Question 1

A muon formed high in Earth's atmosphere travels at a speed 0.960c for a distance (as we see it)of 4500 m before it decays.How far does the muon travel as measured in its frame?
A)2520 m
B)16,100 m
C)4320 m
D)1260 m
E)​4500 m

Accepted Answer

The distance traveled by the muon in its own frame can be found using the length contraction formula: $L = L_0\sqrt{1-v^2/c^2}$, where $L_0$ is the length measured in the stationary frame (4500 m), $v$ is the velocity of the muon (0.960c), and $c$ is the speed of light. Plugging in the values, we find that the length contracts in the muon's frame, resulting in a shorter distance traveled than what is observed from Earth. The correct calculation leads to approximately 1260 m, making option D the correct answer.

Question 2

How fast would a rocket have to move past a ground observer if the latter were to observe a 9.4% length shrinkage in the rocket length? (c = 3.00 × 10⁸ m/s) A)2.98 × 10⁸ m/s B)2.72 × 10⁸ m/s C)1.27 × 10⁸ m/s D)0.420 × 10⁸ m/s E)0.920 × 10⁸ m/s

Accepted Answer

The answer of How fast would a rocket have to...

Question 3

A tuning fork has a frequency of 300 Hz and hence a period of 3.33 × 10−³ s.If the tuning fork is in an inertial frame of reference moving by the observer at speed of 0.620 c,what is the frequency of the fork as measured by the observer? (Assume that measurements are strictly by optical means and that the speed of sound waves in air is not pertinent here). A)235 Hz B)382 Hz C)185 Hz D)185 Hz. E)487 Hz

Accepted Answer

This is a problem of relativistic Doppler effect. We need to use the formula:
f' = fγ(1 ± v/c)
where f is the frequency of the source (tuning fork), f' is the frequency observed by the observer, v is the relative velocity (0.620 c), c is the speed of light, and γ is the Lorentz factor given by:
γ = 1/√(1 - v^2/c^2)

Substituting the given values, we get:
γ = 1/√(1 - (0.620 c)^2/c^2) = 1.5
f' = 300 × 1.5(1 + 0.620) = 300 × 2.79 = 837 Hz (when moving towards the observer)
or
f' = 300 × 1.5(1 - 0.620) = 300 × 0.705 = 211.5 Hz (when moving away from the observer)

Since the tuning fork is moving towards the observer, the frequency observed will be higher than the actual frequency, and the correct answer is the option A with a frequency of 235 Hz.

Question 4

A meter stick moving in a direction parallel to its length appears to be only 43.6 cm long to an observer.What is the meter stick's speed relative to the observer? (c = 3.00 × 10⁸ m/s) A)2.25 × 10⁸ m/s B)1.69 × 10⁸ m/s C)2.70 × 10⁸ m/s D)1.31 × 10⁸ m/s E)2.95 × 10⁸ m/s

Accepted Answer

The answer of A meter stick moving in a direction...

Question 5

An astronaut at rest has a heart rate of 68 beats/min.What will her heart rate be as measured by an Earth observer when the astronaut's spaceship goes by the Earth at a speed of 0.76c?
A)105 beats/min
B)52 beats/min
C)33 beats/min
D)139 beats/min
E)​44 beats/min

Accepted Answer

The answer of An astronaut at rest has a heart...

Question 6

A mass is bouncing on the end of a spring with a period T when measured by a ground observer.What would the period of oscillation be (as measured by the same observer)if the mass and spring were moving past the ground observer at a speed of 0.600c?
A)0.440 T
B)1.25 T
C)0.600 T
D)1.67 T

Accepted Answer

The period of oscillation as measured by the ground observer when the mass and spring are moving at a speed of 0.600c would be affected by time dilation, a consequence of Einstein's theory of relativity. The formula for time dilation is $T' = T / \sqrt{1 - v^2/c^2}$, where $T'$ is the dilated time period, $T$ is the original time period, $v$ is the velocity of the moving object, and $c$ is the speed of light. Plugging in $v = 0.600c$, we get $T' = T / \sqrt{1 - (0.600)^2} = T / \sqrt{1 - 0.36} = T / \sqrt{0.64} = T / 0.8 = 1.25T$. This shows that the period of oscillation increases due to the relativistic effect of time dilation.

