Question 1

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?

Accepted Answer

We cannot simply add velocities in special relativity as in classical mechanics. Instead, we use the formula for relativistic velocity addition: v = (u + w)/(1 + uw/c^2), where u and w are the velocities and c is the speed of light. 
Let's choose one spaceship as the reference frame (call it spaceship A) and find the velocity of the other spaceship (call it spaceship B) relative to A. Using the given velocities, we have:
v = (u + w)/(1 + uw/c^2)
v = (0.21c + 0.34c)/(1 + (0.21c)(0.34c)/c^2)
v = 0.55c/(1 + 0.07254)
v = 0.55c/1.07254
v ≈ 0.5125c
Therefore, the speed of spaceship B relative to spaceship A is approximately 0.5125c. However, the problem doesn't ask for the exact answer, so we choose the closest option, which is B: 0.14c.

Question 2

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?

Accepted Answer

According to the principle of relativity, both Beta and Alpha observers should see the same relative velocity between Gamma and Beta, but in opposite directions. Therefore, the velocity of Gamma with respect to Alpha would be the vector sum of the velocities of Gamma with respect to Beta and Beta with respect to Alpha. Using the relativistic velocity addition formula, we get:
v = (0.70c + (-0.70c)) / (1 - 0.70c x (-0.70c) / c^2) = 0.94c
So the correct answer is D) 0.94c.

Question 3

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?

Accepted Answer

The answer of A spaceship,traveling at 0.100c away from a...

Question 4

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?

Accepted Answer

The speed of light is always constant and independent of the motion of the source or the observer. Therefore, the light emitted by the star will always be measured by Earth-based astronomers at the speed of light, regardless of the speed of the star.

Question 5

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 6

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?

Accepted Answer

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

Question 7

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?

Accepted Answer

According to the relativistic velocity addition formula: V = (v + u)/(1 + vu/c^2), where v is the relative velocity between the observer and object, u is the velocity of the object, and c is the speed of light.

The captain of ship A observes ship B moving at a relative velocity of 0.50c. Let the velocity of ship C be u.

Relative velocity of ship C with respect to ship B = 0.82c (given)

Using the velocity addition formula, we get:

V = (0.82c + 0.50c)/(1 + (0.82c x 0.50c)/c^2)
V = (1.32c)/(1 + 0.41)
V = 0.94c

Therefore, the speed of ship C as measured by the captain of ship A is 0.94c.

Question 8

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?

Accepted Answer

We will use the relativistic velocity addition formula:
v' = (v + u) / (1 + vu/c^2)

where,
v = velocity of the spaceship relative to the space platform = 0.560c
u = velocity of the secondary rocket relative to the spaceship = 0.100c
c = speed of light

Plugging in the values, we get:
v' = (0.560c + 0.100c) / (1 + (0.560c x 0.100c)/(c^2))
v' = 0.625c

Therefore, the speed of the secondary rocket relative to the space platform is 0.625c, which is option D (0.625c is the closest answer option to our calculation).

Question 9

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?

Accepted Answer

The relative velocity of two objects moving at relativistic speeds, $v_1$ and $v_2$, is given by the formula $\frac{v_1 + v_2}{1 + \frac{v_1v_2}{c^2}}$, where $c$ is the speed of light. Substituting $v_1 = -0.741c$ and $v_2 = 0.543c$ (taking left as negative), the calculation yields a relative speed of approximately $0.916c$, making option A correct.

Question 10

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?

Accepted Answer

According to the principle of relativity, both spaceships are equivalent frames of reference. Therefore, we can expect that the length of the other spaceship will appear to be shorter to the passenger in the first spaceship. Using the formula for length contraction:

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

where L is the rest length of the spaceship (100 m), v is the relative velocity (0.5c) and c is the speed of light (299,792,458 m/s), we can calculate the length of the other spaceship as measured by a passenger in one of the spaceships:

L' = 100(1 - 0.5^2/1^2)^1/2
L' = 100(0.75)^1/2
L' = 86.603 m

Therefore, the length of the other spaceship as measured by a passenger in one of the spaceships is 86.603 m. However, we must remember to account for the fact that the other spaceship is moving towards the passenger, so we need to use the relativistic velocity addition formula:

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

where v is the velocity of the first spaceship (0.5c), u is the velocity of the second spaceship (-0.5c), and v' is the relative velocity between them as measured by the first spaceship:

v' = (0.5c - 0.5c)/(1 + 0.5c*-0.5c/c^2)
v' = 0 m/s

This means that the second spaceship appears to be stationary in the rest frame of the first spaceship, so there is no additional length contraction due to motion in the direction of the first spaceship's velocity. Therefore, the length of the other spaceship as measured by a passenger in the first spaceship is simply the length contraction we calculated earlier:

L' = 86.603 m

Rounded to the nearest 10 m, the answer is B) 60 m.

Question 11

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?

Accepted Answer

We need to use the relativistic velocity addition formula v = (u + w)/(1+uw/c^2), where u is the velocity of the spaceship relative to the asteroid (0.80c), w is the velocity of the missile relative to the spaceship (0.50c), and c is the speed of light. Plugging these values in, we get v = (0.80c + 0.50c)/(1+0.80c*0.50c/c^2) = 0.93c. This is the speed of the missile as measured by astronauts on the asteroid, since they are at rest relative to the asteroid. Therefore, the answer is C.

Question 12

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 13

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?

Accepted Answer

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

Question 14

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?

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 15

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?

Accepted Answer

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

Question 16

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?

Accepted Answer

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

Question 17

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 18

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?

Accepted Answer

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

Question 19

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?

Accepted Answer

Let's denote the speed of light as c, and the speed of the spaceship as 0.7c. 
To find the speed that the shuttlecraft must launch at relative to the ship (let's denote this speed as s), we can use the relativistic velocity addition formula:

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

Where v is the resulting velocity, u is the velocity of the first object, w is the velocity of the second object, and c is the speed of light.

In this case, the first object is the spaceship, moving at u = 0.7c, and the second object is the shuttlecraft, moving at w = s. We want v to be 0.987c (98.7% of the speed of light).

So we have:

0.987c = (0.7c + s)/(1 + 0.7cs/c^2)

Simplifying and solving for s, we get:

s = (0.987c - 0.7c)/(1 - 0.69)

s = 0.93c

Therefore, the answer is (C) 0.93c.

Question 20

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?

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 $v = \frac{0.60c + 0.40c}{1 + \frac{0.60c \cdot 0.40c}{c^2}} = \frac{1.0c}{1 + 0.24} = \frac{1.0c}{1.24} = 0.81c$.

Quiz 26: Special Relativity