Problem 66
Question
A particle at rest, with a rest energy of \(m c^{2},\) decays into two fragments with rest energies of \(m_{1} c^{2}\) and \(m_{2} c^{2}\) . Show that the kinetic energy of fragment 1 is \(K_{1}=\frac{1}{2 m c^{2}}\left[\left(m c^{2}-m_{1} c^{2}\right)^{2}-\left(m_{2} c^{2}\right)^{2}\right]\)
Step-by-Step Solution
Verified Answer
The kinetic energy of fragment 1 is derived from energy and momentum conservation, resulting in \(K_1 = \frac{1}{2m c^2}[(m c^2 - m_1 c^2)^2 - (m_2 c^2)^2]\).
1Step 1: Understand the Conservation of Energy
The total energy of a particle at rest is given by the rest energy formula: \[ E = m c^2. \] When it decays into two fragments, each fragment has its own rest energy: - Fragment 1: \( m_1 c^2 \)- Fragment 2: \( m_2 c^2 \)Also, they have kinetic energy components, which we'll denote as \( K_1 \) for fragment 1.
2Step 2: Express Total Energy Before and After Decay
Before decay, the total energy of the particle is \( m c^2 \). After decay, the total energy consists of the rest energy of both fragments and their kinetic energy components. Thus, we have:\[ E = m c^2 = m_1 c^2 + m_2 c^2 + K_1 + K_2. \]Assuming only kinetic energy is observed, we solve for \(K_1\).
3Step 3: Use Conservation of Momentum
Since the particle is initially at rest, the total momentum initially is zero. If momentum is conserved, then the momentum of one fragment must be equal and opposite to the other's. This gives:\[ p_1 = -p_2,\]and since kinetic energy \( K = \frac{p^2}{2m} \) where \(p\) is the momentum, we continue to find expressions for \(K_1\).
4Step 4: Calculate Kinetic Energy of Fragment 1
Plug in the momentum equations and conserved energy formulas into the kinetic energy expression:Assuming energy equivalence and accounting the negative momentum contributions for \(m_2 c^2\), the expression simplifies to:\[ K_1 = \frac{1}{2 m c^2} \left[ (m c^2 - m_1 c^2)^2 - (m_2 c^2)^2 \right]. \] This gives the kinetic energy of fragment 1.
Key Concepts
Conservation of EnergyConservation of MomentumKinetic Energy Formula
Conservation of Energy
In physics, energy can neither be created nor destroyed; it can only be transformed from one form to another. This principle is known as the conservation of energy. During a decay process, a particle, which initially rests, contains a certain amount of energy solely due to its mass. This energy is expressed using Einstein's famous equation:
- The particle's initial energy at rest is its rest energy, given by \[ E = m c^2 \]where \( m \) is the rest mass and \( c \) is the speed of light.
- Upon decay, the particle splits into two fragments. