Question 1

A 1 000-kg automobile moving with a speed of 24 m/s relative to the road collides with a 500-kg automobile initially at rest.If the two stick together,what is the velocity in m/s of the two cars after the collision according to an observer in a truck moving 10 m/s in the same direction as the moving cars?
A)6
B)11
C)24
D)26
E)31

Accepted Answer

A

Question 2

Captain Jirk reports to headquarters that he left the planet Senesca 1.88 * 10⁴ seconds earlier.Headquarters sends back the message: "Was that spaceship proper time?" It will be spaceship proper time if it was A)measured by one clock fixed at one spot on Senesca. B)measured by one clock fixed at one spot on the spaceship. C)measured by a clock on Senesca at departure and by a clock on the spaceship when reporting. D)measured by a clock on the spaceship when departing and by a clock on Senesca when reporting. E)calculated by dividing the distance from Senesca according to Senesca by the speed of the spaceship.

Accepted Answer

The only way for the time to be spaceship proper is if it is measured by one clock fixed at one spot on the spaceship, as this is the only reference frame that is moving with the spaceship and will experience time dilation effects. The other options involve comparing time measurements from different reference frames, which would not give a proper time measurement for the spaceship.

Question 3

A boat has an initial velocity of 2 m/s in the y direction on a stream which is moving in the x direction at 1 m/s.The boat is accelerating in its direction of motion at 1 m/s².What is the acceleration of the boat (in m/s²)relative to the water? A)1 $\hat { \mathbf { i } }$ B)2 $\hat { \mathbf { i } }$ C)3 $\hat { \mathbf { i } }$ + $\tilde{j}$ D)3 $\hat { \mathbf { i } }$ - $\tilde{j}$ E)1 $\tilde{j}$

Accepted Answer

E

Question 4

A 1 000-kg automobile moving with a speed of 24 m/s collides with a 500-kg car initially at rest.If the two stick together,what is the velocity (in m/s)of the two cars after the collision relative to an automobile moving in the same direction at 15 m/s?
A)14
B)16
C)24
D)48
E)1.0

Accepted Answer

The answer of A 1 000-kg automobile moving with a...

Question 5

Boat 1 goes directly across a stream a distance L and back taking a time t₁.Boat 2 goes down stream a distance L and back taking a time t₂.If both boats had the same speed relative to the water,which of the following statements is true? A)t₂ > t₁ B)t₂ < t₁ C)t₂ = t₁ D)t₂ = 2t₁ E)t₂ = 0.5t₁

Accepted Answer

Let the speed of the boat be Vb and the speed of the stream be Vs. For Boat 1, the time taken to go across the stream and back can be calculated using the formula: t₁ = (L/Vb - Vs) + (L/Vb + Vs) = 2L/Vb For Boat 2, the time taken to go down stream and back can be calculated using the formula: t₂ = (L/Vb + Vs)/(1 - Vs/Vb) + (L/Vb - Vs)/(1 + Vs/Vb) = 2Lv/(Vb^2 - Vs^2) < 2L/Vb Therefore, t₂ < t₁.

Question 6

A spaceship from another galaxy passes over the solar system directly above a radial line from the sun to the Earth.(We measure that distance to be 1.5 *10¹¹ m. )On Earth,the spaceship is observed to be traveling at a speed of 0.80 c,for which $\gamma$= 5/3.As measured on Earth it takes the spaceship 625 seconds to travel from the sun to Earth.When a scientist in the spaceship measures the Earth-sun distance and the time it takes her to travel that distance,she finds the results are respectively A)9.0 *10¹⁰ m;375 s. B)9.0 *10¹⁰m;625 s. C)1.5 * 10¹¹ m;625 s. D)2.5 * 10¹¹ m;625 s. E)2.5 *10¹¹ m;1042 s.

Accepted Answer

From Earth's perspective:

Let the distance between the sun and the spaceship be x.
We know that distance between Earth and spaceship is 1.5x.

Using time dilation formula, we have:
t = t0 / sqrt(1 - v^2 / c^2)
where t0 is the time measured in the spaceship's frame of reference, v is the velocity of the spaceship, and c is the speed of light.

Plugging in values, we have:
625 = t0 / sqrt(1 - (0.80c)^2 / c^2)
t0 = 375 seconds

From spaceship's perspective:

The spaceship is at rest in its own frame of reference, so the distance and time measured by the scientist in the spaceship are the same as the ones measured by Earth observers before the spaceship started moving. Therefore, the distance is x = 9.0 x 10^10 m and the time is 375 seconds.

Thus, the answer is A.

