Content check — University Physics 3, Chapter 5: Relativity

Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 5: Relativity

67 problems · 1 figure

Check Your Understanding 5.1 §5.1 source ↗

Explain how special relativity differs from general relativity.

Answer

Special relativity applies only to objects moving at constant velocity, whereas general relativity applies to objects that undergo acceleration.

Check Your Understanding 5.2 §5.3 source ↗

What is

γ

v = 0.650 c ?

Answer

γ =

\frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} =

\frac{1}{\sqrt{1 - \frac{{(0.650 c)}^{2}}{c^{2}}}} =

1.32

Check Your Understanding 5.3 §5.3 source ↗

a. A particle travels at

1.90 \times 10^{8} m/s

and lives

2.10 \times 10^{−8} s

when at rest relative to an observer. How long does the particle live as viewed in the laboratory?

Answer

Δ t =

\frac{Δ τ}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} =

\frac{2.10 \times 10^{−8} s}{\sqrt{1 - \frac{{(1.90 \times 10^{8} m/s)}^{2}}{{(3.00 \times 10^{8} m/s)}^{2}}}} =

2.71 \times 10^{−8} s .

Check Your Understanding 5.3 (b) §5.3 source ↗

b. Spacecraft A and B pass in opposite directions at a relative speed of

4.00 \times 10^{7} m/s .

An internal clock in spacecraft A causes it to emit a radio signal for 1.00 s. The computer in spacecraft B corrects for the beginning and end of the signal having traveled different distances, to calculate the time interval during which ship A was emitting the signal. What is the time interval that the computer in spacecraft B calculates?

Answer

b. Only the relative speed of the two spacecraft matters because there is no absolute motion through space. The signal is emitted from a fixed location in the frame of reference of A, so the proper time interval of its emission is

τ = 1.00 s .

The duration of the signal measured from frame of reference B is then

Δ t =

\frac{Δ τ}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} =

\frac{1.00 s}{\sqrt{1 - \frac{{(4.00 \times 10^{7} m/s)}^{2}}{{(3.00 \times 10^{8} m/s)}^{2}}}} =

1.01 s.

Check Your Understanding 5.4 §5.4 source ↗

A particle is traveling through Earth’s atmosphere at a speed of 0.750c. To an earthbound observer, the distance it travels is 2.50 km. How far does the particle travel as viewed from the particle’s reference frame?

Answer

L =

L_{0} \sqrt{1 - \frac{v^{2}}{c^{2}}} =

(2.50 km) \sqrt{1 - \frac{{(0.750 c)}^{2}}{c^{2}}} =

1.65 km

Check Your Understanding 5.5 §5.5 source ↗

Show that if a time increment dt elapses for an observer who sees the particle moving with velocity v, it corresponds to a proper time particle increment for the particle of

d τ = γ d t .

Answer

Start with the definition of the proper time increment:

d τ =

\sqrt{- {(d s)}^{2} / c^{2}} =

\sqrt{d t^{2} - (d x^{2} + d x^{2} + d x^{2}) / c^{2}} .

where (dx, dy, dx, cdt) are measured in the inertial frame of an observer who does not necessarily see that particle at rest. This therefore becomes

\begin{matrix} d τ & = & \sqrt{- {(d s)}^{2} / c^{2}} = \sqrt{d t^{2} - [{(d x)}^{2} + {(d y)}^{2} + {(d z)}^{2}] / c^{2}} \\ = & d t \sqrt{1 - [{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2} + {(\frac{d z}{d t})}^{2}] / c^{2}} \\ = & d t \sqrt{1 - v^{2} / c^{2}} \\ d t & = & γ d τ . \end{matrix}

Check Your Understanding 5.6 §5.6 source ↗

Distances along a direction perpendicular to the relative motion of the two frames are the same in both frames. Why then are velocities perpendicular to the x-direction different in the two frames?

Answer

Although displacements perpendicular to the relative motion are the same in both frames of reference, the time interval between events differ, and differences in dt and

d t'

lead to different velocities seen from the two frames.

Check Your Understanding 5.7 §5.7 source ↗

Suppose a space probe moves away from Earth at a speed 0.350c. It sends a radio-wave message back to Earth at a frequency of 1.50 GHz. At what frequency is the message received on Earth?

