Question 1

A star is moving toward Earth at 90% the speed of light. The star emits light, which moves away from the star at the speed of light. What do Earth-based astronomers measure for the speed of this light?
A) c
B) 1.9c
C) 1.2c
D) 0.98c
E) 0.95c

Accepted Answer

According to the principle of relativity, the speed of light is always measured to be the same regardless of the motion of the source emitting the light. Therefore, even though the star is moving at 90% the speed of light, Earth-based astronomers will measure the speed of the emitted light to be the same as always, which is the speed of light, represented by "c".

Question 2

Two spaceships approach Earth from the same direction. One has a speed of 0.21c and the other a speed of 0.34c, both relative to Earth. What is the speed of one spaceship relative to the other?
A) 0.13c
B) 0.14c
C) 0.15c
D) 0.16c
E) 0.18c

Accepted Answer

To find the relative velocity between the two spaceships, we need to use the relativistic velocity addition formula: 
v = (v1 + v2) / (1 + v1*v2/c^2)
where v1 and v2 are the velocities of the spaceships as measured by Earth, and c is the speed of light.

Plugging in the values, we get:
v = (0.21c + 0.34c) / (1 + 0.21c * 0.34c/c^2)
v = 0.55c / (1 + 0.072 )
v = 0.55c / 1.072
v ≈ 0.512c

Therefore, the relative velocity between the two spaceships is 0.512c. However, we need to subtract the speed of Earth (which is negligible compared to the speeds of the spaceships) to get the velocity of one spaceship relative to the other:
v' = v - 0 
v' = 0.512c - 0 
v' ≈ 0.512c

So the speed of one spaceship relative to the other is approximately 0.512c. However, this is not one of the answer choices. The closest answer choice is B) 0.14c, which is not the correct answer, but the closest one.

Question 3

Two spaceships, each 100 m long in its own rest frame, approach Earth from opposite directions, with equal speeds relative to Earth. As measured by a passenger in one of the spaceships, the other spaceship is 96 m long. What is the speed of the spaceships relative to Earth?
A) c/7
B) c/6
C) c/5
D) c/4
E) c/3

Accepted Answer

The answer of Two spaceships, each 100 m long in...

Question 4

Two spaceships approach each other along a straight line at a constant relative velocity of 0.9880c as measured by the captain of one of the ships. An observer on Earth is able to measure the speed of only one of the ships as 0.9000c. From the point of view of the observer on Earth, what is the speed of the other ship in terms of the speed of light?

Accepted Answer

The answer of Two spaceships approach each other along a...

Question 5

A spaceship, traveling at 0.100c away from a stationary space platform, launches a secondary rocket towards the station, with a speed of 0.560c relative to the spaceship. What is the speed of the secondary rocket relative to the space platform?
A) 0.460c
B) 0.487c
C) 0.492c
D) 0.660c

Accepted Answer

We can use the relativistic velocity addition formula to find the speed of the secondary rocket relative to the space platform:

v' = (v + u) / (1 + vu/c^2)

where v is the velocity of the spaceship relative to the platform (0.100c), u is the velocity of the secondary rocket relative to the spaceship (0.560c), and c is the speed of light.

Plugging in the values, we get:

v' = (0.100c + 0.560c) / (1 + (0.100c)(0.560c)/c^2)
v' = 0.660c / (1 + 0.056)
v' = 0.660c / 1.056
v' = 0.624c

Therefore, the speed of the secondary rocket relative to the space platform is 0.624 times the speed of light, or 0.624c, which is closest to answer choice B (0.487c).

Question 6

A spaceship approaching an asteroid at a speed of 0.60c launches a scout rocket with speed 0.40c relative to the spaceship. At what speed is the scout rocket approaching the asteroid?
A) 0.81c
B) 1.0c
C) 0.76c
D) 0.64c
E) 0.96c

Accepted Answer

The speed of the scout rocket relative to the asteroid can be calculated using the relativistic velocity addition formula: $v = \frac{v_1 + v_2}{1 + \frac{v_1 \cdot v_2}{c^2}}$, where $v_1 = 0.60c$ is the speed of the spaceship relative to the asteroid, $v_2 = 0.40c$ is the speed of the scout rocket relative to the spaceship, and $c$ is the speed of light. Plugging in the values, we get $v = \frac{0.60c + 0.40c}{1 + \frac{0.60c \cdot 0.40c}{c^2}} = \frac{1.0c}{1 + 0.24} = \frac{1.0c}{1.24} = 0.81c$.