Question 7

A satellite moves east,taken as the positive x axis direction,at a speed of 0.5 c and a spaceship moves toward it (to the west)at a speed of 0.8 c as measured by an observer on the Earth.The speed of the satellite measured by an observer in the spaceship is
A)8.7 c
B)0.21 c
C)-0.21 c
D)0.93 c
E)-0.93 c

Accepted Answer

Using the relativistic velocity addition formula, the speed of the satellite as measured by the observer in the spaceship is 0.93c.

Question 8

The first intergalactic spaceship is headed toward the Magellanic Clouds at a speed of 0.8c.The spaceship is 1 000 m long.Clocks at the front and the rear of the spaceship both read 3:00 P.M.Can it be 3:00 P.M.simultaneously at the front and the back of the spaceship?
A)No,because $\Delta t = \gamma \beta \frac { \Delta x } { c }$  ,where $\Delta$x is the length of the spaceship.
B)No,because one clock has to move after being synchronized with the other.
C)Yes,because $\Delta$x in (a)is zero for different points of the same spaceship.
D)Yes,because two clocks at rest relative to each other can be synchronized by means of light signals when the distance between them is known.
E)The question cannot be answered unless we know the object relative to which the spaceship's velocity is 0.8c.

Accepted Answer

The answer of The first intergalactic spaceship is headed toward...

Question 9

The half-life of a muon is 2.2$\mu$s.How fast is it moving relative to an observer who says its half-life is 4.4 $\mu$s?
A)0.87 c
B)0.75 c
C)0.97 c
D)0.72 c
E)0.50 c

Accepted Answer

The observed half-life is longer due to time dilation, which occurs at relativistic speeds. Using the time dilation formula, the speed can be calculated as 0.87c.

Question 10

In a classroom on the first spaceship to an extrasolar planet--there are children because the trip will take 200 years--a teacher is showing charge Q uniformly distributed along a conducting rod of length L₀ to produce linear charge density $\lambda$₀.As observed on Pluto when the spaceship passes it at 0.800 c,the linear charge density $\lambda ^ { \prime }$ is A)0.360 $\lambda$₀. B)0.600 $\lambda$₀. C)0.800 $\lambda$₀. D)1.00 $\lambda$₀. E)1.67 $\lambda$₀.

Accepted Answer

The linear charge density $\lambda ^ { \prime }$ is affected by length contraction due to relativistic effects. The length of the rod as observed from Pluto is contracted by a factor of $\sqrt{1 - v^2/c^2}$, where $v = 0.800c$. This increases the linear charge density by a factor of $1/\sqrt{1 - (0.800)^2}$, which is approximately 1.67.

Question 11

A spaceship moving past the Earth with a speed of 0.800 c signals to the Earth with pulsed laser photons emitted at 10.0-second intervals according to the spaceship's clock.According to observers on Earth who see the flashes,the time interval they measure is
A)13.4 s
B)16.7 s
C)12.5 s
D)9.70 s
E)6.03 s

Accepted Answer

The time dilation effect described by the theory of relativity explains that time intervals measured in a moving frame (the spaceship) appear longer in the stationary frame (Earth). The formula for time dilation is $t' = \frac{t}{\sqrt{1 - v^2/c^2}}$, where $t'$ is the dilated time, $t$ is the proper time (time interval in the moving frame), $v$ is the velocity of the moving frame, and $c$ is the speed of light. Plugging in the given values ($t = 10.0$ s, $v = 0.800c$), we find $t' = 10.0 / \sqrt{1 - (0.800)^2}$ which calculates to approximately 16.7 seconds.

Question 12

When at rest,a spacecraft has the same height as length.When moving at relativistic speeds in the direction of its length,which one of the following describes its length and height to an observer at rest?
A)Its length is decreased and its height is increased.
B)Its length is decreased and its height is decreased.
C)Its length is decreased and its height is unchanged.
D)Its length is unchanged and its height is increased.
E)Its length is increased and its height is decreased.

Accepted Answer

Lorentz contraction, a consequence of Einstein's theory of relativity, states that an object moving at relativistic speeds will appear shorter in the direction of motion to a stationary observer, while dimensions perpendicular to the motion (like height) remain unchanged.

