All Solutions
Page 972: Assessment
b) The answer a) is due to second postulate of relativity as explained in $textbf{A}$.
$Delta t_o = 4.1$ s
$$
v = 0.84 c
$$
To be able to compute for the time elapses of the flashes as observed by the astronaut, we use the time dilation formula as follows:
$$
begin{align*}
Delta t &= dfrac{Delta t_o}{sqrt{1 – dfrac{v^2}{c^2}}} \
&= dfrac{4.1}{sqrt{1 – dfrac{(0.84 c)^2}{c^2}}} \
&= 7.56 text{ s}
end{align*}
$$
The astronaut will be able to observe that the interval between flashes will be $boxed{7.56 text{ seconds}}$
7.56 text{ seconds}
$$
$$
Delta t = frac{4.1text{ s}}{sqrt{1-0.84^2}} =7.55text{ s}.
$$
$Delta t_o = 7.5$ s
$Delta t = 15$ s
To be able to compute for the speed of the moving spaceship, we use the time dilation formula and solve for v:
$$
begin{align*}
Delta t &= dfrac{Delta t_o}{sqrt{1 – dfrac{v^2}{c^2}}} \
v &= c sqrt{1 – dfrac{(Delta t_o)^2}{(Delta t)^2}} \
&= c sqrt{1 – dfrac{(7.5 text{ s})^2}{(15 text{ s})^2}} \
&= boxed{0.87c}
end{align*}
$$
v = 0.87c
$$
$$
15text{ s} = frac{7.5text{ s}}{sqrt{1-frac{v^2}{c^2}}}.
$$
Solving for $v$ and using the known value for $c$ we get
$$
v=2.6times10^8text{ m/s}.
$$
$v = 0.7705 c$
$Delta t = 15.03$ s
$textbf{Find :} Delta t_o$
To be able to compute for $Delta t_o$, we use the time dilation formula as follows:
$$
begin{align*}
Delta t_o &= Delta t sqrt{1 – dfrac{v^2}{c^2}} \
&= 15.03 text{ s} times sqrt{1 – dfrac{(0.7705 text{ c})^2}{c^2}} \
&= boxed{9.58 text{ s}}
end{align*}
$$
Usain Bolt was able to set a world record time of 9.58 seconds.
$$
15.03text{ s} = frac{Delta t_{Usain Bolt}}{sqrt{1-0.7705^2}}
$$
which yields
$$
Delta t_{Usain Bolt} = 15.03text{ s}sqrt{1-0.7705^2} =9.58text{ s}.
$$
You can check on wikipedia if this is correct!
$L = 150$ m
$L_o = 250$ m
$textbf{Find:}$ the speed of the spaceship, $v$
Using the length contraction formula, we can compute for $v$ as follows:
$$
begin{align*}
L &= L_o sqrt{1 – dfrac{v^2}{c^2}} \
v &= c sqrt{1 – dfrac{L^2}{L_{0}^{2}}} \
&= c sqrt{1 – dfrac{(150 text{ m})^2}{(250 text{ m})^2}} \
&= boxed{ 0.8c}
end{align*}
$$
Hence, the spaceship is moving at the speed of 0.8c
v= 0.8c
$$
$$
150text{ m} = 250text{ m}sqrt{1-frac{v^2}{c^2}}.
$$
Solving for $v$ we get
$$
v=2.4times 10^8text{ m/s}.
