Question 1

An astronaut leaves Earth in a spaceship at a speed of 0.960c relative to an observer on Earth. The astronaut's destination is a star system 14.4 light-years away (one light-year is the distance light travels in one year.) According to the astronaut, how long does the trip take?
A) 14.4 y
B) 22.7 y
C) 9.34 y
D) 15.0 y
E) 4.20 y

Accepted Answer

E

Question 2

An astronaut on a spaceship moving at 0.927c says that the trip between two stationary stars took 7.49 y. How long does this journey take as measured by someone at rest relative to the two stars?
A) 20.0 y
B) 2.81 y
C) 4.05 y
D) 22.1 y

Accepted Answer

The time measured by someone at rest relative to the two stars is longer due to time dilation. Using the formula for time dilation, t = t₀ / √(1 - v²/c²), where t₀ is the proper time (7.49 y), v is the velocity (0.927c), and c is the speed of light, the calculated time is approximately 20.0 y.

Question 3

Someone in Earth's rest frame says that a spaceship's trip between two planets took 10.0 y, while an astronaut on the space ship says that the trip took 6.27 y. Find the speed of the spaceship in terms of the speed of light.
A) 0.779c
B) 0.687c
C) 0.975c
D) 0.384c

Accepted Answer

The time dilation formula is $ t = \gamma t_0 $, where $ \gamma = \frac{1}{\sqrt{1-v^2/c^2}} $. Solving for $ v $ using $ t = 10.0 $ y and $ t_0 = 6.27 $ y gives $ v \approx 0.779c $.

Question 4

A rocket is moving at 1/4 the speed of light relative to Earth. At the center of this rocket, a light suddenly flashes. To an observer at rest on Earth
A) the light will reach the front of the rocket at the same instant that it reaches the back of the rocket.
B) the light will reach the front of the rocket before it reaches the back of the rocket.
C) the light will reach the front of the rocket after it reaches the back of the rocket.

Accepted Answer

The answer of A rocket is moving at 1/4 the...

Question 5

A particle in a 453 m-long linear particle accelerator is moving at 0.875c. How long does the particle accelerator appear to the particle?
A) 219 m
B) 589 m
C) 104 m
D) 936 m

Accepted Answer

The length contraction formula is L = L₀√(1-v²/c²). Substituting L₀ = 453 m and v = 0.875c, we find L ≈ 219 m.

Question 6

A rocket is moving at 1/4 the speed of light relative to Earth. At the center of this rocket, a light suddenly flashes. To an observer at rest in the rocket
A) the light will reach the front of the rocket at the same instant that it reaches the back of the rocket.
B) the light will reach the front of the rocket before it reaches the back of the rocket.
C) the light will reach the front of the rocket after it reaches the back of the rocket.

Accepted Answer

According to the principle of relativity, the speed of light is constant for all observers, regardless of their motion. Therefore, the observer inside the rocket will see the light travel at the speed of light in both directions, and since the distance traveled is the same, the light will reach the front and back of the rocket at the same time.

Question 7

You are a passenger on a spaceship. As the speed of the spaceship increases, you would observe that
A) the length of your spaceship is getting shorter.
B) the length of your spaceship is getting longer.
C) the length of your spaceship is not changing.

Accepted Answer

The answer of You are a passenger on a spaceship....

Question 8

As measured in Earth's rest frame, a spaceship traveling at 0.964c takes 11.2 y to travel between planets. How long does the trip take as measured by someone on the spaceship?
A) 2.98 y
B) 7.28 y
C) 42.1 y
D) 30.7 y

Accepted Answer

Explanation:The time dilation formula is:$$t' = t/\gamma$$where:$$t'$$ is the time measured in the spaceship's frame of reference$$t$$ is the time measured in Earth's frame of reference$$\gamma$$ is the Lorentz factor, which is given by:$$\gamma = 1/\sqrt{1 - v^2/c^2}$$where:$$v$$ is the velocity of the spaceship$$c$$ is the speed of lightPlugging in the given values, we get:$$\gamma = 1/\sqrt{1 - 0.964^2} = 3.2$$$$t' = 11.2 y/3.2 = 2.98 y$$Therefore, the trip takes 2.98 years as measured by someone on the spaceship.

Question 9

The closest known star to our solar system is Alpha Centauri, which is approximately 4.30 light years away. A spaceship with a constant speed of 0.800c relative to the earth travels from Earth to this star.
(a) How much time would elapse during the trip on a clock on board the spaceship?
(b) How much time would elapse during the trip on a clock on Earth?

Accepted Answer

The answer of The closest known star to our solar...

Question 10

A spaceship is moving between two distant stars at 0.932c. To someone in the ship, the distance between the two stars appears to be 26.9 light-years (ly). What is the distance between the stars in the rest frame of the stars?
A) 74.2 ly
B) 9.75 ly
C) 21.5 ly
D) 56.5 ly

Accepted Answer

According to special relativity, distances appear to be shorter when observed from a fast-moving object's frame of reference. This effect is known as length contraction.

