Problem 31 - College Physics

Question

\bullet 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 speed of the proton?

Step-by-Step Solution

Answer

(a) 4.509 × 10^{-10} J, (b) 3.558 × 10^{-19} kg m/s, (c) 2.83 × 10^8 m/s.

Step 1: Understand Rest and Total Energy

The rest energy of a proton is given by the formula \(E_0 = m_0 c^2\), where \(m_0\) is the rest mass and \(c\) is the speed of light. Given the rest mass \(m_0 = 1.67 \times 10^{-27} \mathrm{kg}\) and \(c = 3 \times 10^8 \mathrm{m/s}\), calculate the rest energy.

Step 2: Calculate Rest Energy

Using the equation \(E_0 = m_0 c^2\):\[ E_0 = 1.67 \times 10^{-27} \times (3 \times 10^8)^2 = 1.503 \times 10^{-10} \mathrm{J}.\]

Step 3: Determine Total Energy

The total energy \(E\) is given as 4.00 times the rest energy. Therefore, \(E = 4.00 \times E_0 = 4.00 \times 1.503 \times 10^{-10} = 6.012 \times 10^{-10} \mathrm{J}.\)

Step 4: Find Kinetic Energy

The kinetic energy \(K\) can be found from the total energy minus the rest energy: \(K = E - E_0\). Put the values calculated before:\[ K = 6.012 \times 10^{-10} - 1.503 \times 10^{-10} = 4.509 \times 10^{-10} \mathrm{J}. \]

Step 5: Use Relativistic Energy-Momentum Relation

The relationship between energy, momentum, and mass is given by the equation:\[ E^2 = (pc)^2 + (m_0 c^2)^2. \]Using this equation to find the momentum \(p\), solve for \(p\):\[ p = \frac{\sqrt{E^2 - (m_0 c^2)^2}}{c}. \]

Step 6: Calculate Momentum

Using values from previous steps, we calculate:\[ p = \frac{\sqrt{(6.012 \times 10^{-10})^2 - (1.503 \times 10^{-10})^2}}{3 \times 10^8}. \]Solving gives:\[ p = 3.558 \times 10^{-19} \mathrm{kg \, m/s}. \]

Step 7: Determine the Speed of Proton

The speed \(v\) can be found from the relation between energy and momentum in relativistic terms:\[ v = \frac{pc^2}{E}. \]Substitute in the calculated values:\[ v = \frac{3.558 \times 10^{-19} \times (3 \times 10^8)^2}{6.012 \times 10^{-10}}. \]The calculation yields \(v = 2.83 \times 10^8 \mathrm{m/s}\).

Key Concepts

Proton Kinetic Energy

In the realm of relativistic physics, kinetic energy isn't simply the result of applying classical formulas. Instead, it takes on a more complex form when an object, like a proton, is moving at speeds that approach the speed of light. The kinetic energy of a proton is the difference between its total energy and its rest energy. To find this, first compute the rest energy using the equation \(E_0 = m_0 c^2\). Here, \(m_0 = 1.67 \times 10^{-27} \, \mathrm{kg}\), and \(c\), the speed of light, is \(3 \times 10^8 \, \mathrm{m/s}\). Insert these into the formula to obtain \(E_0 = 1.503 \times 10^{-10} \, \mathrm{J}\).

Given that the total energy of the proton is 4 times its rest energy, calculate the total energy by multiplying \(E_0\) by 4, resulting in \(6.012 \times 10^{-10} \, \mathrm{J}\). Thus, the kinetic energy \(K\) is calculated as \(K = E - E_0\). Substituting the known values gives \(K = 4.509 \times 10^{-10} \, \mathrm{J}\). This energy represents the proton's capacity to perform work due to its motion at relativistic speed.

Momentum Calculation

Momentum in relativistic physics requires a different approach than the ordinary long-known formula \(p = mv\), especially when dealing with particles moving at high speeds. The relationship between the energy, momentum \(p\), and rest mass for a proton moving near the speed of light uses the equation \(E^2 = (pc)^2 + (m_0 c^2)^2\). By rearranging, you can solve for momentum: \(p = \frac{\sqrt{E^2 - (m_0 c^2)^2}}{c}\).

Inserting the earlier-calculated energies, \(E = 6.012 \times 10^{-10} \, \mathrm{J}\) and \(E_0 = 1.503 \times 10^{-10} \, \mathrm{J}\), into this equation, results in a calculation of \(p = 3.558 \times 10^{-19} \, \mathrm{kg \, m/s}\).
Remember, this value represents the product of the proton's relativistic mass and its velocity, depicting how the particle's momentum increases with speed.

Relativistic Speed

Understanding relativistic speed means recognizing that as an object, like a proton, nears the speed of light, its behavior significantly diverges from classical predictions. The velocity \(v\) of a high-speed proton can be derived using \(v = \frac{pc^2}{E}\), where \(p\) is momentum and \(E\) is total energy.

This equation accounts for the large values of the momentum and energy at relativistic speeds. Substitute \(p = 3.558 \times 10^{-19} \, \mathrm{kg \, m/s}\) and \(E = 6.012 \times 10^{-10} \, \mathrm{J}\) into the formula to find \(v = 2.83 \times 10^8 \, \mathrm{m/s}\).
Despite this speed being enormously high, it's still less than the speed of light, illustrating one of the fundamental aspects of relativistic physics that speeds can approach but not reach light speed.