Question 1

A spaceship carrying a light clock moves at a speed of 0.960c relative to an observer on Earth.If the clock on the ship advances by 1.00 s as measured by the space travelers aboard the ship,how long did that advance take as measured by the observer on Earth?

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Question 2

A spaceship visits a star that is 4.5 light-years from Earth,and the spaceship travels at one-half the speed of light for the entire trip.
(a)How long did the trip take according to an observer on Earth?
(b)How long did the trip take according to an observer on the spaceship?

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Question 3

A spaceship with a constant speed of 0.800c relative to Earth travels to a star that is 4.30 light-years from Earth.How much time for this trip would elapse on a clock on board the spaceship?

Accepted Answer

The time elapsed on a clock on board the spaceship can be calculated using the time dilation formula in special relativity, $ t' = \frac{t}{\sqrt{1 - v^2/c^2}} $, where $ t $ is the time in the Earth frame, $ v $ is the velocity of the spaceship, and $ c $ is the speed of light. First, calculate the time in the Earth frame: $ t = \frac{distance}{speed} = \frac{4.30 \, \text{light-years}}{0.800c} = 5.375 \, \text{years} $. Then, apply the time dilation formula: $ t' = \frac{5.375}{\sqrt{1 - (0.800)^2}} = 3.23 \, \text{years} $.

Question 4

A particle physicist observes cosmic rays creating a new particle high in the atmosphere,and the speed of this particle is measured at 99.7% the speed of light.It is unstable,and is observed to decay in an average 37.0 μs.If this particle were at rest in the laboratory,what would be its average lifetime?

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Question 5

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 $5.78 y$ What is the speed of the spaceship,assuming the planets are at essentially at rest?

Accepted Answer

The time dilation formula is $\Delta t' = \Delta t / \sqrt{1 - v^2/c^2}$, where $\Delta t'$ is the time interval measured by the moving observer (5.78 years), $\Delta t$ is the time interval measured by the stationary observer (10.0 years), $v$ is the velocity of the moving observer, and $c$ is the speed of light. Rearranging for $v$, we get $v = c \sqrt{1 - (\Delta t'/\Delta t)^2}$. Substituting $\Delta t' = 5.78$ years and $\Delta t = 10.0$ years, we find $v \approx 0.816 c$.

Question 6

How fast (in terms of the speed of light)should a moving clock travel if it is to be observed by a stationary observer as running at one-half its normal rate?

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Question 7

An astronaut on a spaceship moving at 0.927c says that the trip between two stationary stars took $4.26 \mathrm { y }$ How long does this journey appear to take to someone in the rest frame of the two stars?

Accepted Answer

The time dilation formula, $t = \frac{t_0}{\sqrt{1 - v^2/c^2}}$, where $t_0$ is the proper time (time measured by the moving observer, 4.26 years), $v$ is the velocity of the spaceship (0.927c), and $c$ is the speed of light, can be used to calculate the time observed in the rest frame. Plugging in the values, we find the time to be approximately 11.4 years.

Question 8

A spaceship with a constant speed of 0.800c relative to Earth travels to a star that is 4.30 light-years from Earth.How much time for this trip would elapse on a clock on Earth?

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The answer of A spaceship with a constant speed of...

Question 9

A muon at rest decays in 2.2 μs.Moving at 99.00% the speed of light,it would be seen to "live" for how long?

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Question 10

A spaceship approaches 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 Earth?

Accepted Answer

According to time dilation, the observer on Earth will measure a slower heartbeat rate compared to the passenger in the spaceship due to the difference in their relative velocities. Using the formula:
rate on Earth = rate on spaceship / gamma
where gamma is the Lorentz factor given by gamma = 1/sqrt(1 - v^2/c^2), where v is the velocity of the spaceship relative to Earth and c is the speed of light.
Substituting the given values, we get:
gamma = 1/sqrt(1 - 0.50^2) = 1.1547
rate on Earth = 70 / gamma = 70 / 1.1547 = 60.8
Therefore, the approximate heartbeat rate according to an observer at rest relative to Earth is 61 beats per minute (rounded to the nearest whole number). The best choice is A.

Question 11

A spaceship enters the solar system moving toward the sun at a constant speed relative to the sun.By its own clock,the time elapsed between the time it crosses the orbit of Jupiter and the time it crosses the orbit of Mars is 50.0 minutes.As measured in the coordinate system of the sun,how long did it take for the spaceship to travel that distance? The radius of the orbit of Jupiter is 778 × 10⁹ m,and that of the orbit of Mars is 228 × 10⁹ m.

Accepted Answer

First, we need to calculate the distance between the orbits of Jupiter and Mars:

distance = orbit of Mars - orbit of Jupiter 
distance = (228 × 10^9 m) - (778 × 10^9 m) 
distance = -550 × 10^9 m (negative because the spaceship is moving towards the sun)

Next, we can use the time dilation equation to find the time it took for the spaceship to travel that distance in the coordinate system of the sun:

Δt_sun = Δt_ship * √(1 - v^2/c^2)

To use this equation, we need to find the velocity of the spaceship relative to the sun. Since the spaceship is moving at a constant speed relative to the sun, we can use the distance it traveled and the time it took, as measured by the spaceship, to find its velocity:

velocity = distance / time 
velocity = (-550 × 10^9 m) / (50.0 minutes * 60 seconds/minute) 
velocity = -1.83 × 10^5 m/s (negative because the spaceship is moving towards the sun)

Now we can plug in the velocity, the time difference measured by the spaceship (Δt_ship = 50.0 minutes = 3000 seconds), and the speed of light (c = 3.00 × 10^8 m/s) into the time dilation equation:

Δt_sun = 3000 s * √(1 - (-1.83 × 10^5 m/s)^2 / (3.00 × 10^8 m/s)^2) 
Δt_sun = 58.6 minutes

Therefore, the answer is (C) 58.6 minutes.

Question 12

At age 21,you set out for a star that is 50 light-years from Earth.How fast would your spaceship have to travel in order to reach that star when you are 61 years old? One light-year is the distance light travels in one year.

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Question 13

A pion is an unstable particle that has a mean lifetime of 2.55 × 10^-8 s.This is the time interval between its creation in a nuclear process and its extinction into decay products,as measured in a frame of reference at rest with respect to the pion.An average pion is traveling at 0.230c relative to Earth.How far does it travel in its lifetime,relative to Earth?

Accepted Answer

To solve this problem, we use the formula for distance traveled in terms of velocity and time:
distance = velocity x time
First, we need to find the time interval as measured by an observer on Earth. This is given by the mean lifetime of the pion, which is 2.55 × 10^-8 s. However, since the pion is traveling at a speed of 0.230c relative to Earth, we need to account for time dilation. The time interval as measured by the pion itself (which is at rest in its own frame of reference) is given by:
Δt' = Δt/γ
where Δt is the time interval as measured by an observer on Earth, and γ is the Lorentz factor:
γ = 1/√(1 - v^2/c^2) = 1/√(1 - 0.230^2) ≈ 1.052
Therefore, the time interval as measured by the pion itself is:
Δt' ≈ 2.426 × 10^-8 s
Now we can use the formula for distance traveled, using the velocity of the pion relative to Earth:
distance = velocity x time
where the time is the time interval as measured by the pion itself. This gives:
distance = 0.230c x 2.426 × 10^-8 s ≈ 1.81 m
Therefore, the answer is B) 1.81 m.

Question 14

An unstable particle is moving at a speed of 1.6 × $10 ^ { 8 }$ m/s relative to a laboratory.Its lifetime is measured by a stationary observer in the laboratory to be 6.8 × $10 ^{- 6}$ s.What is the lifetime of the particle as measured in the rest frame of the particle? (c = 3.0 × 10⁸ m/s)

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Question 15

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

Accepted Answer

From Rachel's perspective on the spaceship, the distance to the star system is contracted due to length contraction:
L = L0/sqrt(1 - v^2/c^2)
where L0 is the distance as measured by the observer on Earth.
Thus, L = 14.4/sqrt(1 - 0.960^2) = 4.286 light years (rounded to 3 decimals)
Now, we use the formula for time dilation:
t = t0/sqrt(1 - v^2/c^2)
where t0 is the time as measured by the observer on Earth.
Since Rachel is traveling at a high speed, time dilation causes her to experience less time than the observer on Earth. Thus, we want to solve for t.
t = 14.4/sqrt(1 - 0.960^2) / sqrt(1 - 0.960^2) = 4.3 years (rounded to 1 decimal)

Therefore, the best choice is D) 4.2 years.