$$
$L_o = 1.89$ m
$c = 25 text{ km/h} times left(dfrac{1000 text{ m}}{1 text{ km}}right) times left(dfrac{3600 text{ s}}{1 text{ km}}right) = 9.0 times 10^7$ m/s
$v = 20 text{ km/h} = 7.2 times 10^7$ m/s
$textbf{Find:}$ Length of the bicycle, $L$ measured by an observer
$textbf{Calculation: }$
We use the length contraction formula to be able to solve for $L$, such that:
$$
begin{align*}
L &= L_o sqrt{1 – dfrac{v^2}{c^2}} \
&= 1.89 text{ m} sqrt{1 – dfrac{(7.2 times 10^7 text{ m/s})^2}{7.2 times 10^7 text{ m/s})^2}} \
&= 1.134 text{ m}
end{align*}
$$
Hence, the observer will see that the bicycle will be $boxed{text{1.134 m long}}$
1.134 text{m}
$$
$L_o = 1.24$ m
$L = 0.805$ m
$textbf{Find: } v$
$textbf{Calculation: }$
We use the length contraction formula to be able to solve for $v$, such that:
$$
begin{align*}
L &= L_o sqrt{1 – dfrac{v^2}{c^2}} \
v &= c sqrt{1 – dfrac{L^2}{L_{o}^{2}}} \
&= c sqrt{1 – dfrac{(0.805 text{ m})^2}{(1.24 text{ m})^2}} \
&= boxed{0.76c}
end{align*}
$$
Hence, the painting has to move $0.76c$ to appear to be square.
v = 0.76c
$$
$$
80.05text{ cm} = 124text{ cm}sqrt{1-frac{v^2}{c^2}}
$$
which yields
$$
v=2.29times10^8text{ m/s}.
$$
$omega_{ship} = 0.29$ rad/s
$v = 0.82c$
$textbf{Find: } omega_{observer}$
$textbf{Calculation: }$
We use the time dilation formula to be able to solve for $omega_{observer}$, such that:
$$
begin{align*}
Delta t &= dfrac{Delta t_o}{sqrt{1 – dfrac{v^2}{c^2}}} \
omega_{observer} &= dfrac{omega_{ship}}{sqrt{1 – dfrac{v^2}{c^2}}} \
&= dfrac{0.29 text{ rad/s}}{sqrt{1 – dfrac{(0.82c)^2}{c^2}}} \
&= boxed{0.51 text{ rad/s}}
end{align*}
$$
Hence, the observer will be seeing that the angular speed of the antenna is $textbf{0.51 rad/s}$
$$
T_{ship} = frac{T_{earth}}{sqrt{1-frac{v^2}{c^2}}}
$$
Where $T$ is the period of the antenna. Angular velocities are
$$
omega_{earth} = frac{2pi}{T_{earth}}quad omega_{ship}=frac{2pi}{T_{ship}}.
$$
Dividing last two equalities we get
$$
omega_{ship} = omega_{earth}sqrt{1-0.82^2} = 0.166text{ rad/s}.
$$
b) Time dilation formula yields
$$
t=frac{1text{ y}}{sqrt{1-0.99^2}} =7.1text{ y}.
$$
c)You are $22$ and your sister is $24$.
$$
begin{align*}
E=frac{mc^2}{sqrt{1-frac{v^2}{c^2}}}
end{align*}
$$
where $v$ is the speed of the object and $c$ is the speed of light in vacuum. We see that
$$
begin{align*}
E rightarrow infty, text{when} v rightarrow c
end{align*}
$$
This result has no physical sence. We also see that, if $v>c$, the relativistic energy of the object $E$ would be an imaginary number, which also has no physical sence.
If $v>c$, the energy $E$ would be an imaginary number, which doesn’t make physical sence.
b) The formula for relativistic energy can be written as
$$
E=frac{E_0}{sqrt{1-frac{v^2}{c^2}}}
$$
where $E_0$ denotes the rest mass. Solving for $v$ we get
$$
v=csqrt{1-left(frac{E_0}{E}right)^2}.
$$
We can see from this expression that as ratio $E_0/E$ decreases velocity increases, so reading this ratio from the table we see that the order of increasing velocity is: A,D,C,B.