Answer

We can substitute the data directly into the equation for relativistic Doppler frequency:

f_{obs} =

f_{s} \sqrt{\frac{1 - \frac{v}{c}}{1 + \frac{v}{c}}} =

(1.50 GHz) \sqrt{\frac{1 - \frac{0.350 c}{c}}{1 + \frac{0.350 c}{c}}} =

1.04 GHz.

Check Your Understanding 5.8 §5.8 source ↗

What is the momentum of an electron traveling at a speed 0.985c? The rest mass of the electron is

9.11 \times 10^{−31} kg .

Answer

Substitute the data into the given equation:

p =

γ m u =

\frac{m u}{\sqrt{1 - \frac{u^{2}}{c^{2}}}} =

\frac{(9.11 \times 10^{−31} kg) (0.985) (3.00 \times 10^{8} m/s)}{\sqrt{1 - \frac{{(0.985 c)}^{2}}{c^{2}}}} =

1.56 \times 10^{−21} kg-m/s.

Check Your Understanding 5.9 §5.9 source ↗

What is the kinetic energy of an electron if its speed is 0.992c?

Answer

\begin{matrix} K_{rel} & = (γ - 1) m c^{2} = (\frac{1}{\sqrt{1 - \frac{u^{2}}{c^{2}}}} - 1) m c^{2} \\ = (\frac{1}{\sqrt{1 - \frac{{(0.992 c)}^{2}}{c^{2}}}} - 1) (9.11 \times 10^{- 31} kg) {(3.00 \times 10^{8} m/s)}^{2} = 5.67 \times 10^{- 13} J \end{matrix}

5-1 §5.1 source ↗

Which of Einstein’s postulates of special relativity includes a concept that does not fit with the ideas of classical physics? Explain.

Answer

the second postulate, involving the speed of light; classical physics already included the idea that the laws of mechanics, at least, were the same in all inertial frames, but the velocity of a light pulse was different in different frames moving with respect to each other

5-3 §5.1 source ↗

When you are flying in a commercial jet, it may appear to you that the airplane is stationary and Earth is moving beneath you. Is this point of view valid? Discuss briefly.

Answer

yes, provided the plane is flying at constant velocity relative to the Earth; in that case, an object with no force acting on it within the plane has no change in velocity relative to the plane and no change in velocity relative to the Earth; both the plane and the ground are inertial frames for describing the motion of the object

5-5 §5.3 source ↗

To whom does the elapsed time for a process seem to be longer, an observer moving relative to the process or an observer moving with the process? Which observer measures the interval of proper time?

Answer

The observer moving with the process sees its interval of proper time, which is the shortest seen by any observer.

5-7 §5.4 source ↗

To whom does an object seem greater in length, an observer moving with the object or an observer moving relative to the object? Which observer measures the object’s proper length?

Answer

The length of an object is greatest to an observer who is moving with the object, and therefore measures its proper length.

5-9 §5.4 source ↗

Suppose an astronaut is moving relative to Earth at a significant fraction of the speed of light. (a) Does he observe the rate of his clocks to have slowed? (b) What change in the rate of earthbound clocks does he see? (c) Does his ship seem to him to shorten? (d) What about the distance between two stars that lie in the direction of his motion? (e) Do he and an earthbound observer agree on his velocity relative to Earth?

Answer

a. No, not within the astronaut’s own frame of reference. b. He sees Earth clocks to be in their rest frame moving by him, and therefore sees them slowed. c. No, not within the astronaut’s own frame of reference. d. Yes, he measures the distance between the two stars to be shorter. e. The two observers agree on their relative speed.

5-11 §5.7 source ↗

What happens to the relativistic Doppler effect when relative velocity is zero? Is this the expected result?

Answer

There is no measured change in wavelength or frequency in this case. The relativistic Doppler effect depends only on the relative velocity of the source and the observer, not any speed relative to a medium for the light waves.

5-13 §5.7 source ↗

All galaxies farther away than about

50 \times 10^{6} ly

exhibit a red shift in their emitted light that is proportional to distance, with those farther and farther away having progressively greater red shifts. What does this imply, assuming that the only source of red shift is relative motion?

Answer

It shows that the stars are getting more distant from Earth, that the universe is expanding, and doing so at an accelerating rate, with greater velocity for more distant stars.]

5-15 §5.8 source ↗

Is it possible for an external force to be acting on a system and relativistic momentum to be conserved? Explain.

Answer

Yes. This can happen if the external force is balanced by other externally applied forces, so that the net external force is zero.