Question 7

A rocket ship is 87.0 m in length when measured before leaving the launching pad.What would its velocity be if a ground observer measured its length as 62.0 m while it is in flight? (c = 3.00 × 10⁸ m/s) A)1.61 × 10⁸ m/s B)2.14 × 10⁸ m/s C)1.48 × 10⁸ m/s D)2.10 × 10⁸ m/s E)1.91 × 10⁸ m/s

Accepted Answer

The answer of A rocket ship is 87.0 m in...

Question 8

An astronaut is traveling at a high rate of speed from the Earth to a distant star system.Observers here on Earth note the relativistic effect that the astronaut's clocks run slow.What relativistic effect,if any,does the astronaut observe?
A)Her pulse rate slows.
B)The distance from the Earth to the star system has decreased.
C)Her clocks run faster.
D)There is no relativistic effect to be observed by the astronaut since she is the one doing the moving.

Accepted Answer

The astronaut observes length contraction, a relativistic effect where the distance between two points (in this case, from the Earth to the star system) appears shorter for an observer moving at a significant fraction of the speed of light relative to those points. This is a direct consequence of the principles of special relativity.

Question 9

A knight on horseback holds a 9.0-m lance.The horse can run at 0.75c.(It wins most of its races!)How long will the lance appear to a person that is standing still on the ground as the horse runs past?
A)6.0 m
B)14 m
C)9 m
D)6.8 m
E)​4.5 m

Accepted Answer

The length contraction formula in special relativity is $L = L_0\sqrt{1-v^2/c^2}$, where $L_0$ is the rest length (9.0 m in this case), $v$ is the velocity of the moving object (0.75c), and $c$ is the speed of light. Plugging in the values, we get $L = 9.0\sqrt{1-0.75^2} = 9.0\sqrt{1-0.5625} = 9.0\sqrt{0.4375} \approx 6.0$ m.

Question 10

If astronauts could travel at v = 0.95c,we on Earth would say it takes (4.2/0.95)= 4.4 years to reach Alpha Centauri,4.2 lightyears away.The astronauts disagree.How much time passes on the astronaut's clocks?
A)1.4 years
B)14.1 years
C)4.0 years
D)1.3 years
E)​0.4 years

Accepted Answer

Time dilation: t' = t / γ, where γ = 1 / sqrt(1 - v^2/c^2)t' = 4.4 years / γ = 4.4 years / sqrt(1 - 0.95^2) = 1.4 years

Question 11

A space probe has a 24.7-m length when measured at rest.What length does an observer at rest measure when the probe is going by at a speed of 0.860c?
A)25.2 m
B)6.43 m
C)9.19 m
D)12.6 m

Accepted Answer

The length contraction formula is L = L₀√(1-v²/c²). Substituting L₀ = 24.7 m and v = 0.860c, we find L ≈ 12.6 m.

Question 12

Our best measurements from Earth indicate that the star system Alpha Centauri is 4.2 lightyears away.Suppose some of our astronauts traveled there at a speed v = 0.80c.What would the astronauts measure as the distance to Alpha Centauri?
A)3.1 lightyears
B)7.0 lightyears
C)8.8 lightyears
D)2.5 lightyears
E)​2.7 lightyears

Accepted Answer

Due to the effects of length contraction in special relativity, the distance measured by astronauts moving at a significant fraction of the speed of light (in this case, 0.80c) would be shorter than the distance measured by observers at rest relative to the stars. The formula for length contraction is L = L0 * sqrt(1 - v^2/c^2), where L0 is the rest length (4.2 lightyears), v is the velocity of the astronauts (0.80c), and c is the speed of light. Plugging in the values, we get L = 4.2 * sqrt(1 - (0.80)^2) = 4.2 * sqrt(1 - 0.64) = 4.2 * sqrt(0.36) = 4.2 * 0.6 = 2.52 lightyears, which rounds to 2.5 lightyears.

Question 13

A space probe is 15.6 m in length when measured at rest.What length does an observer at rest measure when the probe is going by at a speed of 0.740c?
A)23.2 m
B)10.5 m
C)8.0 m
D)11.5 m
E)​21.1 m

Accepted Answer

We can use the formula for length contraction:

L' = L√(1-v^2/c^2)

where L is the rest length (15.6 m), v is the velocity (0.740 c), c is the speed of light, and L' is the length measured by the observer.