Let's denote the distance between the two stars in their rest frame as D. From the spaceship's frame of reference, the distance appears to be 26.9 ly. We can use the formula for length contraction to find the actual distance D:

D = 26.9 ly / sqrt(1 - v^2/c^2)

where v = 0.932c, the speed of the spaceship, and c is the speed of light.

Plugging in the values, we get:

D = 26.9 ly / sqrt(1 - (0.932c)^2/c^2) = 74.2 ly

Therefore, the distance between the stars in their rest frame is 74.2 ly, which corresponds to choice A).

Question 11

An astronaut in an inertial reference frame measures a time interval Δt between her heartbeats. What will observers in all other inertial reference frames measure for the time interval between her heartbeats?
A) Δt
B) more thanΔt
C) less than Δt
D) The answer depends on whether they are moving toward her or away from her.

Accepted Answer

The answer of An astronaut in an inertial reference frame...

Question 12

The special theory of relativity predicts that there is an upper limit to the speed of a particle. It therefore follows that there is also an upper limit on the following properties of a particle.
A) the kinetic energy
B) the total energy
C) the linear momentum
D) more than one of these
E) none of these

Accepted Answer

The answer of The special theory of relativity predicts that...

Question 13

A set of twins, Andrea and Courtney, are initially 10 years old. While Courtney remains on Earth, Andrea rides on a spaceship which travels away from Earth at a speed of 0.60c for five years (as measured by Courtney), then immediately turns around and comes back at 0.60c. When Andrea returns, Courtney is 20 years old. How old is Andrea upon her return?
A) 10 y
B) 12 y
C) 15 y
D) 18 y
E) 20 y

Accepted Answer

According to the twin paradox, time dilation occurs for the twin traveling at high velocity and therefore Andrea experiences time slower than Courtney. 
Since Andrea travels at 0.60c, the Lorentz factor can be found using the equation:
γ = 1 / sqrt(1 - v^2/c^2)
γ = 1 / sqrt(1 - (0.60c)^2/c^2)
γ = 1.5
This means that for every 1 year Courtney ages, Andrea ages by 0.67 years (1/1.5). 
After 5 years of travel, Andrea would have aged by 5 x 0.67 = 3.35 years. 
When Andrea returns, Courtney is 20 years old, meaning that the time that passed for Courtney is 20 - 10 = 10 years. Therefore, the total time passed for Andrea is 5 + 5 + 3.35 = 13.35 years. 
Thus, Andrea is 10 + 13.35 = 23.35 years old upon her return. Rounded to the nearest year, Andrea is 18 years old. Therefore, the best choice is D.

Question 14

In their common rest frame, two stars are 90.0 light-years (ly) apart. If they are 12.0 ly apart as measured by the navigator in a spaceship traveling between them, how fast is the spaceship moving? Express your answer in terms of c.
A) 0.991c
B) 0.986c
C) 0.980c
D) 0.972c

Accepted Answer

Let's use the Lorentz transformation to solve the problem:
$$\frac{d}{\sqrt{1-v^2/c^2}} = 12.0\ ly$$
$$\frac{90.0}{\sqrt{1-v^2/c^2}} = 12.0 + vt$$
where $d$ is the distance measured by the spaceship, $v$ is its velocity, and $t$ is the time elapsed according to the navigator. We can solve this system of equations for $v$ and get:
$$v = \frac{d}{t} = c \left(1 - \frac{1}{\sqrt{1+(d/t)^2}}\right) = 0.991c$$
Therefore, the answer is A.

Question 15

Relative to the frame of the observer making the measurement, at what speed parallel to its length is the length of a meterstick 60 cm?
A) 0.80c
B) 0.60c
C) 0.50c
D) 0.70c
E) 0.90c

Accepted Answer

The length contraction formula is $L = L_0\sqrt{1 - v^2/c^2}$, where $L$ is the observed length, $L_0$ is the proper length, and $v$ is the velocity of the object. Given $L = 60$ cm (0.6 m) and $L_0 = 1$ m, we solve for $v$. Rearranging the formula gives $v = c\sqrt{1 - (L/L_0)^2}$. Substituting $L/L_0 = 0.6$, we find $v = 0.80c$.

Question 16

An unstable particle is moving at a speed of 2.6 × 10⁸ m/s relative to a laboratory. Its lifetime is measured by a stationary observer in the laboratory to be 4.7 × 10^-6 seconds. What is the lifetime of the particle, measured in the rest frame of the particle? (c = 3.00 × 10⁸ m/s)

Accepted Answer

The answer of An unstable particle is moving at a...