5-17 §5.9 source ↗

What happens to the mass of water in a pot when it cools, assuming no molecules escape or are added? Is this observable in practice? Explain.

Answer

Because it loses thermal energy, which is the kinetic energy of the random motion of its constituent particles, its mass decreases by an extremely small amount, as described by energy-mass equivalence.

5-19 §5.9 source ↗

The mass of the fuel in a nuclear reactor decreases by an observable amount as it puts out energy. Is the same true for the coal and oxygen combined in a conventional power plant? If so, is this observable in practice for the coal and oxygen? Explain.

Answer

Yes, in principle there would be a similar effect on mass for any decrease in energy, but the change would be so small for the energy changes in a chemical reaction that it would be undetectable in practice.

5-21 §5.9 source ↗

Given the fact that light travels at c , can it have mass? Explain.

Answer

Not according to special relativity. Nothing with mass can attain the speed of light.

5-23 §5.3 source ↗

(a) What is

γ

v = 0.250 c ?

(b) If

v = 0.500 c ?

Answer

a. 1.0328; b. 1.15

5-25 §5.3 source ↗

Particles called

π

-mesons are produced by accelerator beams. If these particles travel at

2.70 \times 10^{8} m/s

and live

2.60 \times 10^{−8} s

when at rest relative to an observer, how long do they live as viewed in the laboratory?

Answer

5.96 \times 10^{−8} s

5-27 §5.3 source ↗

A neutral

π

-meson is a particle that can be created by accelerator beams. If one such particle lives

1.40 \times 10^{−16} s

as measured in the laboratory, and

0.840 \times 10^{−16} s

when at rest relative to an observer, what is its velocity relative to the laboratory?

Answer

0.800c

5-29 §5.3 source ↗

If relativistic effects are to be less than 1%, then

γ

must be less than 1.01. At what relative velocity is

γ = 1.01 ?

Answer

0.140c

5-31 §5.4 source ↗

A spaceship, 200 m long as seen on board, moves by the Earth at 0.970c. What is its length as measured by an earthbound observer?

Answer

48.6 m

5-33 §5.4 source ↗

(a) How far does the muon in Example 5.3 travel according to the earthbound observer? (b) How far does it travel as viewed by an observer moving with it? Base your calculation on its velocity relative to the Earth and the time it lives (proper time). (c) Verify that these two distances are related through length contraction

γ = 3.20 .

Example 5.3 Calculating ΔtΔt for a Relativistic Event in the book ↗

Answer

Using the values given in Example 5.3: a. 2.00 km; b. 0.625 km; c. 2.00/3.20 km

5-35 §5.4 source ↗

Unreasonable Results A spaceship is heading directly toward Earth at a velocity of 0.800c. The astronaut on board claims that he can send a canister toward the Earth at 1.20c relative to Earth. (a) Calculate the velocity the canister must have relative to the spaceship. (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

Answer

a. 10.0c; b. The resulting speed of the canister is greater than c, an impossibility. c. It is unreasonable to assume that the canister will move toward the earth at 1.20c.

5-37 §5.5 source ↗

Describe what happens to the angle

α = tan (v / c),

and therefore to the transformed axes in Figure 5.17, as the relative velocity v of the S and

S'

frames of reference approaches c.

The space time diagram has axes x and c t. The v=c line is a line at 45 degrees. A second set of axes, x prime and c t prime, are also shown. These axes share the same origin as the x c t axes. The x prime axis is an angle alpha = inverse tangent (v/c) above the x axis. The c t prime axis is the same angle alpha to the right of the c t axis. A set of dashed lines parallel to the x prime and c t prime axes are also shown. — Figure 5.17 in the book ↗

Answer

The angle α approaches

45 °,

and the

t' -

and

x' -axes

rotate toward the edge of the light cone.

5-39 §5.5 source ↗

A man standing still at a train station watches two boys throwing a baseball in a moving train. Suppose the train is moving east with a constant speed of 20 m/s and one of the boys throws the ball with a speed of 5 m/s with respect to himself toward the other boy, who is 5 m west from him. What is the velocity of the ball as observed by the man on the station?

Answer

15 m/s east

5-41 §5.5 source ↗

A man is running on a straight road perpendicular to a train track and away from the track at a speed of 12 m/s. The train is moving with a speed of 30 m/s with respect to the track. What is the speed of the man with respect to a passenger sitting at rest in the train?

