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Magazine Chapter 26: Problem 30
A proton is traveling at a speed of \(0.55 c\). What are its (a) total energy, (b) kinetic energy, and (c) momentum?
Short Answer
Expert verified
The proton's total energy is approximately \( 1.80 \times 10^{-10} \text{ J} \), its kinetic energy is \( 0.297 \times 10^{-10} \text{ J} \), and its momentum is \( 3.30 \times 10^{-19} \text{ kg m/s} \).
Step by step solution
01
Identify Given Quantities
We are given the speed of the proton as \[ v = 0.55c \] where \( c \) is the speed of light. The rest mass of a proton \( m_0 \) is approximately \( 1.67 \times 10^{-27} \) kilograms. These quantities will be used to calculate total energy, kinetic energy, and momentum.
02
Calculate Lorentz Factor
The Lorentz factor \( \gamma \) is calculated using the formula: \[ \gamma = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \]Substitute \( v = 0.55c \):\[ \gamma = \frac{1}{\sqrt{1 - (0.55)^2}} = \frac{1}{\sqrt{1 - 0.3025}} \]\[ \gamma \approx \frac{1}{\sqrt{0.6975}} \approx 1.20 \]
03
Compute Total Energy
The total energy \( E \) of the proton is given by: \[ E = \gamma m_0 c^2 \]Calculate:\[ E = 1.20 \cdot 1.67 \times 10^{-27} \cdot (3 \times 10^8)^2 = 1.20 \cdot 1.67 \times 10^{-27} \cdot 9 \times 10^{16} \]\[ E \approx 1.80 \times 10^{-10} \text{ Joules} \]
04
Compute Kinetic Energy
Kinetic energy \( KE \) is the difference between total energy and rest energy:Rest energy \( E_0 = m_0 c^2 = 1.67 \times 10^{-27} \cdot 9 \times 10^{16} \approx 1.503 \times 10^{-10} \text{ Joules} \)Thus, kinetic energy is:\[ KE = E - E_0 = 1.80 \times 10^{-10} - 1.503 \times 10^{-10} \approx 0.297 \times 10^{-10} \text{ Joules} \]
05
Compute Momentum
Momentum \( p \) is given by the formula:\[ p = \gamma m_0 v \]Calculate:\[ p = 1.20 \cdot 1.67 \times 10^{-27} \cdot 0.55 \cdot 3 \times 10^8 \]\[ p \approx 3.30 \times 10^{-19} \text{ kg m/s} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Proton Speed
The speed of a proton plays a crucial role in understanding various relativistic effects, especially those outlined in Einstein's theory of relativity. In the given exercise, the proton is traveling at a speed of \(0.55c\), where \(c\) represents the speed of light, approximately \(3 \times 10^8\) meters per second. This indicates that the proton is moving at 55% of the speed of light, a speed at which relativistic effects become significant. The speed impacts not only the proton鈥檚 momentum but also its energy calculations.
- At such speeds, the classic formulas of Newtonian physics (like \(KE = \frac{1}{2}mv^2\)) are no longer sufficient.
- Relativistic formulas must be used to accurately determine quantities like kinetic energy, total energy, and momentum.
Total Energy
In relativistic physics, the total energy of a particle is an important concept, encompassing both its rest energy and its kinetic energy. To compute the total energy of a proton moving at relativistic speeds, we use the formula:
\[ E = \gamma m_0 c^2 \] where \( \gamma \) is the Lorentz factor and \( m_0 \) is the rest mass of the proton. The total energy includes:
\[ E = \gamma m_0 c^2 \] where \( \gamma \) is the Lorentz factor and \( m_0 \) is the rest mass of the proton. The total energy includes:
- Rest Energy: This is the energy a particle has due to its mass when it is not moving. It is given by \( m_0 c^2 \).
- Kinetic Energy: At relativistic speeds, it becomes more accurate to consider the difference between total energy and rest energy.
Kinetic Energy
The kinetic energy in relativistic physics is calculated differently than in classical physics. It is given as the difference between the total energy and the rest energy. This accounts for the increased "relativistic mass" and velocity effects:
\[ KE = E - E_0 \] where \( E \) is the total energy and \( E_0 = m_0 c^2 \) is the rest energy of the proton.
For a proton traveling at a relativistic speed, this additional energy is what contributes to its motion. In this exercise, the rest energy is calculated as approximately \(1.503 \times 10^{-10}\) Joules, while the kinetic energy turns out to be approximately \(0.297 \times 10^{-10}\) Joules.
\[ KE = E - E_0 \] where \( E \) is the total energy and \( E_0 = m_0 c^2 \) is the rest energy of the proton.
For a proton traveling at a relativistic speed, this additional energy is what contributes to its motion. In this exercise, the rest energy is calculated as approximately \(1.503 \times 10^{-10}\) Joules, while the kinetic energy turns out to be approximately \(0.297 \times 10^{-10}\) Joules.
- This additional kinetic energy comes into play because the proton鈥檚 speed approaches the speed of light, necessitating relativity-based calculations.
Lorentz Factor
The Lorentz factor \( \gamma \) is a key component in relativistic physics, which modifies how we calculate energy and momentum when objects move at speeds close to the speed of light. It is defined by the following formula:
\[ \gamma = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \]By inserting \( v = 0.55c \) into the formula, we derive \( \gamma \approx 1.20 \). This factor tells us how much time, length, and relativistic mass change relative to different reference frames, such as when protons travel at high speeds.
\[ \gamma = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \]By inserting \( v = 0.55c \) into the formula, we derive \( \gamma \approx 1.20 \). This factor tells us how much time, length, and relativistic mass change relative to different reference frames, such as when protons travel at high speeds.
- The larger \( \gamma \), the more pronounced the relativistic effects, reflecting significant increases in energy and momentum compared to when at rest.
Momentum
Momentum in relativistic physics extends the classical notion by incorporating the Lorentz factor. The relativistic momentum is defined as:
\[ p = \gamma m_0 v \]By using \( \gamma \), relativistic momentum accounts for changes at high speeds which aren't reflected in classical calculations.
\[ p = \gamma m_0 v \]By using \( \gamma \), relativistic momentum accounts for changes at high speeds which aren't reflected in classical calculations.
- For the proton in our example, this led to a calculation where momentum is roughly \(3.30 \times 10^{-19}\) kg m/s.
- Relativistic momentum increases at a faster rate compared to classical momentum as speed grows closer to the speed of light.
