 x 10^8 s? At
rest, their lifetime is 2.60 x 10^8 s.
Use
the following equation:
Dt = Dt_{0 }/Ö1  (v^{2}/c^{2})
And solve for v to get
V = c Ö1  (Dt_{0
}^{2}/Dt ^{2})
Now replace the variables in the equation and find v
Dt = 4.10 x 10^8 s
Dt_{0 }= 2.60 x 10^8 s
v = (3.00 x 10^8) Ö1 – ((2.60 x 10^8 s)^2/(4.10 x 10^8 s)^2) = (2.13
x 10^8 m/s) / (3 x 10^8) = .773c
8.
At what speed do the relativistic formulas for length and time
intervals differ from classical values by 1.00 percent?
Use
the following equation:
Dt = Dt_{0 }/Ö1  (v^{2}/c^{2})
And solve for v to get
V = c Ö1  (Dt_{0
}^{2}/Dt ^{2})
Now replace the variables in the equation and find v
Where Dt
= 100 and Dt_{0
}= 99
V = (3.00 x 10^8) Ö1 – ((99)^2/(100)^2) = .196 m/s
12.
How fast must a pion be moving, on average, to travel 10.0 m before it
decays? The average lifetime, at
rest, is 2.60 x 10^8 s.
Use
the following equations:
V
= s/t and
Dt = Dt_{0 }/Ö1  (v^{2}/c^{2})
Put these to equations together to get
S
= vt = vDt_{0
}/Ö1  (v^{2}/c^{2})
Now solve for v to get
V = s / Ö Dt_{0}^{2 }+ (s^{2
}/ c^{2 })
Now replace the variables in the equation and find v
S = 10.0 m
Dt_{0 }= 2.60E8 s
V = s / Ö Dt_{0}^{2 }+ (s^{2
}/ c^{2 }) = 10.0 / Ö
(2.60E8 )^{2 }+ (10.0^{2 }/ c^{2
}) =
2.3655E8 m/s
(2.3655E8 m/s) / (3E8) = .789c
2.
A
spaceship passes you at a speed of .850c. You
measure its length to be 48.2 m. How
long would it be at rest?
Use the following equation:
L
= L_{0}/g
Where
g = 1/Ö1  (v^{2}/c^{2})
Put these together and you have
L
= L_{0 }Ö1  (v^{2}/c^{2})
Solve
for L_{0
}to get
L_{0
}= L /_{ }Ö1  (v^{2}/c^{2})
Now replace the variables in the equation and solve
for L_{0
}
L = 48.2 m
V = .850c
= 2.55 x 10^8 m/s
L_{0
}= L /_{ }Ö1  (v^{2}/c^{2})=
48.2
/ Ö1 – ((2.55 x 10^8 m/s)^2)/(c^{2}))
= 91.5 m
4.
If you were to travel to a star 100 lightyears from Earth at a speed of
2.60 x 10^8 m/s, what would you measure this distance to be?
Use the following equation from #2:
L
= L_{0 }Ö1  (v^{2}/c^{2})
Now replace the variables in the equation and solve
for L_{0
}
L = 100 ly
V = 2.60 x 10^8 m/s
L
= L_{0 }Ö1  (v^{2}/c^{2})=
100
Ö1 – ((2.60 x 10^8 m/s)^2 /c^2)) = 49.9 ly
11.
A friend of yours travels by you in her fast sports vehicle at a speed of
.580c. You measure it to be 5.80 m long and 1.20 m high.
(a) What will be its length and height at rest?
(b) How many seconds would you say elapsed on your friend’s watch when
20.0 s passed on yours? (c) How
fast did you appear to be traveling to your friend?
(d) How many seconds would she say elapsed on your watch when she saw
20.0 s pass on hers?
(a) Use
the following equation:
L = L_{0 }Ö1  (v^{2}/c^{2})
Solve
for L_{0
}to get
L_{0
}= L /_{ }Ö1  (v^{2}/c^{2})
Now replace the variables in the equation and solve
for L_{0}
L = 5.80 m
V = .58c
L_{0
}= L /_{ }Ö1  (v^{2}/c^{2})=
5.80 /_{ }Ö1  (.58^{2})= 7.12 m
The height hasn’t changed so it will still be
1.2 m high
(b) Use the following formula:
Dt = Dt_{0 }/Ö1  (v^{2}/c^{2})
Solve for
Dt_{0
}to get
Dt_{0 }= Dt
Ö1  (v^{2}/c^{2})
Now replace the variables in the equation and solve for Dt_{0 }
_{
}Dt
= 20 s
V = .58c
Dt_{0 }= Dt
Ö1  (v^{2}/c^{2})= 20
Ö1  (.58^{2})=
16.3 s
(c) .580c
(d) 16.3 s
12.