Question 13

Fireworks go off at the same time according to Earth clocks in two cities,Alum and Boron,that are 300 km apart.The people in a spaceship that is flying in a straight line from Alum to Boron at 0.80 c also observe the fireworks.Do they see the fireworks in the two cities simultaneously? If the people in the spaceship say the fireworks were not simultaneous in Alum and Boron,how long before or after the fireworks flashed at Alum did the fireworks flash at Boron according to their calculations? (The spaceship is directly over Alum when the fireworks flash. ) A)Yes;0 B)Before;1.0 *10 ^- ³ s C)After;1.0 *10 ^- ³ s D)Before;1.3 *10 ^- ³ s E)After;1.3 *10 ^- ³ s

Accepted Answer

From the perspective of the people in the spaceship, the distance between Alum and Boron is contracted due to length contraction. The contracted distance can be calculated as follows:

L' = L / γ

where L is the proper distance between Alum and Boron (300 km) and γ is the Lorentz factor for the velocity of the spaceship (0.8 c).

γ = 1 / sqrt(1 - v^2/c^2)
   = 1 / sqrt(1 - 0.8^2)
   = 1.67

L' = 300 km / 1.67
   = 180 km

Since the spaceship is directly over Alum when the fireworks are observed, the observed time difference between the fireworks in Alum and Boron can be calculated using the following formula:

Δt' = Δx' / v

where Δx' is the contracted distance between Alum and Boron (180 km) and v is the velocity of the spaceship (0.8 c).

Δt' = 180 km / (0.8 c)
     = 562.5 s
     ≈ 9.4 min

This means that according to the people in the spaceship, the fireworks in Boron will be observed about 9.4 minutes after the fireworks in Alum. Therefore, the best answer is D: Before; 1.3 s.

Question 14

A jet plane travels around the world at 2 000 mi/hr (894 m/s).Two accurate atomic clocks measure the times of flight,one on board the plane and the second on Earth.If it takes 12 hours to complete the journey,what will the time difference (in $\mu$s)be?
A)0.75
B)0.50
C)0.25
D)1.0
E)0.19

Accepted Answer

The answer of A jet plane travels around the world...

Question 15

A 30-year-old woman takes a trip on a rocket,leaving her 20-year-old brother behind.She travels at a speed of 0.8 c,and is gone 20 years,according to the younger brother.When she returns,how many years older/younger is she than her brother?
A)2 years younger
B)2 years older
C)3 years older
D)10 years older
E)8 years older

Accepted Answer

The answer of A 30-year-old woman takes a trip on...

Question 16

A spaceship moves at a speed of 0.95 c away from the Earth.It shoots a star wars torpedo toward the Earth at a speed of 0.90 c relative to the ship.What is the velocity of the torpedo relative to the Earth? (The direction in which the spaceship moves is the positive direction. )
A)-0.34 c
B)0.34 c
C)0.060 c
D)-0.060 c
E)0

Accepted Answer

The velocity of the torpedo relative to the Earth can be calculated using the relativistic velocity addition formula: $ v = \frac{u + v'}{1 + \frac{uv'}{c^2}} $. Substituting $ u = 0.95c $ and $ v' = -0.90c $, we get $ v \approx 0.34c $.

Question 17

An astronaut traveling with a speed v = 0.90 c holds a meterstick in his hand.If he measures its length,he will obtain a value of
A)1.0 m
B)2.3 m
C)0.19 m
D)0.43 m
E)0.81 m

Accepted Answer

When the astronaut measures the length of the meterstick, he will obtain its proper length, which is the length measured in the rest frame of the object. Since the meterstick is at rest relative to the astronaut, he will measure it to be 1.0 m, its proper length.

Question 18

The half-life of a muon is 2.20 $\mu$s as measured in a stationary reference frame.What is the half life of the muon (in $\mu$s)when it is moving with a speed of v = 0.800 c?
A)8.13
B)2.75
C)3.67
D)15.8
E)1.32

Accepted Answer

The half-life of a moving muon is longer due to time dilation. Using the formula for time dilation, t' = t / sqrt(1 - v^2/c^2), where t is the proper half-life (2.20 μs) and v = 0.800c, the dilated half-life is approximately 3.67 μs.

Question 19

The 500 m-long spaceship Springbrake is at rest on the planet Hitest for refueling.Another spaceship,Summerbrake,passes parallel to Springbrake at 0.600c.The crew on Springbrake measure the length of Summerbrake as it passes and find that the length they measure is exactly the same as the known 500 m rest length of Springbrake.If Summerbrake were at rest next to Springbrake,its length would then be measured to be
A)180 m.
B)320 m.
C)500 m.
D)625 m.
E)781 m.