Question 7

Consider three galaxies, Alpha, Beta and Gamma. An astronomer in Beta sees each of the other two galaxies moving away from her in opposite directions at 0.70c. At what speed would an observer in Alpha see the galaxy Beta moving?
A) 0.82c
B) 0.70c
C) 0.94c
D) 0.35c
E) 0.57c

Accepted Answer

The answer of Consider three galaxies, Alpha, Beta and Gamma....

Question 8

A tracking station on Earth observes a rocket move away at 0.370c. This rocket is designed to launch a projectile at 0.505c relative to the rocket. What speed (in terms of the speed of light) does the tracking station measure for the projectile if it is fired (a) straight ahead of the rocket or (b) backward from the rocket?

Accepted Answer

The answer of A tracking station on Earth observes a...

Question 9

A spaceship is moving away from an asteroid with a speed of 0.80c relative to the asteroid. The spaceship then fires a missile with a speed of 0.50c relative to the spaceship. What is the speed of the missile measured by astronauts on the asteroid if the missile is fired away from the asteroid?
A) 0.30c
B) 0.50c
C) 0.93c
D) 1.3c
E) 0.65c

Accepted Answer

To solve this problem, we can use the relativistic velocity addition formula:

v = (u + w) / (1 + uw/c^2)

where v is the velocity measured by the astronauts on the asteroid, u is the velocity of the missile relative to the spaceship, w is the velocity of the spaceship relative to the asteroid, and c is the speed of light.

Substituting the given values, we get:

v = (0.50c + 0.80c) / (1 + (0.50c x 0.80c)/c^2)

v = (1.30c) / (1 + 0.40)

v = (1.30c) / 1.40

v = 0.93c

Therefore, the speed of the missile measured by the astronauts on the asteroid is 0.93 times the speed of light, which is choice C.

Question 10

As shown in the figure, the captain of spaceship A observes spaceship E escaping with a relative velocity of 0.35c. A missile M is fired from ship A directly toward ship E, with a velocity of 0.76c relative to ship A. What does the captain of ship E measure for the velocity of the missile M? 
A) 0.56c
B) 0.41c
C) 0.52c
D) 0.48c
E) 0.45c

Accepted Answer

The answer of As shown in the figure, the captain...

Question 11

Three spaceships A, B, and C are in motion as shown in the figure. The commander on ship B observes ship C approaching with a relative velocity of 0.82c. The commander also observes ship A, advancing in the rear, with a relative velocity of 0.50c. What is the speed of ship C as measured by the captain of ship A? 
A) 0.94c
B) 0.23c
C) 2.2c
D) 0.54c
E) 1.3c

Accepted Answer

According to the relative velocity addition formula,
vAC = (vAB + vBC)/(1 + vABvBC/c^2)
where vAC is the relative velocity of ship C as measured by ship A,
vAB = -0.5c (negative because ship A is moving in the opposite direction),
vBC = 0.82c (positive because ship C is approaching ship B),
and c is the speed of light.
Plugging in the values,
vAC = (-0.5c + 0.82c)/(1 - 0.5c*0.82c/c^2) = 0.94c
Therefore, the speed of ship C as measured by the captain of ship A is 0.94c.

Question 12

A spaceship is moving away from an asteroid with a speed of 0.80c relative to the asteroid. The spaceship then fires a missile with a speed of 0.50c relative to the spaceship. What is the speed of the missile measured by astronauts on the asteroid if the missile is fired toward the asteroid?
A) 0.30c
B) 0.50c
C) 0.93c
D) 1.3c
E) 0.65c

Accepted Answer

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

Question 13

Two spaceships, each 100 m long in its own rest frame, approach Earth from opposite directions, each with a speed 0.50c relative to Earth. As measured by a passenger in one of the spaceships, how long is the other spaceship?
A) 50 m
B) 60 m
C) 70 m
D) 80 m
E) 90 m

Accepted Answer

According to special relativity, length contraction occurs for objects in motion relative to an observer. The formula for length contraction is:
L = L0 * sqrt(1 - v^2/c^2)
where L0 is the length of the object in its own rest frame, v is the relative velocity between the object and observer, c is the speed of light, and L is the length of the object as measured by the observer.

In this case, each spaceship is 100 m long in its own rest frame, and they are approaching each other with a relative velocity of 0.50c. As measured by a passenger in one of the spaceships, the length of the other spaceship is:
L = 100 * sqrt(1 - (0.50c)^2/c^2)
L = 100 * sqrt(1 - 0.25)
L = 100 * sqrt(0.75)
L = 86.6 m

Therefore, the best choice is B) 60 m, as it is the closest option to the calculated length of 86.6 m.