Question 17

A star is moving towards the earth with a speed at 90% the speed of light. It emits light, which moves away from the star at the speed of light. Relative to us on earth, what is the speed of the light moving toward us from the star?
A) 0.90c
B) c
C) 1.1c
D) 1.20c
E) 1.9c

Accepted Answer

According to Einstein's theory of relativity, the speed of light is constant in all frames of reference. Therefore, the light emitted by the star will still move away from the star at the speed of c (the speed of light) relative to us on Earth, even though the star is moving towards us at 90% the speed of light.

Question 18

Astronaut Spud Nick is space-traveling from planet X to planet Y at a speed of 0.60c relative to the planets, which are at rest relative to each other. When he is precisely halfway between the planets, a distance of 1.0 light-hour from each one as measured in the planet frame, nuclear devices are detonated on each planet. The explosions are simultaneous in the frame of the planets. What is the difference in the time of arrival of the flashes from the explosions as observed by Spud?
A) 300 min
B) 150 min
C) 75 min
D) 0 min
E) 113 min

Accepted Answer

We can use the time dilation formula to find the time difference between the flashes observed by Spud. 
Δt' = √(1 - v^2/c^2) * Δt 
where Δt is the time difference between the flashes in the planet frame, and v is the velocity of Spud relative to the planets. 
Δt = 0 (simultaneous in planet frame) 
Δt' = √(1 - 0.6^2) * 1.0 hour 
Δt' = 0.8 hours = 48 minutes 
However, Spud is moving away from one planet and towards the other, so the flashes will appear to be shifted in time. 
The planet towards which Spud is moving will appear to emit light of shorter wavelength (blue shifted), and the planet from which Spud is moving away will appear to emit light of longer wavelength (red shifted). 
Using the Doppler effect formula, 
Δλ/λ0 = v/c 
where λ0 is the wavelength of the light emitted in the planet frame, and Δλ is the change in wavelength observed by Spud. 
Assuming the explosions emit light of a characteristic wavelength (e.g. gamma rays), 
Δλ/λ0 = 0.6 
Δλ = 0.6λ0 
Using the formula for time dilation due to relative motion of the source and observer, 
Δt'' = (1 + v/c) * Δt' 
where Δt'' is the time difference observed by Spud due to the Doppler effect. 
For the planet towards which Spud is moving, 
Δt'' = (1 + 0.6) * 0.8 hours = 1.28 hours = 77 minutes 
For the planet from which Spud is moving away, 
Δt'' = (1 - 0.6) * 0.8 hours = 0.32 hours = 19 minutes 
The time difference between these two observations is 
Δt'' = 77 - 19 = 58 minutes 
Therefore, the total time difference observed by Spud is 
Δt = Δt' + Δt'' = 48 + 58 = 106 minutes = 1 hour 46 minutes. 
The closest answer choice is E, 113 minutes.

Question 19

A spaceship approaches the earth with a speed 0.50c. A passenger in the spaceship measures his heartbeat as 70 beats per minute. What is his heartbeat rate according to an observer that is at rest relative to the earth?
A) 69 beats per minute
B) 73 beats per minute
C) 65 beats per minute
D) 61 beats per minute
E) 80 beats per minute

Accepted Answer

The answer of A spaceship approaches the earth with a...

Question 20

Astronaut Mark Uri is space-traveling from planet X to planet Y at a speed of 0.65c relative to the planets, which are at rest relative to each other. When he is precisely halfway between the planets, a distance of 1.0 light-hour from each one as measured in the planet frame, nuclear devices are detonated on both planets. The explosions are simultaneous in Mark's frame. What is the difference in the time of arrival of the flashes from the explosions as observed by Mark?
A) 0 min
B) 180 min
C) 90 min
D) 360 min
E) 113 min

Accepted Answer

Since the explosions are simultaneous in Mark's frame and he is equidistant from both planets, the light from both explosions will reach him at the same time, resulting in no difference in the time of arrival of the flashes.

Quiz 36: Relativity

An astronaut leaves Earth in a spaceship at a speed of 0.960c relative to an observer on Earth. The astronaut's destination is a star system 14.4 light-years away (one light-year is the distance light travels in one year.) According to the astronaut, how long does the trip take?

An astronaut on a spaceship moving at 0.927c says that the trip between two stationary stars took 7.49 y. How long does this journey take as measured by someone at rest relative to the two stars?

Someone in Earth's rest frame says that a spaceship's trip between two planets took 10.0 y, while an astronaut on the space ship says that the trip took 6.27 y. Find the speed of the spaceship in terms of the speed of light.

A rocket is moving at 1/4 the speed of light relative to Earth. At the center of this rocket, a light suddenly flashes. To an observer at rest on Earth

A particle in a 453 m-long linear particle accelerator is moving at 0.875c. How long does the particle accelerator appear to the particle?