Answer

32 m/s

5-43 §5.5 source ↗

In a frame at rest with respect to the billiard table, a billiard ball of mass m moving with speed v strikes another billiard ball of mass m at rest. The first ball comes to rest after the collision while the second ball takes off with speed v in the original direction of the motion of the first ball. This shows that momentum is conserved in this frame. (a) Now, describe the same collision from the perspective of a frame that is moving with speed v in the direction of the motion of the first ball. (b) Is the momentum conserved in this frame?

Answer

a. The second ball approaches with velocity −v and comes to rest while the other ball continues with velocity −v; b. This conserves momentum.

5-45 §5.5 source ↗

In a frame S, two events are observed: event 1: a pion is created at rest at the origin and event 2: the pion disintegrates after time

τ

. Another observer in a frame

S'

is moving in the positive direction along the positive x-axis with a constant speed v and observes the same two events in his frame. The origins of the two frames coincide at

t = t' = 0 .

(a) Find the positions and timings of these two events in the frame

S'

(a) according to the Galilean transformation, and (b) according to the Lorentz transformation.

Answer

\begin{matrix} t_{1}' = 0; x_{1}' = 0; \\ t_{2}' = τ; x_{2}' = - v t \end{matrix};

\begin{matrix} t_{1}' = 0; x_{1}' = 0; \\ t_{2}' = \frac{τ}{\sqrt{1 - v^{2} / c^{2}}}; x_{2}' = \frac{- v τ}{\sqrt{1 - v^{2} / c^{2}}} \end{matrix}

5-47 §5.6 source ↗

Two planets are on a collision course, heading directly toward each other at 0.250c. A spaceship sent from one planet approaches the second at 0.750c as seen by the second planet. What is the velocity of the ship relative to the first planet?

Answer

0.615c

5-49 §5.6 source ↗

What is the relative velocity of two spaceships if one fires a missile at the other at 0.750c and the other observes it to approach at 0.950c?

Answer

0.696c

5-51 §5.6 source ↗

Show that for any relative velocity v between two observers, a beam of light projected by one directly away from the other will move away at the speed of light (provided that v is less than c, of course).

Answer

(Proof)

5-53 §5.8 source ↗

Find the momentum of a helium nucleus having a mass of

6.68 \times 10^{−27} kg

that is moving at 0.200c.

Answer

4.09 \times 10^{−19} kg \cdot m/s

5-55 §5.8 source ↗

(a) Find the momentum of a

1.00 \times 10^{9} -kg

asteroid heading towards Earth at 30.0 km/s. (b) Find the ratio of this momentum to the classical momentum. (Hint: Use the approximation that

γ =

1 +

(1 / 2) v^{2} / c^{2}

at low velocities.)

Answer

3.000000015 \times 10^{13} kg \cdot m/s;

b. 1.000000005

5-57 §5.8 source ↗

What is the velocity of an electron that has a momentum of

3.04 \times 10^{−21} kg \cdot m/s

? Note that you must calculate the velocity to at least four digits to see the difference from c.

Answer

2.988 \times 10^{8} m/s

5-59 §5.9 source ↗

What is the rest energy of an electron, given its mass is

9.11 \times 10^{- 31} kg ?

Give your answer in joules and MeV.

Answer

0.512 MeV according to the number of significant figures stated. The exact value is closer to 0.511 MeV.

5-61 §5.9 source ↗

If the rest energies of a proton and a neutron (the two constituents of nuclei) are 938.3 and 939.6 MeV, respectively, what is the difference in their mass in kilograms?

Answer

2.3 \times 10^{- 30} kg;

to two digits because the difference in rest mass energies is found to two digits

5-63 §5.9 source ↗

A supernova explosion of a

2.00 \times 10^{31} kg

star produces

1.00 \times 10^{44} J

of energy. (a) How many kilograms of mass are converted to energy in the explosion? (b) What is the ratio

Δ m / m

of mass destroyed to the original mass of the star?

Answer

1.11 \times 10^{27} kg;

5.56 \times 10^{- 5}

5-65 §5.9 source ↗

(a) Using data from this table, calculate the amount of mass converted to energy by the fusion of 1.00 kg of hydrogen. (b) What is the ratio of mass destroyed to the original mass,

Δ m / m

? (c) How does this compare with

Δ m / m

for the fission of 1.00 kg of uranium?

Answer

7.1 \times 10^{−3} kg;

7.1 \times 10^{- 3} =

7.1 \times 10^{- 3};

\frac{Δ m}{m}

is greater for hydrogen

5-67 §5.9 source ↗

A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. (a) If all the lost mass is converted into the electron’s kinetic energy, find

γ

for the electron. (b) What is the electron’s velocity?

