Problem
Set 26: 3
| 6 | 8 | 12 | 2
| 4 | 11 | 13 | 14
| 17 | 45 | 47 | 18
| 21 | 23 | 26 | 28
| 31 | Go up
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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
= gDt0
Where
g = 1/Ö1 - (v2/c2)
Put
these together and you have
Dt
= Dt0
/Ö1 - (v2/c2)
Now replace
the variables in the equation and solve for Dt
V = 2.70 x
10^8 m/s
Dt0
= 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/(c2)) = 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 = Dt0 /Ö1 - (v2/c2)
And solve for v to get
V = c Ö1 - (Dt0
2/Dt 2)
Now replace the variables in the equation and find v
Dt = 4.10 x 10^-8 s
Dt0 = 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 = Dt0 /Ö1 - (v2/c2)
And solve for v to get
V = c Ö1 - (Dt0
2/Dt 2)
Now replace the variables in the equation and find v
Where Dt
= 100 and Dt0
= 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 = Dt0 /Ö1 - (v2/c2)
Put these to equations together to get
S
= vt = vDt0
/Ö1 - (v2/c2)
Now solve for v to get
V = s / Ö Dt02 + (s2
/ c2 )
Now replace the variables in the equation and find v
S = 10.0 m
Dt0 = 2.60E-8 s
V = s / Ö Dt02 + (s2
/ c2 ) = 10.0 / Ö
(2.60E-8 )2 + (10.02 / c2
) =
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
= L0/g
Where
g = 1/Ö1 - (v2/c2)
Put these together and you have
L
= L0 Ö1 - (v2/c2)
Solve
for L0
to get
L0
= L / Ö1 - (v2/c2)
Now replace the variables in the equation and solve
for L0
L = 48.2 m
V = .850c
= 2.55 x 10^8 m/s
L0
= L / Ö1 - (v2/c2)=
48.2
/ Ö1 – ((2.55 x 10^8 m/s)^2)/(c2))
= 91.5 m
4.
If you were to travel to a star 100 light-years 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
= L0 Ö1 - (v2/c2)
Now replace the variables in the equation and solve
for L0
L = 100 ly
V = 2.60 x 10^8 m/s
L
= L0 Ö1 - (v2/c2)=
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 = L0 Ö1 - (v2/c2)
Solve
for L0
to get
L0
= L / Ö1 - (v2/c2)
Now replace the variables in the equation and solve
for L0
L = 5.80 m
V = .58c
L0
= L / Ö1 - (v2/c2)=
5.80 / Ö1 - (.582)= 7.12 m
The height hasn’t changed so it will still be
1.2 m high
(b) Use the following formula:
Dt = Dt0 /Ö1 - (v2/c2)
Solve for
Dt0
to get
Dt0 = Dt
Ö1 - (v2/c2)
Now replace the variables in the equation and solve
for Dt0
Dt
= 20 s
V = .58c
Dt0 = Dt
Ö1 - (v2/c2)= 20
Ö1 - (.582)=
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
vto/Ö(1-v2/c2)= 10.0 m
And this gets a bit messy, but not really too bad. First, square both sides, yielding
v2to2/(1 - v2/c2) = L2, where L = 10.0 m
Then cross multiply and distribute
v2to2 = L2(1 - v2/c2) = L - Lv2/c2
v2to2 = L2 - L2v2/c2
Get terms of v on the same side:
v2to2 + L2v2/c2 = L2
Factor out v2:
v2(to2 + L2/c2) = L2
Divide both sides by the factor in the ():
v2 = L2/(to2 + L2/c2)
Finally, square root both sides:
v = L/Ö(to2 + L2/c2), L = 10.0, to = 2.6 x 10-8 s, c = 3.00 x 108 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
= gm0
where g = 1/Ö1 - (v2/c2)
Put these together and you have
m
= m0 / Ö1 - (v2/c2)
Now replace the variables in the equation and solve
for m
m0
= 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
= m0 / Ö1 - (v2/c2)=
(1.6726 /
Ö1 - (2.7 x 10^8 m/s)/ (c2)= 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
= m0 / Ö1 - (v2/c2)
Solve for v to get
V = c Ö1 – (m0 2/m
2)
And replace the variables in the equation and solve
for v
m
= 2
m0
= 1
V = c
Ö1 – (m0 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
= m0 / Ö1 - (v2/c2)
Solve for v to get
V = c
Ö1 – (m0 2/m
2)
And replace the variables in the equation and solve