$m = 1.675 times 10^{-27}$ kg
$$
v = 0.99c
$$
a) We compute for the relativistic energy of the neutron as follows:
$$
begin{align*}
E &= dfrac{mc^2}{sqrt{1 – dfrac{v^2}{c^2}}} \
&= dfrac{(1.675 times 10^{-27} text{ kg}) times c^2}{sqrt{1 – dfrac{(0.99c)^2}{c^2}}} \
&= 1.06 times 10^{-9 } text{ J} times left(dfrac{1 text{ MeV}}{1.60 times 10^{-19} text{ J}}right) \
&= boxed{6.68 times 10^9 text{ eV}}
end{align*}
$$
$$
begin{align*}
E_o &= mc^2 \
&= (1.675 times 10^{-27} text{ kg}) times (3.0 times 10^8 text{ m/s})^2 \
&= boxed{9.44 times 10^{8} text{ eV}}
end{align*}
$$
b) $9.44 times 10^8 text{ eV}$
textbf{Given:}
$$
$k = 574$ N/m
$x = 0.39$ m
$textbf{Find: }$ Increase in mass of the spring, $Delta m$
$$
textbf{Formula: }
$$
$$
begin{gather}
E = dfrac{1}{2}kx^2 \
Delta E = Delta m c^2
end{gather}
$$
We combine Eq (1) and Eq (2), to be able to calculate $Delta m$ as follows:
$$
begin{align*}
Delta m c^2 &= dfrac{1}{2} kx^2 \
Delta m&= dfrac{1}{2} dfrac{kx^2}{c^2} \
&= boxed{4.85 times 10^{-16} text{ kg}}
end{align*}
$$
Delta m =4.85 times 10^{-16} text{ kg}
$$
$$
Delta m c^2 = frac{1}{2}kx^2Leftrightarrow Delta m=frac{frac{1}{2}kx^2}{c^2}=4.85times10^{-16}text{ kg}.
$$
$Delta E = 8.3 times 10^-16$ kg
Using the mass-energy equivalence formula, we can calculate for the change in energy, $Delta m$ as follows:
$$
begin{align*}
Delta E &= Delta m c^2 \
&= (8.3 times 10^{-16} text{ kg}) times c^2 \
&= boxed{74.7 text{ J}}
end{align*}
$$
$$
E=Delta m c^2=74.7text{ J}.
$$
$$
E_{prest}+E_{aprest} = E_{gamma 1}+E_{gamma 2}Rightarrow 2mc^2 = E_{gamma 1}+E_{gamma 2}.
$$
Using momentum conservation law we see that the net momentum of produced $gamma$ rays has to be zero since both the proton and antiproton were at rest prior to annihilation. This is possible only if those $gamma$ rays move in opposite directions with the same momentum magnitude. Since the momentum magnitude of a photon is uniquely determined by its’ energy we conclude that the energies of those two $gamma$ photons are the same i.e. $E_{gamma 1} = E_{gamma2}=E_{gamma}.$
Returning to the first equation we see that both photons have the energy of
$$
E_{gamma} = mc^2 =1.503times 10^{-10}text{ J} = 938.3text{ MeV}.
$$
$E = 2.7 times 10^{23}$ J
$m = 2.7 times 10^6$ kg
Using the relativistic energy formula, we can calculate the speed of the rocket, $v$ as follows:
$$
begin{align*}
E &= dfrac{mc^2}{sqrt{1 – dfrac{v^2}{c^2}}} \
v &= c sqrt{1 – left(dfrac{mc^2}{E}right)^2 } \
&= c sqrt{1 – left(dfrac{(2.7 times 10^6) times c^2}{2.7 times 10^{23}}right)^2 } \
&= boxed{0.44c}
end{align*}
$$
v = 0.44c
$$
$$
E = frac{mc^2}{sqrt{1-frac{v^2}{c^2}}}.
$$
Solving for $v$ former equation we obtain
$$
v=csqrt{1-left(frac{mc^2}{E}right)^2}=1.31times 10^8 text{ m/s}.
$$
$E = 5.5 E_o$
$textbf{Find:}$ speed, $v$
$textbf{Formula:}$
$$
begin{gather}
E_o = m_oc^2 \
E = dfrac{m_oc^2}{sqrt{1 – dfrac{v^2}{c^2}}}
end{gather}
$$
We use Eq(1) and Eq(2) to find $v$ such that:
$$
begin{align*}
v &= c sqrt{1 – dfrac{1}{5.5^2}} \
&= boxed{0.98c}
end{align*}
$$
v = 0.98 c
$$
$$
5.5mc^2 = frac{mc^2}{sqrt{1-frac{v^2}{c^2}}}
$$
and that yields
$$
v=csqrt{1-frac{1}{5.5^2}}=2.95times 10^8 text{ m/s}.
$$
$P = 1.0 times 10^9$ W
$time = 1 text{ year} = 31536000 text{ s}$
$textbf{Find:} Delta m$
$textbf{Formula:}$
$$
begin{gather}
P = E/t \
Delta E = Delta m c^2
end{gather}
$$
Using Eq(1) and Eq(2) we can calculate $Delta m$ as follows:
$$
begin{align*}
Delta E = P times t = Delta m c^2 \
Delta m &= dfrac{P times t}{c^2} \
&= dfrac{1.0 times 10^9 text{ W}}{(3 times 10^8)^2} \
&=boxed{ 0.35 text{ kg}}
end{align*}
$$
$$
E=Ptimes1text{ y}=mc^2
$$
and from there we get
$$
m=frac{Ptimes 31536000text{ s} }{c^2} =0.35text{ kg}.
$$
$$
Delta t_{dilated} = frac{Delta t_{proper}}{sqrt{1-frac{v^2}{c^2}}}
$$
since
$$
sqrt{1-frac{v^2}{c^2}}
$$
is always less than one and dividing something with number less than one gives greater number than that something.
$$
begin{align*}
E=mc^2
end{align*}
$$
where $c$ is the speed of light in vacuum.
On the other hand, when the temperature $T$ of that object increases, its energy $E$ also increases. Thus, we see from the equation above that its mass $m$ also increases. So, when we apply some force $F$ to that object, its acceleration $a$ will decrease when the temperature $T$ increases. We conclude that the correct answer is: B.
$L = 2.4$ m
$$
v = 0.84c
$$
$textbf{Find:} T_{observed}$
$textbf{Formula:}$
$$
begin{gather}
T = 2pi sqrt{dfrac{L}{g}} \
T_{observed} = dfrac{T}{sqrt{1 – dfrac{v^2}{c^2}}}
end{gather}
$$
To calculate for $T_{observed}$, we use Eq(1) and Eq(2), such that:
$$
begin{align*}
T &= 2 pi sqrt{dfrac{2.4 text{ m}}{9.8 text{ m/s}}} \
&= 3.1 text{ s} \
T_{observed} &= dfrac{3.1 text{ s}}{sqrt{1 – dfrac{(0.84c)^2}{c^2}}} \
&= boxed{5.73 text{ s}}
end{align*}
$$
$$
T=2pisqrt{frac{l}{g}}= 3.1text{ s}.
$$
The boy can only measure the dilated period
$$
T’=frac{T}{sqrt{1-0.84^2}} =5.73text{ s}.
$$
b)It can be done like this: the “astronaut minute” is dilated so we measure 72 beats per
$$
t=frac{1text{ min}}{sqrt{1-0.65^2}} = 1.31text{ min}
$$
and that gives around $55$ bpm.
the gps system need relativistic correction to give us accurate position.
Magnetism is an effect of relativity, which we use in our every day
life. In the cathode ray tube (used in some old tvs, and other crt
monitors) the electron moves very fast almost half of the speed of
light. Hence the relativistic effect need to considered. There are
several other example of using relativity in daily life.
The satellite in the gps system moves at the speed of 14000 km/hr,
though this is much slower than speed of light, it gives a time delay
of 7 textmu s. Also since the satellite is about 20000 km above
the ground, the gravity experienced by it is much smaller, so because
of this the satellite is about 45 textmu s ahead of time. Hence
total about $45-7=38$ s difference in one day. If we do not take
kare of this time, the position shown by gps will be 11 km off every
day. That’s why we need relativistic correction in gps.
