Question

Two protons (each with rest mass \(M\) = 1.67 \(\times\) 10\(^{-27}\) kg) are initially moving with equal speeds in opposite directions. The protons continue to exist after a collision that also produces an \(\eta_0\) particle (see Chapter 44). The rest mass of the \(\eta_0\) is m = 9.75 \(\times\) 10\(^{-28}\) kg. (a) If the two protons and the \(\eta_0\) are all at rest after the collision, find the initial speed of the protons, expressed as a fraction of the speed of light. (b) What is the kinetic energy of each proton? Express your answer in MeV. (c) What is the rest energy of the \(\eta_0\), expressed in MeV? (d) Discuss the relationship between the answers to parts (b) and (c). 37.39 .

Short answer

Initial speed: approximately 0.367c; each proton's kinetic energy: 457.4 MeV; \(\eta_0\) rest energy: 914.8 MeV.

Step-by-step solution

Understand the Problem

We need to find the speed and kinetic energy of protons before collision and verify the energy balance after producing an \eta_0 particle. The rest masses are given: for protons, \( M = 1.67 \times 10^{-27} \) kg; for \eta_0, \( m = 9.75 \times 10^{-28} \) kg.

Calculate Initial Total Energy

The total initial energy is the rest energy of both protons plus their kinetic energy. Each proton has rest energy of \( E_0 = M c^2 \), where \( c \) is the speed of light. Hence, \( 2M c^2 \) is the initial energy contributed by their rest masses.

Use Conservation of Energy

According to energy conservation, the initial energy should equal the final energy, which includes rest masses of both protons and the \( \eta_0 \) particle. Thus, \( 2T + 2M c^2 = 2M c^2 + mc^2 \). Here, \( T \) represents the kinetic energy of each proton.

Simplify and Solve for Kinetic Energy

The equation simplifies to \( 2T = mc^2 \). Solving for \( T \), we get \( T = \frac{1}{2}mc^2 \). This value represents the kinetic energy of each proton before collision.

Calculate Proton's Initial Speed

Use the relativistic energy-momentum relation \( \gamma Mc^2 = T + Mc^2 \). Thus \( \gamma = 1 + \frac{T}{Mc^2} \). From \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \), solve for \( v \) to find the initial speed of the protons.

Convert Kinetic Energy to MeV

Rest energy of \( m = 9.75 \times 10^{-28} \) kg is \( mc^2 \). Convert this to MeV using the conversion factor \( 1 kg = 5.61 \times 10^{29} \, MeV/c^2 \): \( mc^2 \approx 914.8 \, MeV \).

Discussion on Energy Conversion

Compare the kinetic energy of the protons in MeV and the rest energy of \( \eta_0 \). Discuss the idea of energy conservation and how some kinetic energy converts to form the rest energy of new particles.

Key concepts

energy conservation

Energy conservation is a fundamental principle in physics, stating that energy cannot be created or destroyed, only transformed. In the context of proton collisions, this means that the total energy before the collision must equal the total energy after the collision. Before the collision, the energy consists of the rest mass energy and the kinetic energy of the protons. After the collision, the energy comprises the rest mass energies of the protons and the newly formed \( \eta_0 \) particle.
In this exercise, the initial energy is the combined rest mass energy and kinetic energy of two protons moving in opposite directions. After they collide and come to rest, the energy includes the rest mass energies of both protons and the \( \eta_0 \) particle.
This energy conservation leads to the equation \( 2T + 2Mc^2 = 2Mc^2 + mc^2 \), where \( T \) is the kinetic energy of one proton, \( M \) is the rest mass of a proton, \( m \) is the rest mass of the \( \eta_0 \), and \( c \) is the speed of light. Simplifying this equation shows that the kinetic energy of the protons initially converts into the rest energy of the newly formed particle.

relativistic motion

Relativistic motion refers to scenarios where objects move at speeds comparable to the speed of light. At such high speeds, the classical Newtonian laws of motion become less accurate, and Einstein's theory of relativity provides a more precise framework.
In our exercise, the protons are initially moving at relativistic speeds before their collision, requiring us to consider relativistic effects. The Lorentz factor, \( \gamma \), becomes significant, reflecting the amount by which time, length, and relativistic mass are altered at high speeds. To find the speed \( v \) of protons as a fraction of the speed of light, the relativistic energy-momentum relation \( \gamma Mc^2 = T + Mc^2 \) is used, which incorporates the rest mass energy and kinetic energy.

• The factor \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \) accounts for these relativistic changes.
• Using this, the initial speed \( v \) of the protons can be determined, providing insight into their motion close to the speed of light.

Understanding relativistic motion is crucial for solving physics problems involving particles moving at substantial fractions of the speed of light.

kinetic energy

Kinetic energy is the energy that an object possesses due to its motion. In classical mechanics, it's given by \( KE = \frac{1}{2}mv^2 \). However, at relativistic speeds, the formula needs to be adjusted using the Lorentz factor \( \gamma \).
For the protons in our exercise, the kinetic energy before collision is crucial in determining both the initial speed and the energy dynamics post-collision. It's represented as \( T \), and from the energy balance equation, we find \( T = \frac{1}{2}mc^2 \). This equation arises from setting the initial and final energy equal, according to energy conservation.
The conversion of this kinetic energy into the rest energy of the new \( \eta_0 \) particle is a real-life example of how kinetic energy in particles can become mass, demonstrating the equivalence of mass and energy as highlighted in Einstein's famous equation \( E = mc^2 \).

• Calculating kinetic energy ensures understanding of how much energy was initially available for other conversion processes, like particle creation.
• This relationship exemplifies the conversion of motion-related energy into intrinsic energy of new particles.

rest mass energy

Rest mass energy is the energy contained in an object due to its mass when it is not in motion. This concept is rooted in Einstein's relativity, expressed by the equation \( E = mc^2 \). It shows the intrinsic energy that an object has purely because of its mass.
In the proton collision exercise, the rest mass energy of each proton is \( E_0 = Mc^2 \), and for the \( \eta_0 \) particle, it is \( mc^2 \). The rest mass energy constitutes a part of the total energy in the problem both before and after the collision.
Understanding rest mass energy is pivotal because it helps explain why, after the collision, the energy stored in the kinetic motion of protons can "disappear" into the mass of a new particle, the \( \eta_0 \).

• Rest mass energy provides a stockpile of energy before motion, which contributes alongside kinetic energy to total energy.
• The synthesis of new particles, like \( \eta_0 \), demonstrates conversion from kinetic to rest mass energy, a cornerstone of particle physics.

Exploring exercises like this enhances comprehension of mass-energy equivalence and its implications in particle transformations and conservation laws.