Answer

a. 208; b. 0.999988c; six digits used to show difference from c

5-69 §5.9 source ↗

(a) Calculate the relativistic kinetic energy of a 1000-kg car moving at 30.0 m/s if the speed of light were only 45.0 m/s. (b) Find the ratio of the relativistic kinetic energy to classical.

Answer

6.92 \times 10^{5} J;

b. 1.54

5-71 §5.9 source ↗

(a) Beta decay is nuclear decay in which an electron is emitted. If the electron is given 0.750 MeV of kinetic energy, what is its velocity? (b) Comment on how the high velocity is consistent with the kinetic energy as it compares to the rest mass energy of the electron.

Answer

a. 0.914c; b. The rest mass energy of an electron is 0.511 MeV, so the kinetic energy is approximately 150% of the rest mass energy. The electron should be traveling close to the speed of light.

5-73 §5.AP source ↗

(a) At what relative velocity is

γ = 2.00 ?

(b) At what relative velocity is

γ = 10.0 ?

Answer

a. 0.866c; b. 0.995c

5-75 §5.AP source ↗

(a) How long does it take the astronaut in Example 5.5 to travel 4.30 ly at

0.99944 c

(as measured by the earthbound observer)? (b) How long does it take according to the astronaut? (c) Verify that these two times are related through time dilation with

γ = 30.00

as given.

Example 5.5 View it in the book ↗

Answer

a. 4.303 y to four digits to show any effect; b. 0.1434 y; c.

1 / \sqrt{(1 - v^{2} / c^{2})} =

29.88 .

5-77 §5.AP source ↗

(a) Find the value of

γ

for the following situation. An astronaut measures the length of their spaceship to be 100 m, while an earthbound observer measures it to be 25.0 m. (b) What is the speed of the spaceship relative to Earth?

Answer

a. 4.00; b.

v = 0.968 c

5-79 §5.AP source ↗

An astronaut has a heartbeat rate of 66 beats per minute as measured during his physical exam on Earth. The heartbeat rate of the astronaut is measured when he is in a spaceship traveling at 0.5c with respect to Earth by an observer (A) in the ship and by an observer (B) on Earth. (a) Describe an experimental method by which observer B on Earth will be able to determine the heartbeat rate of the astronaut when the astronaut is in the spaceship. (b) What will be the heartbeat rate(s) of the astronaut reported by observers A and B?

Answer

a. A sends a radio pulse at each heartbeat to B, who knows their relative velocity and uses the time dilation formula to calculate the proper time interval between heartbeats from the observed signal. b.

(66 beats/min) \sqrt{1 - v^{2} / c^{2}} =

57.1 beats/min

5-81 §5.AP source ↗

Same two observers as in the preceding exercise, but now we look at two events occurring in spaceship A. A photon arrives at the origin of A at its time

t = 0

and another photon arrives at

(x = 1.00 m, 0, 0)

t = 0

in the frame of ship A. (a) Find the coordinates and times of the two events as seen by an observer in frame B. (b) In which frame are the two events simultaneous and in which frame are they are not simultaneous?

Answer

a. first photon:

(0, 0, 0)

t = t^{'};

second photon:

\begin{matrix} t' & = & \frac{- v x / c^{2}}{\sqrt{1 - v^{2} / c^{2}}} = \frac{- (c / 2) (1.00 m) / c^{2}}{\sqrt{0.75}} = - \frac{0.577 m}{c} = 1.93 \times 10^{−9} s \\ x' & = & \frac{x}{\sqrt{1 - v^{2} / c^{2}}} = \frac{1.00 m}{\sqrt{0.75}} = 1.15 m \end{matrix}

b. simultaneous in A, not simultaneous in B

5-83 §5.AP source ↗

An observer at origin of inertial frame S sees a flashbulb go off at

x =

150 km,

y =

15.0 km,

and

z = 1.00 km

at time

t =

4.5 \times 10^{−4} s .

At what time and position in the S

'

system did the flash occur, if S

'

is moving along shared x-direction with S at a velocity

v = 0.6 c ?