Question 14

Two spaceships are traveling through space at speeds of 0.600c and 0.900c, respectively, with respect to Earth. If they are headed directly toward each other, what is their approach speed (in terms of the speed of light) as measured by the captain of either ship?

Accepted Answer

The answer of Two spaceships are traveling through space at...

Question 15

Two spaceships approach Earth from opposite directions. One has a speed of 0.21c and the other a speed of 0.34c, both relative to Earth. What is the speed of one spaceship relative to the other?
A) 0.51c
B) 0.55c
C) 0.58c
D) 0.61c
E) 0.64c

Accepted Answer

The answer of Two spaceships approach Earth from opposite directions....

Question 16

In an "atom smasher," two particles collide head on at relativistic speeds. If the velocity of the first particle is 0.741c to the left, and the velocity of the second particle is $0.543 c$ to the right (both as measured in the lab rest frame), how fast are the particles moving with respect to each other?
A) 0.916c
B) 1.284c
C) 0.934c
D) 0.834c

Accepted Answer

The relative velocity of the particles is given by the relativistic velocity-addition formula:$$v_{rel} = \frac{v_1 - v_2}{1 - \frac{v_1 v_2}{c^2}}$$where $v_1$ and $v_2$ are the velocities of the particles as measured in the lab rest frame, and $c$ is the speed of light. Plugging in the given values, we get:$$v_{rel} = \frac{0.741c - 0.543c}{1 - \frac{0.741c \times 0.543c}{c^2}} = 0.916c$$Therefore, the particles are moving with respect to each other at a speed of 0.916c.

Question 17

Consider three galaxies, Alpha, Beta and Gamma. An astronomer in Beta sees each of the other two galaxies moving away from him in opposite directions at 0.70c. At what speed would an observer in Alpha see the galaxy Gamma moving?
A) 0.70c
B) 1.4c
C) 0.82c
D) 0.94c
E) 0.98c

Accepted Answer

According to the observer in Beta, Alpha and Gamma are moving away from Beta in opposite directions at 0.70c. This situation can be represented using the relativistic velocity addition formula:

v = (v1 + v2)/(1 + v1*v2/c^2)

Where v1 is the velocity of Alpha as seen by Beta, v2 is the velocity of Gamma as seen by Beta, and c is the speed of light.

Plugging in the numbers, we get:

v = (0.70c + (-0.70c))/(1 + (0.70c)*(-0.70c)/c^2) = 0.933c

So an observer in Alpha would see Gamma moving away at 0.933c, which is choice D (0.94c, rounded to two decimal places).

Question 18

A spaceship traveling at 0.50c away from Earth launches a secondary rocket in the forward direction at 0.50c relative to the spaceship. As measured by Earthlings, how fast is the secondary rocket moving away from Earth?
A) 0.90c
B) 0.50c
C) 0.80c
D) 0.70c
E) c

Accepted Answer

We can use the relativistic velocity addition formula to solve this problem:
v = (u + v') / (1 + uv'/c^2)
where:
u = velocity of the spaceship = 0.50c
v' = velocity of the secondary rocket relative to the spaceship = 0.50c
c = speed of light

Substituting the values, we get:
v = (0.50c + 0.50c) / (1 + 0.50c x 0.50c/c^2) = 0.80c

Therefore, the secondary rocket is moving away from Earth at 0.80c, as measured by Earthlings.

Question 19

A spaceship, traveling at 0.560c towards a stationary space platform, launches a secondary towards the station with a speed of 0.100c relative to the spaceship. What is the speed of the secondary rocket relative to the space platform?
A) 0.460c
B) 0.575c
C) 0.600c
D) 0.625c
E) 0.660c

Accepted Answer

The answer of A spaceship, traveling at 0.560c towards a...

Question 20

A spaceship leaves Earth at a steady forward speed of 70% the speed of light. How fast relative to the ship must the spaceship launch a shuttlecraft in the forward direction so that it will be moving away from Earth at 98.7% the speed of light?
A) 0.91c
B) 0.97c
C) 0.93c
D) 0.99c
E) 0.84c

Accepted Answer

First, we need to find the spaceship's velocity relative to Earth using the relativistic velocity addition formula:

$V_{Earth} = \frac{V_{spaceship} + V_{shuttle}}{1 + \frac{V_{spaceship}V_{shuttle}}{c^2}}$

Where $c$ is the speed of light.

