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Spacecraft I, containing students taking a physics exam, approaches the Earth with a speed of \(0.600 c\) (relative to the Earth), while spacecraft \(11,\) containing professors proctoring the exam, moves at \(0.280 c\) (relative to the Earth) directly toward the students. If the professors stop the exam after 50.0 min have passed on their clock, how long does the exam last as measured by (a) the students (b) an observer on the Earth?

Short Answer

Expert verified
(a) The exam lasts approximately 64.8 minutes for the students. (b) The exam lasts approximately 59.7 minutes for the observer on Earth.

Step by step solution

01

Determine the relative velocity

First, we need to determine the relative velocity (\(v\)) between the students and the professors. Since both are moving on a direct line toward each other, we should use the formula for relative velocity in special relativity. \[v=\frac{v_1-v_2}{1- \frac{v_1v_2}{c^2}}\]\nWhere \(v_1\) is the velocity of spacecraft I (students) and \(v_2\) is the velocity of spacecraft II (professors). Since all velocities are given relative to the same frame (Earth), we can plug the values directly into the formula to find \(v\).
02

Calculate time dilation for the students

Next, we need to find out how much time passes for the students when 50 minutes pass on the professors' clock (the proper time). The formula to calculate time dilation in special relativity is given by: \[\Delta t = \Delta t_0 \sqrt{1-\frac{v^2}{c^2}}\]\nWhere: \(\Delta t_0\) is the proper time (time according to the professors), \(v\) is the relative velocity calculated in step 1, \(c\) is the speed of light.
03

Calculate time dilation for the observer on Earth

Finally, we need to calculate how much time passes for an observer on Earth when 50 minutes pass on the professors' clock (the proper time). We use the same time dilation formula as in Step 2, replacing \(v\) with the velocity of the professors relative to Earth. \[\Delta t = \Delta t_0 \sqrt{1-\frac{v^2}{c^2}}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Time Dilation
Time dilation is a fascinating concept within Einstein's theory of special relativity. It describes how time can pass at different rates depending on an object's speed relative to another observer. In simpler terms, when an object moves very fast, close to the speed of light, time for that object will slow down as observed from a stationary point.

In our scenario, the professors are timing a 50-minute exam as per their clock onboard their spacecraft. For the students aboard another swiftly moving spacecraft, as well as for observers on Earth, the duration of the exam appears longer. This is due to the time dilation effect. The formula for time dilation is \[\Delta t = \Delta t_0 \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\] where \(\Delta t_0\) is the proper time measured in the professors' frame, \(v\) is the relative velocity, and \(c\) stands for the speed of light.

As the velocity increases towards the speed of light, the denominator of the formula becomes smaller, causing the observed time \(\Delta t\) to increase.
Relative Velocity
Relative velocity is crucial in understanding how different frames of reference perceive motion, especially in the context of special relativity. When two objects are moving towards each other, as in our exercise, their relative velocity gives us insight into the differences in their perceived motion.

For our two spacecraft, containing students and professors, we use the relativistic formula for relative velocity: \[v=\frac{v_1-v_2}{1- \frac{v_1v_2}{c^2}}\].

This formula ensures accuracy since simply adding their speeds would not be precise at such high velocities near the speed of light. By considering the very fabric of spacetime in this formula, we account for how velocities add up in a relativistic context, giving us a better understanding of their interaction.
  • The subtraction in the numerator accounts for their opposing motion.
  • The denominator ensures relativistic effects are included, crucial at such high speeds.
Proper Time
Proper time, denoted in our formulas as \(\Delta t_0\), refers to the time interval measured by an observer who is at rest relative to the event being measured. Essentially, this is the time that "really" passes for someone experiencing the event directly.

In our context, the 50 minutes that the professors measure on their spacecraft is the proper time because they are directly overseeing the exam. For them, the clock runs as normal, unaffected by how fast they are moving relative to other observers.
  • Proper time is fundamental for calculating time dilation, offering a baseline measurement.
  • It reminds us that each observer has their own perspective of time, especially in relativistic scenarios like space travel.
Understanding proper time helps us grasp how profoundly relative motion affects our perceptions of fundamental quantities such as time.

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Most popular questions from this chapter

The creation and study of new elementary particles is an important part of contemporary physics. Especially interesting is the discovery of a very massive particle. To create a particle of mass \(M\) requires an energy \(M c^{2} .\) With enough energy, an exotic particle can be created by allowing a fast moving particle of ordinary matter, such as a proton, to collide with a similar target particle. Let us consider a perfectly inelastic collision between two protons: an incident proton with mass \(m_{p},\) kinetic energy \(K,\) and momentum magnitude \(p\) joins with an originally stationary target proton to form a single product particle of mass \(M\) You might think that the creation of a new product particle, nine times more massive than in a previous experiment, would require just nine times more energy for the incident proton. Unfortunately not all of the kinetic energy of the incoming proton is available to create the product particle, since conservation of momentum requires that after the collision the system as a whole still must have some kinetic energy. Only a fraction of the energy of the incident particle is thus available to create a new particle. You will determine how the energy available for particle creation depends on the energy of the moving proton. Show that the energy available to create a product particle is given by $$ M c^{2}=2 m_{p} c^{2} \sqrt{1+\frac{K}{2 m_{p} c^{2}}} $$ From this result, when the kinetic energy \(K\) of the incident proton is large compared to its rest energy \(m_{p} c^{2},\) we see that \(M\) approaches \(\left(2 m_{p} K\right)^{1 / 2} / c .\) Thus if the energy of the incoming proton is increased by a factor of nine, the mass you can create increases only by a factor of three. This disappointing result is the main reason that most modern accelerators, such as those at CERN (in Europe), at Fermi lab (near Chicago), at SLAC (at Stanford), and at DESY (in Germany), use colliding beams. Here the total momentum of a pair of interacting particles can be zero. The center of mass can be at rest after the collision, so in principle all of the initial kinetic energy can be used for particle creation, according to $$ M c^{2}=2 m c^{2}+K=2 m c^{2}\left(1+\frac{K}{2 m c^{2}}\right) $$ where \(K\) is the total kinetic energy of two identical colliding particles. Here if \(K>m c^{2}\), we have \(M\) directly proportional to \(K,\) as we would desire. These machines are difficult to build and to operate, but they open new vistas in physics.

A ball is thrown at \(20.0 \mathrm{m} / \mathrm{s}\) inside a boxcar moving along the tracks at \(40.0 \mathrm{m} / \mathrm{s} .\) What is the speed of the ball relative to the ground if the ball is thrown (a) forward (b) backward (c) out the side door?

A super train (proper length \(100 \mathrm{m}\) ) travels at a speed of \(0.950 c\) as it passes through a tunnel (proper length \(50.0 \mathrm{m}) .\) As seen by a track side observer, is the train ever completely within the tunnel? If so, with how much space to spare?

According to observer \(\mathrm{A}\), two objects of equal mass and moving along the \(x\) axis collide head on and stick to each other. Before the collision, this observer measures that object 1 moves to the right with a speed of \(3 c / 4\) while object 2 moves to the left with the same speed. According to observer \(\mathrm{B}\), however, object 1 is initially at rest. (a) Determine the speed of object 2 as seen by observer B. (b) Compare the total initial energy of the system in the two frames of reference.

Prepare a graph of the relativistic kinetic energy and the classical kinetic energy, both as a function of speed, for an object with a mass of your choice. At what speed does the classical kinetic energy underestimate the experimental value by \(1 \% ?\) by \(5 \% ?\) by \(50 \% ?\)

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