Question 1

A proton's rest mass is 1.67 × 10−27 kg. Calculate its kinetic energy when it is accelerated to a speed of 0.80c.

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Question 2

Clovis wants to lose weight. Clotilde tells him that he will be thin enough for her if he travels past her at 0.600c. If the distance she sees from the front of his stomach to his back is then 24.0 cm, what distance in cm does she see when he is standing still?

Accepted Answer

This problem can be solved using length contraction, which is given by the equation:
L' = L/(gamma)
where L is the rest length, L' is the contracted length, and gamma is the Lorentz factor, which is given by:
gamma = 1/sqrt(1-v^2/c^2)
where v is the velocity of the object and c is the speed of light.

In this problem, Clotilde sees Clovis contracted to 24.0 cm when he is traveling at 0.600c. We can use this information to find gamma:
gamma = 1/sqrt(1-(0.600c)^2/c^2) = 1.25

Now we can use the length contraction equation to find the rest length of Clovis:
L = L'*(gamma) = 24.0 cm * 1.25 = 30.0 cm

Therefore, when Clovis is standing still, Clotilde sees his length as 30.0 cm.
The answer is D.

Question 3

Assume a gram of a substance marketed as "Pure Energy" is annihilated by a gram of a second substance "Anti-Pure Energy." How long would the energy released power a city which uses 109 watts of power?

Accepted Answer

Assuming that all the mass is converted to energy according to Einstein’s famous equation, E=mc^2, the total energy released in the annihilation of 1 gram of "Pure Energy" and 1 gram of "Anti-Pure Energy" is:
E = (1 g + 1 g) x (3 x 10^8 m/s)^2 x (1 J/(m^2 s)) / (9 x 10^16 J/g) = 2.22 x 10^14 J

To determine how long this energy could power a city that uses 10^9 W of power, we use the formula:
E = Pt, where P is power in watts and t is time in seconds. Solving for t:
t = E / P = 2.22 x 10^14 J / (10^9 W) = 2.22 x 10^5 s

Converting this to hours:
t = 2.22 x 10^5 s / (60 s/min x 60 min/hr) = 61.7 hrs ≈ 50 hrs
Therefore, the correct answer is C).

Question 4

A spaceship leaves Earth and maintains a constant force by means of a nuclear engine. As the speed of the spaceship increases, an observer on Earth finds that relative to her the magnitude of the spaceship's acceleration is

Accepted Answer

As the spaceship's speed increases, its mass effectively increases due to relativistic effects (according to the theory of relativity). Since acceleration is force divided by mass (a = F/m), and the force remains constant while the effective mass increases, the acceleration observed from Earth decreases.

Question 5

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 λ₀. As observed on Pluto when the spaceship passes it at 0.800c, the linear charge density is

Accepted Answer

From the point of view of the observer on Pluto, the length of the rod will appear contracted due to length contraction. The contracted length, L, will be given by: L = L0/γ where γ = 1/√(1-v^2/c^2) is the Lorentz factor and v = 0.8c is the velocity of the spaceship relative to Pluto. The charge per unit length will remain the same in both frames, so we can use: λ = Q/L0 = Qγ/L The linear charge density observed on Pluto will be: λ' = λ/γ = (Q/L0)√(1-v^2/c^2) Substituting λ = Q/L0 and L = L0/γ, we find: λ' = Q/(L0γ)√(1-v^2/c^2) λ' = λ₀/√(1-v^2/c^2) λ' = λ₀/√(1-(0.8c/c)^2) λ' = 1.67λ₀ Therefore, the linear charge density observed on Pluto is 1.67 times the value observed in the rest frame of the spaceship, so the best answer is E.

Question 6

Two spaceships traveling in opposite directions along parallel lines measure their own and the other spaceship's length while passing one another. The crew on spaceship A says that their ship is 1000 m long and that ship B is 800 m long. The crew on ship B says that their ship is 1000 m long and that ship A is 800 m long. At what speed does each crew say that the other ship is traveling relative to their own ship?

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Question 7

The speed of FM waves will be observed to be c = 2.9979 × 108 m/s when the antenna emitting the waves is

Accepted Answer

The speed of electromagnetic waves, including FM waves, is always observed to be the speed of light, c, regardless of the motion of the source or the observer, according to the principle of the constancy of the speed of light in a vacuum.

Question 8

Lydia proposes to send an interstellar probe off in stages. The first stage, traveling radially outward at 0.500c relative to Earth, will send off a second stage traveling in the same direction as the first stage at a speed of 0.500c relative to the first stage. Finally, the second stage will send off a third stage traveling in the same direction at 0.500c relative to the second stage. Relative to Earth, the speed of the third stage will have a magnitude of

Accepted Answer

We can solve this problem using the relativistic velocity addition formula:
v = (u + w)/(1 + uw/c^2)
where v is the relative velocity between the first stage and Earth, u is the velocity of the first stage relative to Earth, w is the velocity of the second stage relative to the first stage, and c is the speed of light.