Answer

\begin{matrix} t' & = & \frac{t - v x / c^{2}}{\sqrt{1 - v^{2} / c^{2}}} = \frac{(4.5 \times 10^{−4} s) - (0.6 c) (\frac{150 \times 10^{3} m}{c^{2}})}{\sqrt{1 - {(0.6)}^{2}}} \\ = & 1.88 \times 10^{−4} s \\ x' & = & \frac{x - v t}{\sqrt{1 - v^{2} / c^{2}}} = \frac{150 \times 10^{3} m - (0.60) (3.00 \times 10^{8} m/s) (4.5 \times 10^{−4} s)}{\sqrt{1 - {(0.6)}^{2}}} \\ = & 8.6 \times 10^{4} m = 86 km \\ y & = & y' = 15 km \\ z & = & z' = 1 km \end{matrix}

5-85 §5.AP source ↗

An observer standing by the railroad tracks sees two bolts of lightning strike the ends of a 500-m-long train simultaneously at the instant the middle of the train passes him at 50 m/s. Use the Lorentz transformation to find the time between the lightning strikes as measured by a passenger seated in the middle of the train.

Answer

\begin{matrix} Δ t & = & \frac{Δ t' + v Δ x' / c^{2}}{\sqrt{1 - v^{2} / c^{2}}} \\ 0 & = & \frac{Δ t' + v (500 m) / c^{2}}{\sqrt{1 - v^{2} / c^{2}}}; \end{matrix}

since

v ≪ c,

we can ignore the term

v^{2} / c^{2}

and find

Δ t' =

-

\frac{(50 m/s) (500 m)}{{(3.00 \times 10^{8} m/s)}^{2}} =

−2.78 \times 10^{−13} s

The breakdown of Newtonian simultaneity is negligibly small, but not exactly zero, at realistic train speeds of 50 m/s.

5-87 §5.AP source ↗

Two astronomical events are observed from Earth to occur at a time of 0.30 s apart and a distance separation of

2.0 \times 10^{9} m

from each other. How fast must a spacecraft travel from the site of one event toward the other to make the events occur at the same time when measured in the frame of reference of the spacecraft?

Answer

\begin{matrix} Δ t' & = & \frac{Δ t - v Δ x / c^{2}}{\sqrt{1 - v^{2} / c^{2}}} \\ 0 & = & \frac{(0.30 s) - \frac{(v) (2.0 \times 10^{9} m)}{{(3.00 \times 10^{8} m/s)}^{2}}}{\sqrt{1 - v^{2} / c^{2}}} \\ v & = & \frac{(0.30 s)}{(2.0 \times 10^{9} m)} {(3.00 \times 10^{8} m/s)}^{2} \\ v & = & 1.35 \times 10^{7} m/s \end{matrix}

5-89 §5.AP source ↗

(a) All but the closest galaxies are receding from our own Milky Way Galaxy. If a galaxy

12.0 \times 10^{9} ly

away is receding from us at 0.900c, at what velocity relative to us must we send an exploratory probe to approach the other galaxy at 0.990c as measured from that galaxy? (b) How long will it take the probe to reach the other galaxy as measured from Earth? You may assume that the velocity of the other galaxy remains constant. (c) How long will it then take for a radio signal to be beamed back? (All of this is possible in principle, but not practical.)

Answer

Note that all answers to this problem are reported to five significant figures, to distinguish the results. a. 0.99947c; b.

1.2064 \times 10^{11} y;

1.2058 \times 10^{11} y

5-91 §5.AP source ↗

Repeat the preceding problem with the ship heading directly away from Earth.

Answer

a. –0.400c; b. –0.909c

5-93 §5.AP source ↗

(a) Suppose the speed of light were only 3000 m/s. A jet fighter moving toward a target on the ground at 800 m/s shoots bullets, each having a muzzle velocity of 1000 m/s. What are the bullets’ velocity relative to the target? (b) If the speed of light was this small, would you observe relativistic effects in everyday life? Discuss.

Answer

a. 1.65 km/s; b. Yes, if the speed of light were this small, speeds that we can achieve in everyday life would be larger than 1% of the speed of light and we could observe relativistic effects much more often.

5-95 §5.AP source ↗

A space probe speeding towards the nearest star moves at

0.250 c

and sends radio information at a broadcast frequency of 1.00 GHz. What frequency is received on Earth?

Answer

775 MHz

5-97 §5.AP source ↗

(a) Calculate the speed of a

1.00 - μg

particle of dust that has the same momentum as a proton moving at 0.999c. (b) What does the small speed tell us about the mass of a proton compared to even a tiny amount of macroscopic matter?

