Question

A particle of mass 12 MeV/c \(^{2}\) has a kinetic energy of 1 \(\mathrm{MeV}\). What are its momentum (in MeV/c) and its speed (in units of \(c\) )?

Short answer

Momentum: 5 MeV/c; Speed: 5/13c.

Step-by-step solution

Use Energy-Momentum Relation

To find the momentum, use the energy-momentum relation:\[E^2 = (pc)^2 + (m_0 c^2)^2\]where \(E\) is the total energy, \(p\) is the momentum, \(m_0\) is the rest mass, and \(c\) is the speed of light.Total energy: \(E = K + m_0 c^2 = 1 \text{ MeV} + 12 \text{ MeV} = 13 \text{ MeV}\).Plug in \(E\) and \(m_0 c^2\):\[13^2 = (pc)^2 + 12^2\].

Solve for Momentum

Rearrange the energy-momentum equation to find momentum:\[(p c)^2 = 13^2 - 12^2\]Calculate:\[(p c)^2 = 169 - 144 = 25\].Thus, \(p c = \sqrt{25} = 5 \text{ MeV}\).Therefore, the momentum \(p\) is \(5 \text{ MeV}/c\).

Calculate Speed using Relativistic Relationships

Use the relation:\[p = \frac{m_0 v}{\sqrt{1-v^2/c^2}}\]Substituting the values,\[5 = \frac{12v}{\sqrt{1-v^2}}\].Rearrange the equation and solve for \(v\):\[5\sqrt{1-v^2} = 12v\].Square both sides:\[25(1-v^2) = 144v^2\].Simplify and solve the resulting equation for \(v\).

Solve the Speed Equation

Continuing from the previous step,\[25 - 25v^2 = 144v^2\],\[25 = 169v^2\],\[v^2 = \frac{25}{169}\],\[v = \sqrt{\frac{25}{169}} = \frac{5}{13}\].Thus, the speed \(v\) in units of \(c\) is \(\frac{5}{13}c\).

Key concepts

Energy-Momentum Relation

The energy-momentum relation is a fundamental concept in relativistic mechanics. It connects a particle's total energy, its momentum, and its rest mass through the equation: \[ E^2 = (pc)^2 + (m_0c^2)^2 \] Here, \( E \) represents the total energy of the particle, \( p \) is the momentum, \( m_0 \) is the rest mass, and \( c \) is the speed of light. In simpler terms, this equation says that a particle's total energy is made up of its energy due to motion (momentum) and its energy due to mass (rest energy).

• Total Energy \( E = K + m_0c^2 \) where \( K \) is the kinetic energy.
• The formula helps link energy with momentum for particles moving near light speed.

In our exercise, the total energy was calculated by adding the kinetic energy (1 MeV) to the rest energy (12 MeV), resulting in 13 MeV. This value was then used in the equation to help determine the momentum of the particle. The ability to link and solve for different variables through this relation is crucial in understanding high-speed particle physics.

Relativistic Momentum

Relativistic momentum comes into play when particles move at speeds close to that of light. Unlike classical mechanics, where momentum is simply \( mv \), relativistic momentum is adjusted due to the effects of relativity. It is defined as: \[ p = \frac{m_0 v}{\sqrt{1-v^2/c^2}} \] This formula shows that as a particle’s speed \( v \) approaches the speed of light \( c \), the momentum sharply increases. This is because the denominator \( \sqrt{1-v^2/c^2} \) approaches zero, making the fraction larger.

• Newtonian momentum fails at high speeds, but this formula accounts for relativistic effects.
• Momentum increases infinitely as velocity approaches the speed of light.

In solving our exercise problem, the calculated momentum \( 5 \text{ MeV}/c \) was determined using the energy-momentum relation before being plugged into this relativistic equation to find the particle's speed. This ensures both energy and motion factors are consistently represented in calculations.

Particle Speed Calculation

Calculating the speed of a particle in relativistic mechanics requires revisiting the formula for relativistic momentum: \[ p = \frac{m_0 v}{\sqrt{1-v^2/c^2}} \] To find the speed \( v \), we must solve this equation. This involves rearranging the expression to isolate \( v \). Begin by substituting the known momentum and rest mass into the equation. In this case, \( 5 = \frac{12v}{\sqrt{1-v^2}} \). Solving for \( v \) involves:

• Rearranging: \( 5\sqrt{1-v^2} = 12v \).
• Squaring both sides to eliminate the square root.
• Simplifying and solving the resulting quadratic equation: \( v^2 = \frac{25}{169} \).
• Finally, calculating the square root to find \( v = \frac{5}{13}c \).

This step-by-step approach shows how speed is determined by using known values of momentum and mass, demonstrating the careful balance of relativistic forces. The final speed \( v \) confirms how close the particle is moving in comparison to the speed of light, offering insights into particle dynamics at high velocities.