Question 1

An astronaut at rest has a heart rate of 65 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 0.60 c?
A)39 beats/min
B)52 beats/min
C)108 beats/min
D)81 beats/min

Accepted Answer

Due to time dilation in special relativity, the heart rate measured by the Earth observer will be slower. The time dilation factor is given by the Lorentz factor, which is approximately 0.8 for 0.60c. Thus, the heart rate will be 65 beats/min * 0.8 = 52 beats/min.

Question 2

A muon formed high in Earth's atmosphere travels at a speed 0.990 0 c for a distance (as we see it)of 4 600 m before it decays.How far does the muon travel as measured in its frame?
A)4 554 m
B)2 596 m
C)1 298 m
D)649 m

Accepted Answer

The answer of A muon formed high in Earth's atmosphere...

Question 3

From a stationary position,I observe a moving boxcar,which has a mirror along the front wall,but it is open at the back of the boxcar.I send a flash of light from my flashlight and time the flash of light as it goes to the front of the boxcar and returns to the back of the boxcar.A passenger in the boxcar also times the round trip of the flash of light.Compare the times recorded on our watches.
A)The time recorded on his watch is longer.
B)The time recorded on the two watches is the same.
C)The time recorded on his watch is shorter.
D)The answer depends on the reference system you are in.

Accepted Answer

The answer of From a stationary position,I observe a moving...

Question 4

A space probe has an 18.0-m length when measured at rest.What length does an observer at rest measure when the probe is going by at a speed of 0.700 c?
A)25.2 m
B)12.9 m
C)12.6 m
D)9.18 m

Accepted Answer

The answer of A space probe has an 18.0-m length...

Question 5

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 = 0.95 c.What would the astronauts measure as the distance to Alpha Centauri?
A)4.0 lightyears
B)2.7 lightyears
C)1.9 lightyears
D)1.3 lightyears

Accepted Answer

According to special relativity, length contraction occurs when an object is moving relative to an observer. The equation for length contraction is:

L = L0/γ

where L is the contracted length, L0 is the rest length, and γ is the Lorentz factor, which depends on the relative velocity of the object and the observer.

In this case, the rest distance to Alpha Centauri is 4.2 lightyears. If the astronauts are traveling at 0.95 c, then γ = 1/√(1-(0.95)^2) = 3.2.

Plugging in the values, we get:

L = 4.2/3.2 = 1.3 lightyears

Therefore, the distance to Alpha Centauri as measured by the astronauts would be 1.3 lightyears. The best choice is D.

Question 6

If astronauts could travel at v = 0.95 c,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)1.9 years
C)2.4 years
D)3.0 years

Accepted Answer

Explanation:The time dilation equation is:t' = t / γwhere t' is the time measured by the observer in the moving frame, t is the time measured by the observer in the stationary frame, and γ is the Lorentz factor:γ = 1 / sqrt(1 - v^2/c^2)Plugging in the given values, we get:γ = 1 / sqrt(1 - 0.95^2) = 3t' = 4.4 years / 3 = 1.4 yearsTherefore, the astronauts would measure 1.4 years of time passing on their clocks.

Question 7

A muon formed high in the Earth's atmosphere travels at a speed 0.990 c for a distance of 4.60 km before it decays.What is the muon's lifetime as measured in its reference frame? A)1.55 x 10^-5 s B)2.18 x 10^-6 s C)3.04 x 10^-6 s D)4.65 x 10^-6 s

Accepted Answer

The muon's lifetime in its reference frame is calculated using time dilation. The proper time (muon's lifetime in its frame) is the distance divided by the speed, adjusted by the Lorentz factor for relativistic speeds.

Question 8

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

According to the theory of Special Relativity, the speed of light is constant in all inertial frames of reference. Since the boxcar is moving at a constant speed, the speed of light in both directions is the same relative to the boxcar. Therefore, both flashes of light will take the same amount of time to travel from one end of the boxcar to the other end, and the answer is C – They both took the same time.

Question 9

A spaceship of triangular shape,having a length twice 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)0.500 c
B)0.750 c
C)0.866 c
D)This is not possible.

Accepted Answer

According to the theory of relativity, the length of an object appears to decrease as its speed approaches the speed of light. This is known as length contraction.

Let's assume that the spaceship is moving at a speed v. Its length is given as twice the width, so if its width is w, then its length is 2w.

To a stationary observer, the length of the spaceship appears to be contracted to its width. So, we can set up the following equation:

2w*sqrt(1-v^2/c^2) = w

where c is the speed of light.

Simplifying the equation, we get:

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

Squaring both sides, we get:

1-v^2/c^2 = 1/4

v^2/c^2 = 3/4

v/c = sqrt(3)/2

v = (sqrt(3)/2)*c

This is approximately 0.866 c, which is the speed at which the spaceship would have to move so that its length would appear equal to its width to a stationary observer. Therefore, the correct answer is C.

Question 10

The astronaut whose heart rate on Earth is 60 beats/min increases his velocity to v = 0.80 c.Now what is his heart rate as measured by an Earth observer?
A)36 beats/min
B)48 beats/min
C)75 beats/min
D)100 beats/min

Accepted Answer

Due to time dilation in special relativity, the astronaut's heart rate as observed from Earth will decrease. The time dilation factor is given by the Lorentz factor, which is approximately 0.6 for v = 0.80c. Thus, the heart rate observed from Earth is 60 beats/min * 0.6 = 36 beats/min.

Question 11

A rocket ship is 80.0 m in length when measured before leaving the launching pad.What would its velocity be if a ground observer measured its length as 60.0 m while it is in flight? (c = 3.00 x 10⁸ m/s) A)0.980 x 10⁸ m/s B)1.15 x 10⁸ m/s C)1.33 x 10⁸ m/s D)1.98 x 10⁸ m/s

Accepted Answer

The answer of A rocket ship is 80.0 m in...

Question 12

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 13

A meter stick moving in a direction parallel to its length appears to be only 40.0 cm long to an observer.What is the meter stick's speed relative to the observer? (c = 3.00 x 10⁸ m/s) A)1.19 x 10⁸m/s B)2.52 x 10⁸ m/s C)2.75 x 10⁸ m/s D)2.93 x 10⁸ m/s

Accepted Answer

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

Question 14

An earth observer sees a spaceship at an altitude of 980 m moving downward toward the earth at a speed of 0.800 c.What is the spaceship's altitude as measured by an observer in the spaceship?
A)1 630 m
B)1 270 m
C)893 m
D)588 m

Accepted Answer

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

Question 15

Muons at speed 0.999 4 c are sent round and round a circular storage ring of radius 500 m.If a muon at rest decays into other particles after an average T = 2.2 x 10^-6 s,how many trips around the storage ring do we expect the 0.999 4 c muons to make before they decay? A)0.2 B)2 C)4 D)6

Accepted Answer

The time dilation effect in special relativity means that moving muons will have a longer lifetime than stationary ones. The time dilation factor (gamma) is approximately 29.3 for a speed of 0.9994c. Therefore, the moving muons will last about 29.3 times longer than 2.2 x 10^-6 s, which is about 6.45 x 10^-5 s. The circumference of the storage ring is 2π x 500 m, and at 0.9994c, the muons travel this distance in about 2.1 x 10^-6 s per trip. Dividing the dilated lifetime by the time per trip gives approximately 6 trips.

Question 16

A boxcar without a front or a back is moving toward the right.Two electrons move through the boxcar,one moving from back to front toward the right,the other moving from front to back toward the left.According to me,each electron is moving with a speed of 0.8 c,and the boxcar is moving with a speed of 0.6 c.A passenger in the boxcar records how long it takes each electron to pass from one end of the boxcar to the other end.According to the passenger,which took longer?
A)the electron going from back to front
B)the electron going from front to back
C)They both took the same time.
D)Since nothing can go faster than light,an electron cannot move toward the left with a speed of 0.8 c through a boxcar moving toward the right with a speed of 0.6 c.

Accepted Answer

The time taken by each electron to traverse the boxcar will appear different to the passenger due to relativistic effects. Specifically, the velocities of the electrons relative to the boxcar are not simply additive or subtractive because of the relativistic velocity addition formula. For the electron moving from back to front (toward the right, in the same direction as the boxcar), its speed relative to the boxcar is less than 0.8c when corrected for relativistic effects. For the electron moving from front to back (toward the left, opposite to the boxcar's direction), its speed relative to the boxcar is also adjusted by relativistic effects but in a manner that makes its effective speed in the boxcar's frame slower than the first electron's effective speed. Thus, due to these relativistic adjustments, the electron moving from back to front effectively travels faster relative to the boxcar and takes less time to cross it than the electron moving from front to back, making the latter take longer from the passenger's perspective.

Question 17

At what speed would a clock have to be moving in order to run at a rate that is one-third the rate of a clock at rest?
A)0.79 c
B)0.89 c
C)0.94 c
D)0.97 c

Accepted Answer

According to time dilation, the rate of time on a moving object relative to a stationary object is given by the equation:
t' = t / sqrt(1-v^2/c^2)
where t is the time on the stationary clock, t' is the time on the moving clock, v is the velocity of the moving clock, and c is the speed of light.
If the moving clock is running at one-third the rate of the stationary clock, then t' = (1/3)t. Plugging these values into the time dilation equation and solving for v, we get:
1/3 = 1 / sqrt(1-v^2/c^2)
v^2/c^2 = 0.9444
v/c = sqrt(0.9444) = 0.971
So, the clock would have to be moving at a speed of 0.971c (rounded to two decimal places). The closest answer choice is option C, which is 0.94c.