Plugging in the values:

L' = 15.6 √(1-(0.740)^2/(1)^2) = 10.5 m

Therefore, the observer at rest would measure the length of the space probe as 10.5 m. The correct choice is B.

Question 14

The period of an oscillating weight on a spring in an inertial frame of reference is 2.5 s.What would be its speed if it were to move by an observer who measures its period as 3.4 s? (c = 3.00 × 10⁸ m/s) A)1.5 × l0⁸ m/s B)2.0 × 10⁸ m/s C)2.8 × l0⁸ m/s D)6.8 × 10⁷ m/s E)1.3 × 10⁸ m/s

Accepted Answer

The speed of the observer relative to the inertial frame can be found using the relativistic Doppler effect formula for time dilation: $v = c \sqrt{1 - \left(\frac{T_0}{T}\right)^2}$, where $T_0$ is the period in the inertial frame (2.5 s), $T$ is the observed period (3.4 s), and $c$ is the speed of light. Plugging in the values, we find that the speed that matches the given conditions is $2.0 \times 10^8 m/s$, which corresponds to choice B.

Question 15

An Earth observer sees a spaceship at an altitude of 920 m moving downward toward the Earth at a speed of 0.690c.What is the spaceship's altitude as measured by an observer in the spaceship?
A)1270 m
B)512 m
C)1650 m
D)666 m
E)​635 m

Accepted Answer

The answer of An Earth observer sees a spaceship at...

Question 16

A spaceship of triangular shape,having a length three times its width,is capable of relativistic speeds.How fast would it have to move so that to a stationary observer its length would equal its width?
A)0.333c
B)0.816c
C)0.943c
D)​0.650c
E)​This is not possible.

Accepted Answer

Let the width of the spaceship be x, then its length is 3x. According to special theory of relativity, the length of an object moving with relativistic speed is contracted by a factor of gamma, which is given by:

gamma = 1 / sqrt(1 - (v^2/c^2))

where v is the velocity of the spaceship and c is the speed of light.

Now, when the spaceship is moving with a velocity v, its length along the direction of motion is contracted to x, which is equal to its width. Therefore, we have:

gamma * 3x = x

Simplifying this equation, we get:

gamma = 1 / 3

Substituting this value of gamma in the first equation, we get:

1/3 = 1 / sqrt(1 - (v^2/c^2))

Solving for v, we get:

v = 0.943c

Therefore, the spaceship would have to move at a velocity of 0.943 times the speed of light in order to appear to a stationary observer to have length equal to its width. Hence, the best choice is (C).

Question 17

Which of the following statements is correct? A)Because γ for an object in relative motion is always less than 1,a proper time interval Δt_p is always shorter than the time interval Δt and a proper length L_p is always longer than the length L. B)Because γ for an object in relative motion is always greater than 1,a proper time interval Δt_p is always longer than the time interval Δt and a proper length L_p is always shorter than the length L. C)Because γ for an object in relative motion is always less than 1,a proper time interval Δt_p is always longer than the time interval Δt and a proper length L_p is always shorter than the length L. D)Because γ for an object in relative motion is always greater than 1,a proper time interval Δt_p is always shorter than the time interval Δt and a proper length L_p is always longer than the length L.

Accepted Answer

Because γ for an object in relative motion is always greater than 1, a proper time interval Δt_p is always shorter than the time interval Δt and a proper length L_p is always longer than the length L.

Question 18

A spaceship is moving away from an observer at relativistic speed.Its length at rest is 100 m,but the observer measures its length as 60 m.A time interval of 84 s on the spaceship would be measured by the observer to be which of the following?
A)50.4 s
B)30.2 s
C)134 s
D)140 s
E)​84 s

Accepted Answer

The answer of A spaceship is moving away from an...

Question 19

The short lifetime of muons created in the upper atmosphere of the Earth would not allow them to reach the surface of the Earth unless their lifetime increased by time dilation.From the reference system of the muons,the muons can reach the surface of the Earth because
A)time dilation increases their velocity.
B)time dilation increases their energy.
C)length contraction decreases the distance to the Earth.
D)the relativistic speed of the Earth toward them is added to their velocity.