Answer

1.12 \times 10^{−8} m/s;

b. The small speed tells us that the mass of a protein is substantially smaller than that of even a tiny amount of macroscopic matter.

5-99 §5.AP source ↗

Show that the relativistic form of Newton’s second law is (a)

F =

m \frac{d u}{d t} \frac{1}{{(1 - u^{2} / c^{2})}^{3 / 2}};

(b) Find the force needed to accelerate a mass of 1 kg by 1

{m/s}^{2}

when it is traveling at a velocity of c/2.

Answer

\begin{matrix} F & = \frac{d p}{d t} = \frac{d}{d t} (\frac{m u}{\sqrt{1 - u^{2} / c^{2}}}) \\ = \frac{d u}{d t} (\frac{m}{\sqrt{1 - u^{2} / c^{2}}}) - \frac{1}{2} \frac{m u^{2}}{{(1 - u^{2} / c^{2})}^{3 / 2}} 2 \frac{d u}{d t} \\ = \frac{m}{{(1 - u^{2} / c^{2})}^{3 / 2}} \frac{d u}{d t} \end{matrix};

\begin{matrix} F & = \frac{m}{{(1 - u^{2} / c^{2})}^{3 / 2}} \frac{d u}{d t} \\ = \frac{1 kg}{{(1 - {(\frac{1}{2})}^{2})}^{3 / 2}} (1 {m/s}^{2}) \\ = 1.53 N \end{matrix}

5-101 §5.AP source ↗

What is the kinetic energy in MeV of a π-meson that lives

1.40 \times 1 0^{−16} s

as measured in the laboratory, and

0.840 \times 1 0^{−16} s

when at rest relative to an observer, given that its rest energy is 135 MeV?

Answer

90.0 MeV

5-103 §5.AP source ↗

(a) Show that

{(p c)}^{2} / {(m c^{2})}^{2} =

γ^{2} -

1 .

This means that at large velocities

p c >

>

m c^{2} .

(b) Is

E \approx p c

when

γ = 30.0,

as for the astronaut discussed in the twin paradox?

Answer

γ^{2} - 1;

b. yes

5-105 §5.AP source ↗

What is

γ

for a proton having a mass energy of 938.3 MeV accelerated through an effective potential of 1.0 TV (teravolt)?

Answer

1.07 \times 10^{3}

5-107 §5.AP source ↗

(a) Using data from Potential Energy of a System, find the mass destroyed when the energy in a barrel of crude oil is released. (b) Given these barrels contain 200 liters and assuming the density of crude oil is

750 {kg/m}^{3},

what is the ratio of mass destroyed to original mass,

Δ m / m ?

Answer

6.56 \times 10^{−8} kg;

m =

(200 L) ({1 m}^{3} /1000 L) ({750 kg/m}^{3}) =

150 kg;

therefore,

\frac{Δ m}{m} =

4.37 \times 10^{- 10}

5-109 §5.AP source ↗

A Van de Graaff accelerator utilizes a 50.0 MV potential difference to accelerate charged particles such as protons. (a) What is the velocity of a proton accelerated by such a potential? (b) An electron?

Answer

a. 0.314c; b. 0.99995c (Five digits used to show difference from c)

5-111 §5.AP source ↗

(a) A nuclear power plant converts energy from nuclear fission into electricity with an efficiency of 35.0%. How much mass is destroyed in one year to produce a continuous 1000 MW of electric power? (b) Do you think it would be possible to observe this mass loss if the total mass of the fuel is

10^{4} kg ?

Answer

a. 1.00 kg; b. This much mass would be measurable, but probably not observable just by looking because it is 0.01% of the total mass.

5-113 §5.AP source ↗

The sun produces energy at a rate of

3.85 \times 10^{26}

W by the fusion of hydrogen. About 0.7% of each kilogram of hydrogen goes into the energy generated by the Sun. (a) How many kilograms of hydrogen undergo fusion each second? (b) If the sun is 90.0% hydrogen and half of this can undergo fusion before the sun changes character, how long could it produce energy at its current rate? (c) How many kilograms of mass is the sun losing per second? (d) What fraction of its mass will it have lost in the time found in part (b)?

Answer

6.06 \times 10^{11} kg/s;

4.67 \times 10^{10} y;

4.27 \times 10^{9} kg;

d. 0.32%

Loading chapter 6…

Problems and answers are adapted from University Physics Volume 3 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/university-physics-volume-3/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.