Question 1

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

To solve this problem, we need to use the relativistic velocity addition formula: 
v = (u + w)/(1 + uw/c^2)
where v is the relative velocity between the two objects, u is the velocity of the spaceship, w is the velocity of the secondary rocket relative to the spaceship, and c is the speed of light. 
Plugging in the given values, we get: 
v = (0.100c + 0.560c)/(1 + (0.100c)(0.560c)/c^2) 
v = 0.660c/(1 + 0.056) 
v = 0.487c 
So the speed of the secondary rocket relative to the space platform is 0.487c, which corresponds to choice B.

Question 2

The figure shows a right-angled construction frame ABC.When measured at rest,it has dimensions as follows: AB = 26.0 m,AC = 10.0 m,and BC = 24.0 m.This frame is now given a new velocity of 0.760c,relative to Earth,in a direction parallel to AC.How long is part AC of this frame with the new velocity as measured by Earth-based engineers?

Accepted Answer

The answer of The figure shows a right-angled construction frame...

Question 3

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 4

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 correct answer is A because the relative speed of the two spaceships is given by the formula:v = (v1 + v2) / (1 + v1 * v2 / c^2)where v1 and v2 are the speeds of the two spaceships relative to Earth, and c is the speed of light. Plugging in the given values, we get:v = (0.21c + 0.34c) / (1 + 0.21c * 0.34c / c^2) = 0.51c

Question 5

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

To find the speed of the secondary rocket relative to the space platform, use the relativistic velocity addition formula: $ v = \frac{u + v'}{1 + \frac{uv'}{c^2}} $, where $ u = 0.560c $ and $ v' = 0.100c $. Plugging in the values, the speed is approximately 0.625c.

Question 6

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

The length contraction formula is $L = L_0\sqrt{1 - v^2/c^2}$, where $L_0$ is the rest length, $v$ is the relative velocity, and $c$ is the speed of light. For a spaceship moving at 0.50c, the length as observed from the other spaceship, taking into account the relativistic velocity addition formula for the effective velocity between the spaceships, is contracted to 60 m. This is because the effective velocity is not simply $0.50c + 0.50c$ due to relativistic effects, but the calculation simplifies since the question only requires recognizing that length contraction occurs, and the given options suggest a basic understanding that the length will be contracted, with 60 m being the correct calculation for a length contraction scenario at the given speed.

Question 7

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

Question 8

The figure shows a right-angled construction frame ABC.When measured at rest,it has dimensions as follows: AB = 18.9 m,AC = 10.0 m,and BC = 16.0 m.This frame is now given a new velocity of 0.420c,relative to Earth,in a direction parallel to AC.How long is part BC of this frame with the new velocity as measured by Earth-based engineers?

Accepted Answer

The answer of The figure shows a right-angled construction frame...

Question 9

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

The relative speed between two objects moving in the same direction in special relativity is given by the formula $v_{rel} = \frac{v_2 - v_1}{1 - \frac{v_2v_1}{c^2}}$, where $v_1$ and $v_2$ are the velocities of the two objects (with $v_2 > v_1$) and $c$ is the speed of light. Plugging in $v_1 = 0.21c$ and $v_2 = 0.34c$, the calculation simplifies to $v_{rel} = \frac{0.34c - 0.21c}{1 - (0.34)(0.21)} = \frac{0.13c}{1 - 0.0714} = \frac{0.13c}{0.9286} \approx 0.14c$.

Question 10

A spaceship with a constant velocity of 0.800c relative to Earth travels to the star that is 4.30 light-years from Earth.What distance does the space ship travel as measured by a passenger on the ship?

Accepted Answer

The distance measured by a passenger on the ship is shorter due to length contraction, a phenomenon of special relativity. The formula for length contraction is $L = L_0\sqrt{1 - v^2/c^2}$, where $L_0$ is the rest length (4.30 light-years), $v$ is the velocity of the ship (0.800c), and $c$ is the speed of light. Plugging in the numbers, $L = 4.30\sqrt{1 - (0.800)^2} = 4.30\sqrt{1 - 0.64} = 4.30\sqrt{0.36} = 4.30 \times 0.6 = 2.58$ light-years.

Question 11

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 12

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 
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 answer of In an "atom smasher," two particles collide...

Question 13

How fast must a meter stick move,relative to the frame of the observer making the measurement,so its length will be measured to be 0.60 m?

Accepted Answer

The length contraction formula is $L = L_0\sqrt{1 - \frac{v^2}{c^2}}$, where $L$ is the contracted length, $L_0$ is the rest length, $v$ is the velocity of the object, and $c$ is the speed of light. Rearranging for $v$, given $L_0 = 1$ m and $L = 0.60$ m, we find $v = c\sqrt{1 - \left(\frac{L}{L_0}\right)^2} = c\sqrt{1 - 0.36} = 0.8c$.

Question 14

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

According to the theory of relativity, the speed of light is constant for all observers, regardless of their relative motion. Therefore, Earth-based astronomers will measure the speed of light emitted by the star to be c (the speed of light in a vacuum), regardless of the speed of the star itself. The fact that the star is moving toward Earth at 90% the speed of light does not affect the speed of the light it emits.

Question 15

Astronaut Sarah leaves Earth in a spaceship at a speed of 0.280c relative to Earth.Sarah's destination is a star system 12.5 light-years away (one light-year is the distance light travels in one year).According to Sarah,how far did she travel during the trip?

Accepted Answer

The answer of Astronaut Sarah leaves Earth in a spaceship...

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

Astronaut Hans leaves Earth in a spaceship traveling at a speed of 0.280c relative to the observer Brian at rest on Earth.Hans is holding a pencil at an angle of 30° with the direction of travel,as seen by Hans.What angle does the pencil make with the direction of travel,as seen by Brian on Earth?

Accepted Answer

Due to relativistic effects, such as length contraction, the angle observed by Brian will be slightly larger than 30°.

Question 18

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 use the relativistic velocity addition formula:

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

Where:
u = velocity of the spaceship (0.70c)
v' = velocity of the shuttlecraft relative to the spaceship (what we're looking for)
v = velocity of the shuttlecraft relative to Earth (0.987c)
c = speed of light

Plugging in the values and solving for v', we get:

v' = (v - u) / (1 - uv/c^2) = (0.987c - 0.70c) / (1 - 0.987c * 0.70c / c^2) ≈ 0.93c

So the answer is (C) 0.93c.

Question 19

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

Using the relativistic velocity addition formula:
v = (u + v)/(1 + uv/c^2)
where u=0.50c, v=0.80c, and c is the speed of light
v = (0.50c + 0.80c)/(1 + (0.50c)(0.80c)/c^2) = 1.3c/(1 + 0.4) = 0.93c
Therefore, the speed of the missile measured by astronauts on the asteroid is 0.93c.

Question 20

Astronaut Jill leaves Earth in a spaceship and is now traveling at a speed of 0.280c relative to an observer on Earth.When Jill left Earth,the spaceship was equipped with all kinds of scientific instruments,including a meter stick.Now that Jill is underway,how long does she measure the meter stick to be?

Accepted Answer

Jill, being in the same inertial frame as the meter stick, would measure it to be its proper length, which is 1.00 m. The effects of length contraction are observed from a different inertial frame, not within the same frame as the object.

Quiz 29: Relativity