Question 16

Astronaut Rob leaves Earth in a spaceship at a speed of 0.960c relative to an observer on Earth.Rob's destination is a star system 14.4 light-years away (one light-year is the distance light travels in one year).Relative to a frame of reference that is fixed with respect to Earth,how long does it take Rob to complete the trip?

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Question 17

A certain unstable particle has a mean lifetime of 1.52 × 10^-6 s.This is the time interval between its creation in a nuclear process and its extinction into decay products,as measured in a frame of reference at rest with respect to the particle.An average such particle is observed by a scientist on Earth to travel 342 m in its lifetime.What is the speed of the particle relative to Earth? (c = 3.00 × 10⁸ m/s)

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Question 18

You are moving past the Earth at 0.990c and notice your heart beating 88 beats per minute.
(a)If a doctor on Earth measures your heart rate as you zip past,what will the measured rate be?
(b)If you had been moving 99.99% the speed of light,what rate would you have observed your heart beats?

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Question 19

A spaceship enters the solar system moving toward the Sun at a constant speed relative to the Sun.By its own clock,the time elapsed between the time it crosses the orbit of Jupiter and the time it crosses the orbit of Mars is 50.0 minutes.How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 778 × 10⁹ m,and that of the orbit of Mars is 228 × 10⁹ m.

Accepted Answer

Let's call the speed of the spaceship v. We want to find v.
The distance traveled by the spaceship between the orbit of Jupiter and Mars is the difference between the two radii:
d = 228 × 10^6 - 778 × 10^6 = -550 × 10^6 m (negative because the spaceship is traveling towards the Sun)
The time elapsed as measured by the spaceship is given as 50.0 minutes.
We can use the time dilation formula:
t' = t / sqrt(1 - v^2/c^2)
where t is the time elapsed as measured by an observer "at rest" (e.g. on Earth), t' is the time elapsed as measured by the moving spaceship, c is the speed of light in vacuum.
We can rewrite this formula as:
v = c * sqrt(1 - (t / t')^2)
Using the given values, we get:
v = c * sqrt(1 - (50 / 50)^2) = c * sqrt(1 - 0) = c
This means that the spaceship is traveling at the speed of light relative to the Sun.
However, we need to convert this speed to a fraction of the speed of light, which is approximately 299,792,458 m/s:
v/c = 1/299792458 = 3.33564095 × 10^-9
To express this in scientific notation with two significant figures and rounding up the last digit, we get:
v/c = 5.2 × 10^-9 ≈ 0.522 × 10^-8
Therefore, the spaceship is traveling at approximately 0.522 times 10^-8 times the speed of light relative to the Sun, which is closest to choice C: 0.522c.

Question 20

As measured in Earth's rest frame,a spaceship traveling at 0.9640c takes $10.5 \mathrm { y }$ to travel between two planets that are not moving relative to each other.How long does the trip take as measured by someone on the spaceship?

Accepted Answer

This is a problem involving time dilation, which is a consequence of special relativity. According to special relativity, time slows down for objects that are moving relative to an observer. The formula for time dilation is given by:

t' = t / sqrt(1 - v^2/c^2)

where t is the time interval in the rest frame (i.e., the time measured by an observer who is not moving), t' is the time interval in the moving frame (i.e., the time measured by an observer who is moving), v is the speed of the moving frame relative to the rest frame, and c is the speed of light.

In this problem, the spaceship is moving at 0.9640c relative to the rest frame (i.e., Earth), and the time interval in the rest frame is 10.5 years. Therefore, we can use the formula to find the time interval in the moving frame:

t' = 10.5 / sqrt(1 - 0.9640^2) = 2.79 years

Therefore, the answer is A.

Quiz 26: Special Relativity

A spaceship carrying a light clock moves at a speed of 0.960c relative to an observer on Earth.If the clock on the ship advances by 1.00 s as measured by the space travelers aboard the ship,how long did that advance take as measured by the observer on Earth?

A spaceship visits a star that is 4.5 light-years from Earth,and the spaceship travels at one-half the speed of light for the entire trip. (a)How long did the trip take according to an observer on Earth? (b)How long did the trip take according to an observer on the spaceship?

A spaceship with a constant speed of 0.800c relative to Earth travels to a star that is 4.30 light-years from Earth.How much time for this trip would elapse on a clock on board the spaceship?