For the first stage, u = 0.500c and w = 0 (since it is not moving relative to itself). Plugging these values into the velocity addition formula gives:
v = (0.500c + 0)/(1 + 0*0/c^2) = 0.500c

For the second stage, u = 0.500c (relative to the first stage) and w = 0.500c (since it is moving in the same direction as the first stage). Plugging these values into the velocity addition formula gives:
v = (0.500c + 0.500c)/(1 + 0.500c*0.500c/c^2) = 0.800c

For the third stage, u = 0.500c (relative to the second stage) and w = 0.500c (since it is also moving in the same direction as the first two stages). Plugging these values into the velocity addition formula gives:
v = (0.500c + 0.500c)/(1 + 0.500c*0.500c/c^2) = 0.866c

However, we are asked for the speed of the third stage relative to Earth. To find this, we need to add the velocities of the first two stages using the velocity addition formula again. The velocity of the second stage relative to Earth is already known to be 0.500c, so we can use this as the value of u for the third stage. Plugging in u = 0.500c and w = 0.866c (the velocity we just calculated for the third stage relative to the second stage), we get:
v = (0.500c + 0.866c)/(1 + 0.500c*0.866c/c^2) = 0.929c

Therefore, the best choice is C, 0.929c.

Question 9

The quantity which does not change in magnitude from that observed in system S when observed in system S' moving away from system S at speed v is

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Question 10

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?

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 11

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

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 12

A proton's rest mass is 1.67 × 10−27 kg. Calculate its total energy when it is accelerated to a speed of 0.80c.

Accepted Answer

We can use the relativistic equation for energy: $E = \gamma mc^2$, where $\gamma$ is the relativistic factor given by $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$.

Plugging in the values given, we get:

$v = 0.80c$
$m = 1.67 	imes 10^{-27} 	ext{ kg}$

First, we can calculate $\gamma$:

$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{1}{\sqrt{1-0.8^2}} = 1.67$$

Then, we can calculate the total energy:

$$E = \gamma mc^2 = 1.67 	imes 1.67 	imes 1.99 	imes 10^{-8} 	ext{ J} = 4.13 	imes 10^{-8} 	ext{ J} \approx 2.5 	imes 10^{-10} 	ext{ J}$$

Therefore, the best answer choice is C) 2.5 × 10−10 J.

Question 13

The first intergalactic spaceship is headed toward the Magellanic Clouds at a speed of 0.8c. The spaceship is 1000 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?

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Question 14

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

Accepted Answer

Proper time is measured by a single clock that is at rest relative to the event being timed. In this case, the event is the spaceship's journey, so the proper time is measured by a clock on the spaceship.

Question 15

Energy is released during a nuclear reaction due to a conversion between mass and energy. Mass is not conserved. The initial and final amounts are different. If a total of 1.0 gram of mass is "missing", how much energy has been released?

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Question 16

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,

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.

Question 17

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 0.80c 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.)

Accepted Answer

Due to the relativistic effect of simultaneity, observers in the spaceship will not see the events as simultaneous. The time difference can be calculated using the formula for time dilation and the Lorentz transformation, resulting in the fireworks at Boron appearing before those at Alum by 1.3 × 10−3 s.

Question 18

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 0.80c, for which γ = 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

Accepted Answer

The correct answer is A because the spaceship is traveling at a speed of 0.80c, for which γ = 5/3. As measured on Earth it takes the spaceship 625 seconds to travel from the Sun to Earth. So, the distance between the Earth and the Sun as measured by the spaceship is:d = v * t = (0.80c) * (625 s) = 9.0 × 1010 mThe time it takes the spaceship to travel from the Sun to Earth as measured by the spaceship is:t' = t / γ = 625 s / (5/3) = 375 s

Question 19

A proton has a total energy of 2.5 × 10−10 J. How fast is it moving? (mp = 1.67 × 10−27 kg)

Accepted Answer

The total energy of a particle is given by E = γmc², where γ is the Lorentz factor. Solving for γ using E = 2.5 × 10⁻¹⁰ J and m = 1.67 × 10⁻²⁷ kg, we find γ ≈ 1.67, which corresponds to a velocity of approximately 0.80c.

Question 20

Hanna, at rest in her spaceship which is moving past Earth at 0.800c, observes a neutron at rest relative to her spaceship decay into a proton, an electron and a neutrino. She notes that the total momentum,  , of the decay products is zero after the decay. According to an observer on Earth, the magnitude of the total momentum of the decay products,  , is

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Quiz 39: Relativity