How
fast must a pion be moving, on average to travel 10.0 m before it decays? The
average lifetime at rest is 2.60 x 10^{8} s
Well, the pion will be going at a velocity such that
vt = 10.0 m, which sounds simple enough, but the time is of course, dilated, which makes it a bit tricky. It really is
vt_{o}/Ö(1v^{2}/c^{2})= 10.0 m
And this gets a bit messy, but not really too bad. First, square both sides, yielding
v^{2}t_{o}^{2}/(1  v^{2}/c^{2}) = L^{2}, where L = 10.0 m
Then cross multiply and distribute
v^{2}t_{o}^{2} = L^{2}(1  v^{2}/c^{2}) = L  Lv^{2}/c^{2 }v^{2}t_{o}^{2} = L^{2}  L^{2}v^{2}/c^{2 }Get terms of v on the same side:
v^{2}t_{o}^{2 }+ L^{2}v^{2}/c^{2 }= L^{2}
Factor out v^{2}:
v^{2}(t_{o}^{2 }+ L^{2}/c^{2}) = L^{2}
Divide both sides by the factor in the ():
v^{2} = L^{2}/(t_{o}^{2 }+ L^{2}/c^{2})
Finally, square root both sides:
v = L/Ö(t_{o}^{2 }+ L^{2}/c^{2}), L = 10.0, t_{o} = 2.6 x 10^{8} s, c = 3.00 x 10^{8} m/s, v = 2.37 x 108 m/s or .789 c
13.
What is the mass of a proton traveling at v = 0.90c?
Use the following equation:
m
= gm_{0
}
where g = 1/Ö1  (v^{2}/c^{2})
Put these together and you have
m
= m_{0 }/ Ö1  (v^{2}/c^{2})
Now replace the variables in the equation and solve
for m
m_{0
}= 1.6726
10^27 (the rest mass of a proton found in data packet)
v = 0.90c = 2.7 x 10^8 m/s
m
= m_{0 }/ Ö1  (v^{2}/c^{2})=
(1.6726_{ }/
Ö1  (2.7 x 10^8 m/s)/ (c^{2})= 3.8
x 10^27 kg
14.
At what speed will an object’s mass be twice its rest mass?
Use the following equation from #13:
m
= m_{0 }/ Ö1  (v^{2}/c^{2})
Solve for v to get
V = c Ö1 – (m_{0 }^{2}/m
^{2})
And replace the variables in the equation and solve
for v
m
= 2
m_{0
}= 1
V = c
Ö1 – (m_{0 }^{2}/m
^{2})= c Ö1 – ((1^2)/(2^2)) = .866c
17.
(a) What is the speed of an electron whose mass is 10,000 times its rest
mass? (b) If the electrons travel in the lab through a tube 3.0 km
long, how long is this tube in the electron’s reference frame?
(a) Use
the following equation from #13:
m
= m_{0 }/ Ö1  (v^{2}/c^{2})
Solve for v to get
V = c
Ö1 – (m_{0 }^{2}/m
^{2})
And replace the variables in the equation and solve
for v
m
= 10,000
m_{0
}= 1
V = c
Ö1 – (m_{0 }^{2}/m
^{2})= c Ö1 – (1_{ }^{2}/10,000
^{2})= .99999c
(b) Use the equation:
L
= L_{0 }Ö1  (v^{2}/c^{2})
And replace the variables in the equation and solve
for L
L_{0
}= 3 km =
3,000 m = 30,000 cm
V
= .9999 c
L_{
}= L_{0
}Ö1  (v^{2}/c^{2})=
30,000 Ö1  (.99999^{2}) = 30 cm
45.
A person on a rocket traveling at 0.50c observes a meteor comes from
behind and pass her at a speed she measures as 0.50c. How fast is the meteor moving with respect to the Earth?
Use the equation:
u'_{x}
= (u_{x}  v)/(1  (u_{x}v/c^{2}))
Since the rockets are going in the same direction the
– changes to a +
u'_{x}
= (u_{x} + v)/(1 + (u_{x}v/c^{2}))
Now replace the variables in the equation and solve for u'_{x}
V = .50c
u_{x
}= .50c
u'_{x}
= (u_{x} + v)/(1 + (u_{x}v/c^{2}))= (.50
+ .50)/(1 + (.50 x .50))= .8c
47.
An observer on Earth sees an alien vessel approach at a speed of 0.60c.
The Enterprises comes to the rescue overtaking the aliens while moving
directly toward Earth at a speed of 0.90c relative to Earth.
What is the relative speed of one vessel seen by the other?
Use the equation:
u'_{x}
= (u_{x}  v)/(1  (u_{x}v/c^{2}))
Since the rockets are going in the same direction the
– changes to a +
u'_{x}
= (u_{x} + v)/(1 + (u_{x}v/c^{2}))
Now replace the variables in the equation and solve for u_{x}
u'_{x
}= .90c
v
= .60c
.90
= (u_{x} + .60c)/(1 + (u_{x }x
.60))
u_{x
}= .65c
18.