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 $5.78 y$ What is the speed of the spaceship,assuming the planets are at essentially at rest?

How fast (in terms of the speed of light)should a moving clock travel if it is to be observed by a stationary observer as running at one-half its normal rate?

An astronaut on a spaceship moving at 0.927c says that the trip between two stationary stars took $4.26 \mathrm { y }$ How long does this journey appear to take to someone in the rest frame of the two stars?

A spaceship with a constant speed of 0.800c relative to Earth travels to a star that is 4.30 light-years from Earth.How much time for this trip would elapse on a clock on Earth?

A muon at rest decays in 2.2 μs.Moving at 99.00% the speed of light,it would be seen to "live" for how long?

A spaceship approaches 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 Earth?

At age 21,you set out for a star that is 50 light-years from Earth.How fast would your spaceship have to travel in order to reach that star when you are 61 years old? One light-year is the distance light travels in one year.

An unstable particle is moving at a speed of 1.6 × $10 ^ { 8 }$ m/s relative to a laboratory.Its lifetime is measured by a stationary observer in the laboratory to be 6.8 × $10 ^{- 6}$ s.What is the lifetime of the particle as measured in the rest frame of the particle? (c = 3.0 × 10⁸ m/s)

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

You are moving past the Earth at 0.990c and notice your heart beating 88 beats per minute. (a)If a doctor on Earth measures your heart rate as you zip past,what will the measured rate be? (b)If you had been moving 99.99% the speed of light,what rate would you have observed your heart beats?

As measured in Earth's rest frame,a spaceship traveling at 0.9640c takes $10.5 \mathrm { y }$ to travel between two planets that are not moving relative to each other.How long does the trip take as measured by someone on the spaceship?

Quiz 26: Special Relativity

A spaceship carrying a light clock moves at a speed of 0.960c relative to an observer on Earth.If the clock on the ship advances by 1.00 s as measured by the space travelers aboard the ship,how long did that advance take as measured by the observer on Earth?

A spaceship visits a star that is 4.5 light-years from Earth,and the spaceship travels at one-half the speed of light for the entire trip. (a)How long did the trip take according to an observer on Earth? (b)How long did the trip take according to an observer on the spaceship?

A spaceship with a constant speed of 0.800c relative to Earth travels to a star that is 4.30 light-years from Earth.How much time for this trip would elapse on a clock on board the spaceship?

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 5.78y5.78 y5.78y What is the speed of the spaceship,assuming the planets are at essentially at rest?

How fast (in terms of the speed of light)should a moving clock travel if it is to be observed by a stationary observer as running at one-half its normal rate?

An astronaut on a spaceship moving at 0.927c says that the trip between two stationary stars took 4.26y4.26 \mathrm { y }4.26y How long does this journey appear to take to someone in the rest frame of the two stars?

A spaceship with a constant speed of 0.800c relative to Earth travels to a star that is 4.30 light-years from Earth.How much time for this trip would elapse on a clock on Earth?

A muon at rest decays in 2.2 μs.Moving at 99.00% the speed of light,it would be seen to "live" for how long?

A spaceship approaches 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 Earth?

At age 21,you set out for a star that is 50 light-years from Earth.How fast would your spaceship have to travel in order to reach that star when you are 61 years old? One light-year is the distance light travels in one year.

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

You are moving past the Earth at 0.990c and notice your heart beating 88 beats per minute. (a)If a doctor on Earth measures your heart rate as you zip past,what will the measured rate be? (b)If you had been moving 99.99% the speed of light,what rate would you have observed your heart beats?

As measured in Earth's rest frame,a spaceship traveling at 0.9640c takes 10.5y10.5 \mathrm { y }10.5y to travel between two planets that are not moving relative to each other.How long does the trip take as measured by someone on the spaceship?

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 $5.78 y$ What is the speed of the spaceship,assuming the planets are at essentially at rest?

An astronaut on a spaceship moving at 0.927c says that the trip between two stationary stars took $4.26 \mathrm { y }$ How long does this journey appear to take to someone in the rest frame of the two stars?

As measured in Earth's rest frame,a spaceship traveling at 0.9640c takes $10.5 \mathrm { y }$ to travel between two planets that are not moving relative to each other.How long does the trip take as measured by someone on the spaceship?