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Two particles in a high-energy accelerator experiment approach each other head-on with a relative speed of \(0.890 c .\) Both particles travel at the same speed as measured in the laboratory. What is the speed of each particle, as measured in the laboratory?

Short Answer

Expert verified
The speed of each particle, as measured in the laboratory, is approximately \(0.944 c\).

Step by step solution

01

Identify the given values

It's given that two particles approach each other head-on with a relative speed of \(0.890c\) where \(c = 1\) is the speed of light. Both particles travel at the same speed as measured in the laboratory, let's denote their velocities as \(v1\) and \(v2\). Because they are going in opposite directions, we have \(v1 = -v2\).
02

Apply the formula for adding velocities in relativity

Einstein's addition of velocities formula is \(v = (v1 + v2) / (1 + v1*v2/c^2)\). Here \(v\) represents the relative speed of the two particles. Substituting \(v = 0.890c\), \(v1 = -v2\), and \(c=1\) into this formula, we get: \(0.890 = (v1 - v1) / (1 - v1^2)\). After simplifying, this equation becomes \(0.890 - v1^2 = 0\).
03

Solve the equation for \(v1\)

After solving the equation \(0.890 - v1^2 = 0\) for \(v1\), we get \(v1 = \sqrt{0.890}\).
04

Convert back to the speed of light

As we considered the speed of light \(c = 1\) at the beginning, now we need to multiply our result by \(c\) to get the velocity of each particle as measured in the laboratory. So, the speed of each particle is \(v1 * c = \sqrt{0.890} * c\).

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Key Concepts

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

Velocity Addition
In the realm of special relativity, adding velocities isn't as straightforward as simply adding numbers. When dealing with speeds approaching the speed of light, the classical mechanics approach doesn't work. Instead, we use Einstein's formula for velocity addition to ensure calculations remain consistent with the principles of relativity.

The formula used is:
  • \(v = \frac{v1 + v2}{1 + \frac{v1 \cdot v2}{c^2}}\)
Here, \(v1\) and \(v2\) are the velocities of two objects as measured in a given reference frame, and \(v\) is the resultant velocity. This equation ensures that no matter how high the individual velocities are, the resulting velocity \(v\) will never exceed the speed of light \(c\).

Let's consider the case where two particles approach each other head-on. If they have the same speed in opposite directions—making one velocity the negative of the other—Einstein's formula still helps align their relative motion correctly according to special relativity.
Relative Speed
Relative speed becomes essential when observing objects moving towards or away from each other, especially when dealing with high velocities near the speed of light. In special relativity, the concept of relative speed accounts for how quickly an object appears to move from the perspective of another object.

In the given scenario, two particles are moving towards each other with a relative speed of \(0.890c\). Understanding relative speed means recognizing that the direct addition of velocities seen in everyday experiences isn't applicable here. Instead, the speed as perceived between the particles requires the relativistic velocity addition formula to be applied to get an accurate measure.

Thus, even as the particles race towards one another, the perceived speed between them remains lower than \(c\), adhering to the rules of special relativity and ensuring that the combined speed doesn't exceed the speed of light.
Speed of Light
The speed of light, denoted as \(c\), is a fundamental constant of physics, valued at approximately \(3 \times 10^8\) meters per second. In the world of special relativity, it serves as a cosmic speed limit, an unbeatable velocity that nothing with mass can reach or exceed.

Einstein revolutionized our understanding of motion with the insight that the speed of light remains constant in all inertial frames of reference, regardless of the motion of the light source or observer. This essence is embedded in the Lorentz transformations, which replace the Galilean ones used in classical mechanics.

In practical terms, when calculating situations involving high-speed entities, like particles in a collider, adhering to the speed of light as the ultimate speed ensures no law of physics is violated. It has profound implications for how velocities are combined, as seen with the velocity addition formula, and defines an entirely new way of interpreting how objects move and interact at high velocities in the universe.

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Most popular questions from this chapter

In high-energy physics, new particles can be created by collisions of fast- moving projectile particles with stationary particles. Some of the kinetic energy of the incident particle is used to create the mass of the new particle. A proton-proton collision can result in the creation of a negative kaon \(\left(\mathrm{K}^{-}\right)\) and a positive \(\operatorname{kaon}\left(\mathrm{K}^{+}\right)\) $$ p+p \rightarrow p+p+\mathrm{K}^{-}+\mathrm{K}^{+} $$ (a) Calculate the minimum kinetic energy of the incident proton that will allow this reaction to occur if the second (target) proton is initially at rest. The rest energy of each kaon is \(493.7 \mathrm{MeV},\) and the rest energy of each proton is \(938.3 \mathrm{MeV}\). (Hint: It is useful here to work in the frame in which the total momentum is zero. But note that the Lorentz transformation must be used to relate the velocities in the laboratory frame to those in the zero-total-momentum frame.) (b) How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (c) Suppose that instead the two protons are both in motion with velocities of equal magnitude and opposite direction. Find the minimum combined kinetic energy of the two protons that will allow the reaction to occur. How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (This example shows that when colliding beams of particles are used instead of a stationary target, the energy requirements for producing new particles are reduced substantially.)

If a muon is traveling at \(0.999 c,\) what are its momentum and kinetic energy? (The mass of such a muon at rest in the laboratory is 207 times the electron mass.)

The positive muon \(\left(\mu^{+}\right),\) an unstable particle, lives on average \(2.20 \times 10^{-6} \mathrm{~s}\) (measured in its own frame of reference) before decaying. (a) If such a particle is moving, with respect to the laboratory, with a speed of \(0.900 c\), what average lifetime is measured in the laboratory? (b) What average distance, measured in the laboratory, does the particle move before decaying?

The distance to a particular star, as measured in the earth's frame of reference, is 7.11 light-years ( 1 light-year is the distance that light travels in \(1 \mathrm{y}\) ). A spaceship leaves the earth and takes \(3.35 \mathrm{y}\) to arrive at the star, as measured by passengers on the ship. (a) How long does the trip take, according to observers on earth? (b) What distance for the trip do passengers on the spacecraft measure?

A proton (rest mass \(1.67 \times 10^{-27} \mathrm{~kg}\) ) has total energy that is 4.00 times its rest energy. What are (a) the kinetic energy of the proton; (b) the magnitude of the momentum of the proton; (c) the proton's speed?

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