What is the kinetic energy of an electron whose mass is 3.0 times its
rest mass?
Use the following equation from the book:
KE
= mc^{2}  m_{0}c^{2 }which is really
KE
=(m  m_{0})c^{2}
Now
replace the variables in the equation and solve for KE
m_{0
}= 9.11 x
10^31 (the rest mass of an electron from data packet)
M
= 3(9.11 x
10^31)
KE
=(m  m_{0})c^{2 }= ((3
x 9.11 x 10^31)  9.11 x 10^31))c^{2
}= 1.6
x 10^13 J
21.
Calculate the rest energy of an electron in joules and in MeV (1 MeV =
1.60 x 10^13 J).
Use the following equation:
E
= mc^{2 }
Now
replace the variables in the equation and solve for E
M = 9.11 x 10^31 (the rest mass of an electron from
data packet)
E
= mc^{2 }=
(9.11 x
10^31)(3E8)^2 = 8.20E14
But it also wants it in MeV so divide by 1.60 x
10^13
(8.20E14)/(
1.60 x 10^13) = .511 MeV
23.
The total annual energy consumption in the United States is about 8 x
10^19 J. How much mass would have
to be converted to energy to fuel this need?
Use the following equation:
E
= mc^{2 }
And solve for m to get
M
= E / c^{2 }
Now
replace the variables in the equation and solve for m
E = 8E19
C = 3.00E8
M
= E / c^{2 }=
(8E19)/(3E8)^{ 2 }= 9 x 10^{2 }kg
26.
(a) How much work is required to accelerate a proton from rest up to a
speed of 0.998c? (b) What would be
the momentum of this proton?
(a) Use
the following equation:
m
= m_{0 }/ Ö1  (v^{2}/c^{2})
And
replace the variables in the equation and solve for m
m_{0
}= 1.6726E27
(the rest mass of a proton from data packet)
v = .998c
m
= m_{0 }/ Ö1  (v^{2}/c^{2})=
1.6726E27 / Ö1  (.998^{2})= 2.6459E26
Now use the following equation
KE
=(m  m_{0})c^{2 }
And
replace the variables in the equation and solve for m
M = 2.6459E26
m_{0
}=
1.6726E27
KE
=(m  m_{0})c^{2 }= (2.6459E26
 1.6726E27)c^{2 }= 2.23E9 J
(b) Use
the following equation:
p = mv
And
replace the variables in the equation and solve for p
M = 2.6459E26 (from (a) above)
V = 0.998c = 2.994 x 10^8 m/s
p = mv = (2.6459E26)( 2.994 x 10^8) = 7.91
x 10^18 kgm/s
28.
Calculate the kinetic energy and momentum of proton traveling 2.50 x 10^8
m/s.
Use the following equation:
m
= m_{0 }/ Ö1  (v^{2}/c^{2})
And
replace the variables in the equation and solve for m
m_{0
}= 1.6726E27
(the rest mass of a proton from data packet)
v = 2.5 x 10^8 m/s
m
= m_{0 }/ Ö1  (v^{2}/c^{2})=
1.6726E27 / Ö1  (2.5 x 10^8 m/s^{ 2}/3E8^{2})=
3.0258E27 kg
Now use the equation
KE
=(m  m_{0})c^{2}
And
replace the variables in the equation and solve for KE
KE
=(m  m_{0})c^{2 }= (3.0258
 1.6726E27)c^{2
}= 1.22E10
J
Now to find the momentum use the equation:
P = mv
And
replace the variables in the equation and solve for p
M = 3.0258E27 kg
V = 2.5 x 10^8 m/s
P = mv = (3.0258E27)( 2.5 x 10^8) = 7.55E19
kgm/s
31.
What is the speed of an electron whose KE is 1.00 MeV?
Use the following equation:
KE
= mc^{2}  m_{0}c^{2 }
And solve for m to get
M = KE + m_{0}
c^{2 }_{ / }c^{2}
Now replace
the variables in the equation and solve for m
KE = 1.00 MeV = 1.60 x 10^13 J
m_{0
}= 9.11E31
kg (the rest mass of an electron)
M = KE + m_{0
}= (1.60 x
10^13) + (9.11E31)
c^{2 }/ c^{2 }= 2.6887777E30
Now to find the speed of the electron use the
equation:
m
= m_{0 }/ Ö1  (v^{2}/c^{2})
And solve for v to get
V = c
Ö1 – (m_{0 }^{2}/m
^{2})
Now replace the variables in the equation and solve for v
m_{0
}= 9.11E31
kg (the rest mass of an electron)
m = 2.6887777E30