Problem 28 - College Physics

Question

What is the speed of a particle whose kinetic energy is equal to (a) its rest energy, (b) five times its rest energy?

Step-by-Step Solution

Answer

(a) Speed is \( \frac{\sqrt{3}}{2}c \). (b) Speed is \( \frac{\sqrt{35}}{6}c \).

Step 1: Understanding Rest Energy

The rest energy of a particle is given by the formula \( E_0 = mc^2 \), where \( m \) is the rest mass of the particle and \( c \) is the speed of light.

Step 2: Kinetic Energy and Total Energy Relation

The total energy \( E \) of a particle moving at speed \( v \) is given by \( E = \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} \). The kinetic energy \( KE \) is then \( KE = E - E_0 \).

Step 3: Equation Setup for (a)

For part (a), set the kinetic energy equal to the rest energy: \( KE = E_0 = mc^2 \). Substituting into the kinetic energy formula gives \( \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} - mc^2 = mc^2 \).

Step 4: Simplifying Equation for (a)

Simplify the equation: \( \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} = 2mc^2 \). Cancel \( mc^2 \) and solve for \( v \): \( \sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{2} \).

Step 5: Solving for Speed in (a)

Square both sides: \( 1 - \frac{v^2}{c^2} = \frac{1}{4} \). Solve for \( \frac{v^2}{c^2} \): \( \frac{v^2}{c^2} = \frac{3}{4} \). This implies \( v = \frac{\sqrt{3}}{2}c \).

Step 6: Equation Setup for (b)

For part (b), set the kinetic energy equal to five times the rest energy: \( KE = 5E_0 = 5mc^2 \). Thus, \( \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} - mc^2 = 5mc^2 \).

Step 7: Simplifying Equation for (b)

Rearrange the equation: \( \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} = 6mc^2 \). Cancel \( mc^2 \) and solve for \( v \): \( \sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{6} \).

Step 8: Solving for Speed in (b)

Square both sides: \( 1 - \frac{v^2}{c^2} = \frac{1}{36} \). Solve for \( \frac{v^2}{c^2} \): \( \frac{v^2}{c^2} = \frac{35}{36} \). This means \( v = \frac{\sqrt{35}}{6}c \).

Key Concepts

Understanding Rest Energy

Rest energy is a key concept in physics that describes the energy contained within a particle at rest, relative to an observer. It's represented by the ubiquitous equation:

• \( E_0 = mc^2 \)

Here, \( m \) represents the rest mass of the particle, and \( c \) symbolizes the speed of light. The rest energy is an inherent property of matter, implying that even a particle not in motion has energy due to its mass.
This principle stems from Einstein's theory of relativity, which suggests that mass and energy are interchangeable. This energy can be seen as the intrinsic energy that an object possesses simply by "existing". It is important in understanding how energy is conserved in various systems.

The Role of the Speed of Light

The speed of light, denoted by \( c \), is a universal constant that plays a crucial role in many areas of physics, particularly in relativity. Its approximate value is about 299,792,458 meters per second (m/s).
Light speed acts as a cosmic speed limit for how fast information and matter can travel through the universe.
In the equation \( E_0 = mc^2 \), the speed of light serves not only as a conversion factor between mass and energy but also reinforces the idea that the closer a particle's speed is to \( c \), the more significant relativistic effects become. This influences how we perceive and calculate energies and velocities in high-speed scenarios.

Describing the Lorentz Factor

The Lorentz factor, \( \gamma \), is a component used in calculating relativistic effects when velocities reach a significant fraction of the speed of light. It is given by the equation:

• \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \)

Within this formula, \( v \) is the velocity of the object, and \( c \) is the speed of light. The Lorentz factor approaches infinity as \( v \) nears \( c \), indicating that reaching or exceeding light speed requires infinite energy, which is impossible for massive particles.
The Lorentz factor helps adjust for time dilation and length contraction, two relativistic effects observed at high speeds. In terms of energy, it helps to determine the total energy of moving particles by modifying their rest energy.

Calculating Particle Speed

Calculating the speed of a particle whose kinetic energy is a multiple of its rest energy involves solving equations derived from relativistic principles.
For example, if the kinetic energy equals the rest energy (as described in part (a) of our exercise), the equation becomes:

• \( \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} = 2mc^2 \)

Solving this, we find that \( v = \frac{\sqrt{3}}{2}c \), meaning the particle moves at about 86.6% of the speed of light.
Similarly, when kinetic energy is five times the rest energy (part (b)), the equation becomes:

• \( \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} = 6mc^2 \)

This yields \( v = \frac{\sqrt{35}}{6}c \), or about 98.2% of the speed of light. Calculations like these highlight how energy requirements rapidly increase as velocities approach light speed. This illustrates why, in reality, massive objects can never reach the speed of light.