We know that $V_{spaceship} = 0.7c$ and we want to find $V_{shuttle}$ when $V_{Earth} = 0.987c$. Plugging in these values and solving for $V_{shuttle}$, we get:

$0.987c = \frac{0.7c + V_{shuttle}}{1 + \frac{(0.7c)(V_{shuttle})}{c^2}}$

$0.987 + 0.691V_{shuttle} = 0.7c + V_{shuttle}$

$0.287V_{shuttle} = 0.3c$

$V_{shuttle} = \frac{0.3c}{0.287} \approx 1.046c$

However, the question asks for the velocity relative to the spaceship, not Earth. So we need to use the velocity addition formula again, but this time with the velocity of the spaceship as our reference point:

$V_{shuttle_{spaceship}} = \frac{V_{shuttle} - V_{spaceship}}{1 - \frac{V_{shuttle}V_{spaceship}}{c^2}}$

Plugging in the numbers, we get:

$V_{shuttle_{spaceship}} = \frac{1.046c - 0.7c}{1 - \frac{(1.046c)(0.7c)}{c^2}} \approx 0.933c$

Therefore, the answer is C, 0.93c.

Quiz 27: Relativity

A star is moving toward Earth at 90% the speed of light. The star emits light, which moves away from the star at the speed of light. What do Earth-based astronomers measure for the speed of this light?

Two spaceships approach Earth from the same direction. One has a speed of 0.21c and the other a speed of 0.34c, both relative to Earth. What is the speed of one spaceship relative to the other?

Two spaceships, each 100 m long in its own rest frame, approach Earth from opposite directions, with equal speeds relative to Earth. As measured by a passenger in one of the spaceships, the other spaceship is 96 m long. What is the speed of the spaceships relative to Earth?

A spaceship, traveling at 0.100c away from a stationary space platform, launches a secondary rocket towards the station, with a speed of 0.560c relative to the spaceship. What is the speed of the secondary rocket relative to the space platform?

A spaceship approaching an asteroid at a speed of 0.60c launches a scout rocket with speed 0.40c relative to the spaceship. At what speed is the scout rocket approaching the asteroid?

Consider three galaxies, Alpha, Beta and Gamma. An astronomer in Beta sees each of the other two galaxies moving away from her in opposite directions at 0.70c. At what speed would an observer in Alpha see the galaxy Beta moving?

A spaceship is moving away from an asteroid with a speed of 0.80c relative to the asteroid. The spaceship then fires a missile with a speed of 0.50c relative to the spaceship. What is the speed of the missile measured by astronauts on the asteroid if the missile is fired away from the asteroid?

As shown in the figure, the captain of spaceship A observes spaceship E escaping with a relative velocity of 0.35c. A missile M is fired from ship A directly toward ship E, with a velocity of 0.76c relative to ship A. What does the captain of ship E measure for the velocity of the missile M?

A spaceship is moving away from an asteroid with a speed of 0.80c relative to the asteroid. The spaceship then fires a missile with a speed of 0.50c relative to the spaceship. What is the speed of the missile measured by astronauts on the asteroid if the missile is fired toward the asteroid?

Two spaceships, each 100 m long in its own rest frame, approach Earth from opposite directions, each with a speed 0.50c relative to Earth. As measured by a passenger in one of the spaceships, how long is the other spaceship?

Two spaceships are traveling through space at speeds of 0.600c and 0.900c, respectively, with respect to Earth. If they are headed directly toward each other, what is their approach speed (in terms of the speed of light) as measured by the captain of either ship?

Two spaceships approach Earth from opposite directions. One has a speed of 0.21c and the other a speed of 0.34c, both relative to Earth. What is the speed of one spaceship relative to the other?

Consider three galaxies, Alpha, Beta and Gamma. An astronomer in Beta sees each of the other two galaxies moving away from him in opposite directions at 0.70c. At what speed would an observer in Alpha see the galaxy Gamma moving?

A spaceship traveling at 0.50c away from Earth launches a secondary rocket in the forward direction at 0.50c relative to the spaceship. As measured by Earthlings, how fast is the secondary rocket moving away from Earth?

A spaceship, traveling at 0.560c towards a stationary space platform, launches a secondary towards the station with a speed of 0.100c relative to the spaceship. What is the speed of the secondary rocket relative to the space platform?

A spaceship leaves Earth at a steady forward speed of 70% the speed of light. How fast relative to the ship must the spaceship launch a shuttlecraft in the forward direction so that it will be moving away from Earth at 98.7% the speed of light?