for v
m
= 10,000
m0
= 1
V = c
Ö1 – (m0 2/m
2)= c Ö1 – (1 2/10,000
2)= .99999c
(b) Use the equation:
L
= L0 Ö1 - (v2/c2)
And replace the variables in the equation and solve
for L
L0
= 3 km =
3,000 m = 30,000 cm
V
= .9999 c
L
= L0
Ö1 - (v2/c2)=
30,000 Ö1 - (.999992) = 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
= (ux - v)/(1 - (uxv/c2))
Since the rockets are going in the same direction the
– changes to a +
u'x
= (ux + v)/(1 + (uxv/c2))
Now replace the variables in the equation and solve for u'x
V = .50c
ux
= .50c
u'x
= (ux + v)/(1 + (uxv/c2))= (.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
= (ux - v)/(1 - (uxv/c2))
Since the rockets are going in the same direction the
– changes to a +
u'x
= (ux + v)/(1 + (uxv/c2))
Now replace the variables in the equation and solve for ux
u'x
= .90c
v
= .60c
.90
= (ux + .60c)/(1 + (ux x
.60))
ux
= .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
= mc2 - m0c2 which is really
KE
=(m - m0)c2
Now
replace the variables in the equation and solve for KE
m0
= 9.11 x
10^-31 (the rest mass of an electron from data packet)
M
= 3(9.11 x
10^-31)
KE
=(m - m0)c2 = ((3
x 9.11 x 10^-31) - 9.11 x 10^-31))c2
= 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
= mc2
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
= mc2 =
(9.11 x
10^-31)(3E8)^2 = 8.20E-14
But it also wants it in MeV so divide by 1.60 x
10^-13
(8.20E-14)/(
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
= mc2
And solve for m to get
M
= E / c2
Now
replace the variables in the equation and solve for m
E = 8E19
C = 3.00E8
M
= E / c2 =
(8E19)/(3E8) 2 = 9 x 102 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
= m0 / Ö1 - (v2/c2)
And
replace the variables in the equation and solve for m
m0
= 1.6726E-27
(the rest mass of a proton from data packet)
v = .998c
m
= m0 / Ö1 - (v2/c2)=
1.6726E-27 / Ö1 - (.9982)= 2.6459E-26
Now use the following equation
KE
=(m - m0)c2
And
replace the variables in the equation and solve for m
M = 2.6459E-26
m0
=
1.6726E-27
KE
=(m - m0)c2 = (2.6459E-26
- 1.6726E-27)c2 = 2.23E-9 J
(b) Use
the following equation:
p = mv
And
replace the variables in the equation and solve for p
M = 2.6459E-26 (from (a) above)
V = 0.998c = 2.994 x 10^8 m/s
p = mv = (2.6459E-26)( 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
= m0 / Ö1 - (v2/c2)
And
replace the variables in the equation and solve for m
m0
= 1.6726E-27
(the rest mass of a proton from data packet)
v = 2.5 x 10^8 m/s
m
= m0 / Ö1 - (v2/c2)=
1.6726E-27 / Ö1 - (2.5 x 10^8 m/s 2/3E82)=
3.0258E-27 kg
Now use the equation
KE
=(m - m0)c2
And
replace the variables in the equation and solve for KE
KE
=(m - m0)c2 = (3.0258
- 1.6726E-27)c2
= 1.22E-10
J
Now to find the momentum use the equation:
P = mv
And
replace the variables in the equation and solve for p
M = 3.0258E-27 kg
V = 2.5 x 10^8 m/s
P = mv = (3.0258E-27)( 2.5 x 10^8) = 7.55E-19
kgm/s
31.
What is the speed of an electron whose KE is 1.00 MeV?
Use the following equation:
KE
= mc2 - m0c2
And solve for m to get
M = KE + m0
c2 / c2
Now replace
the variables in the equation and solve for m
KE = 1.00 MeV = 1.60 x 10^-13 J
m0
= 9.11E-31
kg (the rest mass of an electron)
M = KE + m0
= (1.60 x
10^-13) + (9.11E-31)
c2 / c2 = 2.6887777E-30
Now to find the speed of the electron use the
equation:
m
= m0 / Ö1 - (v2/c2)
And solve for v to get
V = c
Ö1 – (m0 2/m
2)
Now replace the variables in the equation and solve for v
m0
= 9.11E-31
kg (the rest mass of an electron)
m = 2.6887777E-30