Accepted Answer

The length contraction formula is $L = L_0\sqrt{1 - v^2/c^2}$, where $L$ is the observed length, $L_0$ is the rest length, and $v$ is the velocity of the object. Given that the observed length $L$ is 500 m and $v = 0.600c$, we can solve for $L_0$. Rearranging the formula to solve for $L_0$, we get $L_0 = L / \sqrt{1 - v^2/c^2}$. Plugging in the values, $L_0 = 500 / \sqrt{1 - 0.600^2}$, which calculates to approximately 625 m. This means that when Summerbrake is not moving, its rest length would be measured as 625 m.

Question 20

Captain Ray,on the Galactic Explorer,sets off bright signal lights at the two ends of his spaceship so that they are seen as flashing simultaneously by General Kay,who is watch the ship go by.General Kay concludes that she was next to the midpoint of the ship at the instant when the lights flashed in her reference frame.This can only happen if Captain Ray,at the center of the Galactic Explorer,
A)sees the lights flash simultaneously in his coordinate system.
B)sees the bow (front)light flash before the aft (rear)light.
C)sees the aft (rear)light flash before the bow(front light.
D)can calculate that the light flashes reach General Kay so that the flash from the bow light reaches her before the flash from the aft light.
E)can calculate that the light flashes reach General Kay so that the flash from the aft light reaches her before the flash from the bow light.

Accepted Answer

According to the theory of relativity, if two events (such as the flashing of lights) are simultaneous in one frame of reference (General Kay's), they are not necessarily simultaneous in another frame of reference that is moving relative to the first (Captain Ray's on the Galactic Explorer). Since General Kay sees the lights flash simultaneously while she is at the midpoint of the ship, it means that in Captain Ray's moving frame, the light from the bow (front) must have been emitted earlier than the light from the aft (rear) to account for the time it takes light to travel to the midpoint from each end of the ship, given the ship's motion relative to General Kay.

Quiz 9: Relativity

Captain Jirk reports to headquarters that he left the planet Senesca 1.88 * 10⁴ seconds earlier.Headquarters sends back the message: "Was that spaceship proper time?" It will be spaceship proper time if it was

A boat has an initial velocity of 2 m/s in the y direction on a stream which is moving in the x direction at 1 m/s.The boat is accelerating in its direction of motion at 1 m/s².What is the acceleration of the boat (in m/s²) relative to the water?

A 1 000-kg automobile moving with a speed of 24 m/s collides with a 500-kg car initially at rest.If the two stick together,what is the velocity (in m/s) of the two cars after the collision relative to an automobile moving in the same direction at 15 m/s?

Boat 1 goes directly across a stream a distance L and back taking a time t₁.Boat 2 goes down stream a distance L and back taking a time t₂.If both boats had the same speed relative to the water,which of the following statements is true?

A satellite moves east,taken as the positive x axis direction,at a speed of 0.5 c and a spaceship moves toward it (to the west) at a speed of 0.8 c as measured by an observer on the Earth.The speed of the satellite measured by an observer in the spaceship is

The first intergalactic spaceship is headed toward the Magellanic Clouds at a speed of 0.8c.The spaceship is 1 000 m long.Clocks at the front and the rear of the spaceship both read 3:00 P.M.Can it be 3:00 P.M.simultaneously at the front and the back of the spaceship?

The half-life of a muon is 2.2 $\mu$ s.How fast is it moving relative to an observer who says its half-life is 4.4 $\mu$ s?

A spaceship moving past the Earth with a speed of 0.800 c signals to the Earth with pulsed laser photons emitted at 10.0-second intervals according to the spaceship's clock.According to observers on Earth who see the flashes,the time interval they measure is

When at rest,a spacecraft has the same height as length.When moving at relativistic speeds in the direction of its length,which one of the following describes its length and height to an observer at rest?

A jet plane travels around the world at 2 000 mi/hr (894 m/s) .Two accurate atomic clocks measure the times of flight,one on board the plane and the second on Earth.If it takes 12 hours to complete the journey,what will the time difference (in $\mu$ s) be?

A 30-year-old woman takes a trip on a rocket,leaving her 20-year-old brother behind.She travels at a speed of 0.8 c,and is gone 20 years,according to the younger brother.When she returns,how many years older/younger is she than her brother?

A spaceship moves at a speed of 0.95 c away from the Earth.It shoots a star wars torpedo toward the Earth at a speed of 0.90 c relative to the ship.What is the velocity of the torpedo relative to the Earth? (The direction in which the spaceship moves is the positive direction. )

An astronaut traveling with a speed v = 0.90 c holds a meterstick in his hand.If he measures its length,he will obtain a value of

The half-life of a muon is 2.20 $\mu$ s as measured in a stationary reference frame.What is the half life of the muon (in $\mu$ s) when it is moving with a speed of v = 0.800 c?