In PET scan we use radioactive material is injected in the body and
then the positron that is emitted by the radioactive material is detected.
So the energy in particle emitted by any radioactive material actually
comes from the conversion of mass to energy. And this conversion is
actually governed by the equation $E=mc^{2}$. Hence the equation
plays a vital role in the PET scan.
In nuclear reactor the energy comes from the radioactive decay. The
energy in the radioactive decay comes from the mass that is decreased
by the radioactive process. That is the mass get converted to energy
in the nuclear reactor. Hence the equation $E=mc^{2}$ become very
important to estimate how much energy we will get from a particular
reactor.
The maximum kinetic energy of the electrons in the electron beam is given by
$K = eV$.
But the kinetic energy of an electron moving with speed $v$ is given by
$K = dfrac{1}{2} m_e v^2$.
where $m_e$ is the mass of electron and its value is $m_e = 9.11 times 10^{-31}:kg$
Substituting this relation in the previous equation we get
$dfrac{1}{2} m_e v^2 = eV$
This equation can be rewritten for velocity of the electrons in the electron beam as
$v = sqrt{dfrac{2 eV}{m_e}}$
Substituting the values,
$v = sqrt{dfrac{2 times 1.6 times 10^{-19} times 25 times 10^3}{9.11 times 10^{-31}}} = sqrt{dfrac{8 times 10^{-15}}{9.11 times 10^{-31}}} = sqrt{0.8782} times 10^8:m/s$
$v = 0.937 times 10^8 = 0.312: c: :m/s$
where $c = 3 times 10^8;m/s$ is the speed of light.
This can be understood by the following reasoning.
The kinetic energy of the electron in television is $eV$. It is determined by the potential difference $V$. This value remains same even when relativistic effects are included.
But when relativistic effects are included, the relativistic mass ($m$) of the electron is larger than its rest mass ($m_0$). That is,
$m = dfrac{m_0}{sqrt{1-dfrac{v^2}{c^2}}}$
So the speed of the electron has to decrease to keep the value of kinetic energy same. Because the kinetic energy of the electron depends only on its mass and speed.
$K = dfrac{m_0 c^2}{sqrt{1 – dfrac{v^2}{c^2}}} – m_0 c^2$
Now we have
$eV = dfrac{m_0 c^2}{sqrt{1 – dfrac{v^2}{c^2}}} – m_0 c^2$
This relation can be rewritten for the velocity $v$ as
$v = c sqrt{1 – dfrac{1}{(1 + dfrac{eV}{m_0 c^2})^2}}$
The value of the expression $dfrac{eV}{m_0 c^2}$ is
$dfrac{eV}{m_0 c^2} = dfrac{1.6 times 10^{-19} times 25 times 10^3}{9.11 times 10^{-31} times (3 times 10^8)^2} = dfrac{40 times 10^{-16}}{81.99 times 10^{-15}} = dfrac{4}{81.99} = 0.0488$
Substituting this value in the expression of the velocity,
$v = c sqrt{1 – dfrac{1}{(1+0.0488)^2}} = 0.301:c::m/s$
$v = c sqrt{1 – dfrac{1}{(1 + dfrac{eV}{m_0 c^2})^2}}$
The accelerating potential difference $V$ is given as
$V = (25 times 10):kV = 250: kV$.
So, $dfrac{eV}{m_0 c^2} = dfrac{1.6 times 10^{-19} times 250 times 10^3}{9.11 times 10^{-31} times (3 times 10^8)^2} = dfrac{400 times 10^{-16}}{81.99 times 10^{-15}} = dfrac{40}{81.99} = 0.488$
Substituting this in the expression of $v$,
$v = c sqrt{1 – dfrac{1}{(1+0.488)^2}} = 0.740:c$