Accepted Answer

From the reference system of the muons, length contraction, a consequence of their high velocity relative to the Earth, effectively decreases the distance they have to travel to reach the Earth's surface. This phenomenon allows them to cover the distance within their short lifetimes, as observed from their own frame of reference.

Question 20

A boxcar without a front or a back is moving toward the right.Two flashes of light move through the boxcar,one moving from back to front toward the right,the other moving from front to back toward the left.A passenger in the boxcar records how long it takes each flash of light to pass from one end of the boxcar to the other end.According to the passenger,which took longer?
A)the flash going from back to front
B)the flash going from front to back
C)They both took the same time.
D)It depends on whether the passenger is sitting at the front or the back of the boxcar.

Accepted Answer

Both flashes of light are moving at the same speed and cover the same distance. Since they are moving in opposite directions, their relative speed is double the speed of either flash of light alone. However, this does not affect the time it takes for them to travel the distance between the front and back of the boxcar – which is the same for both flashes. Therefore, both flashes took the same amount of time.

Quiz 26: Relativity

A muon formed high in Earth's atmosphere travels at a speed 0.960c for a distance (as we see it) of 4500 m before it decays.How far does the muon travel as measured in its frame?

How fast would a rocket have to move past a ground observer if the latter were to observe a 9.4% length shrinkage in the rocket length? (c = 3.00 × 10⁸ m/s)

A meter stick moving in a direction parallel to its length appears to be only 43.6 cm long to an observer.What is the meter stick's speed relative to the observer? (c = 3.00 × 10⁸ m/s)

An astronaut at rest has a heart rate of 68 beats/min.What will her heart rate be as measured by an Earth observer when the astronaut's spaceship goes by the Earth at a speed of 0.76c?

A mass is bouncing on the end of a spring with a period T when measured by a ground observer.What would the period of oscillation be (as measured by the same observer) if the mass and spring were moving past the ground observer at a speed of 0.600c?

A rocket ship is 87.0 m in length when measured before leaving the launching pad.What would its velocity be if a ground observer measured its length as 62.0 m while it is in flight? (c = 3.00 × 10⁸ m/s)

An astronaut is traveling at a high rate of speed from the Earth to a distant star system.Observers here on Earth note the relativistic effect that the astronaut's clocks run slow.What relativistic effect,if any,does the astronaut observe?

A knight on horseback holds a 9.0-m lance.The horse can run at 0.75c.(It wins most of its races!) How long will the lance appear to a person that is standing still on the ground as the horse runs past?

If astronauts could travel at v = 0.95c,we on Earth would say it takes (4.2/0.95) = 4.4 years to reach Alpha Centauri,4.2 lightyears away.The astronauts disagree.How much time passes on the astronaut's clocks?

A space probe has a 24.7-m length when measured at rest.What length does an observer at rest measure when the probe is going by at a speed of 0.860c?

Our best measurements from Earth indicate that the star system Alpha Centauri is 4.2 lightyears away.Suppose some of our astronauts traveled there at a speed v = 0.80c.What would the astronauts measure as the distance to Alpha Centauri?

A space probe is 15.6 m in length when measured at rest.What length does an observer at rest measure when the probe is going by at a speed of 0.740c?

The period of an oscillating weight on a spring in an inertial frame of reference is 2.5 s.What would be its speed if it were to move by an observer who measures its period as 3.4 s? (c = 3.00 × 10⁸ m/s)

An Earth observer sees a spaceship at an altitude of 920 m moving downward toward the Earth at a speed of 0.690c.What is the spaceship's altitude as measured by an observer in the spaceship?

A spaceship of triangular shape,having a length three times its width,is capable of relativistic speeds.How fast would it have to move so that to a stationary observer its length would equal its width?

Which of the following statements is correct?

A spaceship is moving away from an observer at relativistic speed.Its length at rest is 100 m,but the observer measures its length as 60 m.A time interval of 84 s on the spaceship would be measured by the observer to be which of the following?

The short lifetime of muons created in the upper atmosphere of the Earth would not allow them to reach the surface of the Earth unless their lifetime increased by time dilation.From the reference system of the muons,the muons can reach the surface of the Earth because