Question 18

A knight on horseback holds a 10-m lance.The horse can run at 0.70 c.(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)7.1 m
B)10 m
C)14 m
D)15 m

Accepted Answer

Due to the effects of special relativity, specifically length contraction, the lance will appear shorter to a stationary observer. The contracted length L is given by L = L0 * sqrt(1 - v^2/c^2), where L0 is the proper length (10 m), v is the velocity (0.70c), and c is the speed of light. Calculating this gives approximately 7.1 m.

Question 19

How fast would a rocket have to move past a ground observer if the latter were to observe a 4.0% length shrinkage in the rocket length? (c = 3.00 ´ 10⁸ m/s) A)0.12 x 10⁸ m/s B)0.28 x 10⁸ m/s C)0.84 x 10⁸ m/s D)1.2 x 10⁸ m/s

Accepted Answer

We can use the formula for length contraction:

L' = L * sqrt(1 - v^2/c^2)

where L is the original length, L' is the contracted length, v is the velocity of the rocket, and c is the speed of light.

We are given that L' = 0.96L (since there is a 4.0% length shrinkage). Substituting this and c into the equation, we get:

0.96L = L * sqrt(1 - v^2/c^2)

Squaring both sides and isolating v^2/c^2, we get:

v^2/c^2 = 1 - (0.96)^2

v/c = sqrt(1 - (0.96)^2)

v = sqrt(1 - (0.96)^2) * c

v ≈ 0.84 * 10^8 m/s

Therefore, the answer is (C).

Question 20

From a stationary position,I observe a moving boxcar,which has a mirror along the front wall,but it is open at the back of the boxcar.I send a flash of light from my flashlight and time the flash of light as it goes to the front of the boxcar and returns to the back of the boxcar.A passenger in the boxcar also times the round trip of the flash of light.Previously I had measured the time required for the round trip of a flash of light when the boxcar was stationary,and I call this the stationary time.Which two times are the same?
A)the time recorded on my watch and the previous stationary time
B)the time recorded on the passenger's watch and the previous stationary time
C)the time recorded on my watch and the time recorded on the passenger's watch
D)None of the times are the same.

Accepted Answer

The answer of From a stationary position,I observe a moving...

Quiz 26: Relativity

An astronaut at rest has a heart rate of 65 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 0.60 c?

A muon formed high in Earth's atmosphere travels at a speed 0.990 0 c for a distance (as we see it) of 4 600 m before it decays.How far does the muon travel as measured in its frame?

A space probe has an 18.0-m length when measured at rest.What length does an observer at rest measure when the probe is going by at a speed of 0.700 c?

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 = 0.95 c.What would the astronauts measure as the distance to Alpha Centauri?

If astronauts could travel at v = 0.95 c,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 muon formed high in the Earth's atmosphere travels at a speed 0.990 c for a distance of 4.60 km before it decays.What is the muon's lifetime as measured in its reference frame?

A spaceship of triangular shape,having a length twice 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?

The astronaut whose heart rate on Earth is 60 beats/min increases his velocity to v = 0.80 c.Now what is his heart rate as measured by an Earth observer?

A rocket ship is 80.0 m in length when measured before leaving the launching pad.What would its velocity be if a ground observer measured its length as 60.0 m while it is in flight? (c = 3.00 x 10⁸ m/s)

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 meter stick moving in a direction parallel to its length appears to be only 40.0 cm long to an observer.What is the meter stick's speed relative to the observer? (c = 3.00 x 10⁸ m/s)

An earth observer sees a spaceship at an altitude of 980 m moving downward toward the earth at a speed of 0.800 c.What is the spaceship's altitude as measured by an observer in the spaceship?

Muons at speed 0.999 4 c are sent round and round a circular storage ring of radius 500 m.If a muon at rest decays into other particles after an average T = 2.2 x 10^-6 s,how many trips around the storage ring do we expect the 0.999 4 c muons to make before they decay?

At what speed would a clock have to be moving in order to run at a rate that is one-third the rate of a clock at rest?

A knight on horseback holds a 10-m lance.The horse can run at 0.70 c.(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?

How fast would a rocket have to move past a ground observer if the latter were to observe a 4.0% length shrinkage in the rocket length? (c = 3.00 ´ 10⁸ m/s)