Spaceship I, which contains students taking a physics exam, approaches Earth with a speed of 0.640 c, while spaceship II, which contains instructors proctoring the exam, moves away from Earth at 0.280 c, as in Figure P26.27. If the instructors in spaceship II stop the exam after 65 min have passed on their clock, how long does the exam last as measured by (a) the students (b) an observer on earth
I've been trying relative velocity addition followed but time dilation but have not gotten the problem right so far
Ah, the classic physics exam in space scenario! Let's see if I can bring some laughter and clarity to your problem.
(a) First, let's calculate the time dilation for the students on Spaceship I, who are moving at 0.640 c relative to Earth. Using the time dilation equation:
Δt' = Δt / √(1 - (v^2/c^2))
where Δt' is the time experienced by the students and Δt is the time experienced by an observer at rest (the instructors), we can plug in the values:
Δt' = 65 min / √(1 - (0.640^2 / 1^2))
Δt' = 65 min / √(1 - 0.4096)
Δt' = 65 min / √0.5904
Δt' = 65 min / 0.7675
Δt' ≈ 84.68 min
So, according to the students, the exam lasts approximately 84.68 minutes.
(b) Now let's calculate the time experienced by an observer on Earth. We know that Spaceship II is moving away from Earth at 0.280 c. Since the instructors stop the exam after 65 minutes on their clock, we need to find the corresponding time on Earth.
Δt = Δt' * √(1 - (v^2/c^2))
Δt = 65 min * √(1 - (0.280^2 / 1^2))
Δt = 65 min * √(1 - 0.0784)
Δt = 65 min * √0.9216
Δt = 65 min * 0.9594
Δt ≈ 62.36 min
So, according to an observer on Earth, the exam lasts approximately 62.36 minutes.
And there you have it! The students experience the exam for around 84.68 minutes, while an observer on Earth sees it lasting approximately 62.36 minutes. Science fiction and physics can really add a twist to exams, can't they?
To solve this problem, we can use the concepts of time dilation and relative velocity addition. Let's begin by finding the time dilation factor for each spaceship.
The time dilation factor (γ) equation can be expressed as:
γ = 1 / sqrt(1 - v^2 / c^2)
Where:
γ = time dilation factor
v = velocity of the spaceship
c = speed of light
For Spaceship I:
Given the velocity of Spaceship I (v1) = 0.640c, we can calculate its time dilation factor (γ1):
γ1 = 1 / sqrt(1 - (0.640c)^2 / c^2)
= 1 / sqrt(1 - 0.4096)
≈ 1.142
For Spaceship II:
Given the velocity of Spaceship II (v2) = 0.280c, we can calculate its time dilation factor (γ2):
γ2 = 1 / sqrt(1 - (0.280c)^2 / c^2)
= 1 / sqrt(1 - 0.0784)
≈ 1.0284
(a) To find the time of the exam as measured by the students on Spaceship I, we need to consider the time dilation experienced by the students. Let's assume the exam duration as Δt1 (measured by the students).
The time measured by the students is related to the time measured by the instructors through the equation:
Δt1 = γ1 * Δt2
Since the instructors stop the exam after 65 minutes on their clock (Δt2), we can substitute the known values into the equation:
Δt1 = 1.142 * 65 minutes
≈ 74.23 minutes
Therefore, the exam lasts approximately 74.23 minutes as measured by the students on Spaceship I.
(b) To find the time of the exam as measured by an observer on Earth, we can use the concept of relative velocity addition considering Spaceship I moving towards Earth and Spaceship II moving away from Earth.
The time measured by the observer on Earth is related to the time measured by the instructors through the equation:
Δt_observer = Δt2 / γ2
Substituting the known values into the equation:
Δt_observer = 65 minutes / 1.0284
≈ 63.21 minutes
Therefore, the exam lasts approximately 63.21 minutes as measured by an observer on Earth.
To solve this problem, we'll need to apply the principles of time dilation and relativistic velocity addition. Let's break it down step by step:
Step 1: Understand the scenario
Spaceship I (with students) is approaching Earth with a speed of 0.640 c, and Spaceship II (with instructors) is moving away from Earth at 0.280 c. The instructors stop the exam after 65 minutes have passed on their clock.
Step 2: Calculate the time dilation factor
Time dilation occurs due to the relativistic effects of velocity. The time dilation factor, γ (gamma), is given by the formula:
γ = 1 / (√(1 - (v^2 / c^2)))
where v is the velocity of the object relative to a stationary observer and c is the speed of light.
For Spaceship I (approaching Earth):
v1 = 0.640 c
And for Spaceship II (moving away from Earth):
v2 = 0.280 c
Using these values, we can calculate the time dilation factor for both spaceships.
γ1 = 1 / (√(1 - (0.640)^2))
γ1 ≈ 1.25
γ2 = 1 / (√(1 - (0.280)^2))
γ2 ≈ 1.05
Step 3: Calculate the observed time on Earth
We know that the instructors stop the exam after 65 minutes have passed on their clock (Spaceship II). To find the observed time on Earth (t0), we can use the formula:
t0 = γ2 * t
where t is the time recorded on Spaceship II's clock.
Substituting the given values:
t0 = 1.05 * 65
t0 ≈ 68.25 minutes (as observed from Earth)
Therefore, the exam lasts approximately 68.25 minutes as measured by an observer on Earth.
Step 4: Calculate the time for the students (Spaceship I)
To find the time experienced by the students (t1), we can use the formula:
t1 = γ1 * t
where t is the time recorded on Spaceship II's clock.
Substituting the given values:
t1 = 1.25 * 65
t1 ≈ 81.25 minutes (for the students)
Therefore, the exam lasts approximately 81.25 minutes as measured by the students on Spaceship I.
In summary:
(a) The exam lasts approximately 81.25 minutes as measured by the students.
(b) The exam lasts approximately 68.25 minutes as measured by an observer on Earth.
(a) The relative velocity of the students with respect to the proctors can be calculation from the relative velocity addition formula
Vsp = (Vs + Vp)/[1 + VsVp/c^2]
= 0.92c/1.1792 = 0.7802c
Rest-frame exam time as measured by the students will be longer than 65 minutes, as determined by the gamma factor for that relatve velocity.
1/sqrt[1 - (.7802)^2] = 1.599
Which makes the actual exam time 104 minutes.
(b) Apply the gamma factor with a (earth-proctor) relative velocity of
Vep = 0.28 c.
That would give you 65 minutes * 1/sqrt[1-(.28^2)] = 67.7 minutes