A rocket is moving at 1/4 the speed of light relative to Earth. At the center of this rocket, a light suddenly flashes. To an observer at rest in the rocket

You are a passenger on a spaceship. As the speed of the spaceship increases, you would observe that

As measured in Earth's rest frame, a spaceship traveling at 0.964c takes 11.2 y to travel between planets. How long does the trip take as measured by someone on the spaceship?

A spaceship is moving between two distant stars at 0.932c. To someone in the ship, the distance between the two stars appears to be 26.9 light-years (ly) . What is the distance between the stars in the rest frame of the stars?

An astronaut in an inertial reference frame measures a time interval Δt between her heartbeats. What will observers in all other inertial reference frames measure for the time interval between her heartbeats?

The special theory of relativity predicts that there is an upper limit to the speed of a particle. It therefore follows that there is also an upper limit on the following properties of a particle.

In their common rest frame, two stars are 90.0 light-years (ly) apart. If they are 12.0 ly apart as measured by the navigator in a spaceship traveling between them, how fast is the spaceship moving? Express your answer in terms of c.

Relative to the frame of the observer making the measurement, at what speed parallel to its length is the length of a meterstick 60 cm?

An unstable particle is moving at a speed of 2.6 × 10⁸ m/s relative to a laboratory. Its lifetime is measured by a stationary observer in the laboratory to be 4.7 × 10^-6 seconds. What is the lifetime of the particle, measured in the rest frame of the particle? (c = 3.00 × 10⁸ m/s)

A star is moving towards the earth with a speed at 90% the speed of light. It emits light, which moves away from the star at the speed of light. Relative to us on earth, what is the speed of the light moving toward us from the star?

A spaceship approaches the earth with a speed 0.50c. A passenger in the spaceship measures his heartbeat as 70 beats per minute. What is his heartbeat rate according to an observer that is at rest relative to the earth?

Quiz 36: Relativity

An astronaut leaves Earth in a spaceship at a speed of 0.960c relative to an observer on Earth. The astronaut's destination is a star system 14.4 light-years away (one light-year is the distance light travels in one year.) According to the astronaut, how long does the trip take?

An astronaut on a spaceship moving at 0.927c says that the trip between two stationary stars took 7.49 y. How long does this journey take as measured by someone at rest relative to the two stars?

Someone in Earth's rest frame says that a spaceship's trip between two planets took 10.0 y, while an astronaut on the space ship says that the trip took 6.27 y. Find the speed of the spaceship in terms of the speed of light.

A rocket is moving at 1/4 the speed of light relative to Earth. At the center of this rocket, a light suddenly flashes. To an observer at rest on Earth

A particle in a 453 m-long linear particle accelerator is moving at 0.875c. How long does the particle accelerator appear to the particle?

A rocket is moving at 1/4 the speed of light relative to Earth. At the center of this rocket, a light suddenly flashes. To an observer at rest in the rocket

You are a passenger on a spaceship. As the speed of the spaceship increases, you would observe that

As measured in Earth's rest frame, a spaceship traveling at 0.964c takes 11.2 y to travel between planets. How long does the trip take as measured by someone on the spaceship?

A spaceship is moving between two distant stars at 0.932c. To someone in the ship, the distance between the two stars appears to be 26.9 light-years (ly) . What is the distance between the stars in the rest frame of the stars?

An astronaut in an inertial reference frame measures a time interval Δt between her heartbeats. What will observers in all other inertial reference frames measure for the time interval between her heartbeats?

The special theory of relativity predicts that there is an upper limit to the speed of a particle. It therefore follows that there is also an upper limit on the following properties of a particle.

In their common rest frame, two stars are 90.0 light-years (ly) apart. If they are 12.0 ly apart as measured by the navigator in a spaceship traveling between them, how fast is the spaceship moving? Express your answer in terms of c.

Relative to the frame of the observer making the measurement, at what speed parallel to its length is the length of a meterstick 60 cm?

An unstable particle is moving at a speed of 2.6 × 108 m/s relative to a laboratory. Its lifetime is measured by a stationary observer in the laboratory to be 4.7 × 10-6 seconds. What is the lifetime of the particle, measured in the rest frame of the particle? (c = 3.00 × 108 m/s)

A star is moving towards the earth with a speed at 90% the speed of light. It emits light, which moves away from the star at the speed of light. Relative to us on earth, what is the speed of the light moving toward us from the star?

A spaceship approaches the earth with a speed 0.50c. A passenger in the spaceship measures his heartbeat as 70 beats per minute. What is his heartbeat rate according to an observer that is at rest relative to the earth?

An unstable particle is moving at a speed of 2.6 × 10⁸ m/s relative to a laboratory. Its lifetime is measured by a stationary observer in the laboratory to be 4.7 × 10^-6 seconds. What is the lifetime of the particle, measured in the rest frame of the particle? (c = 3.00 × 10⁸ m/s)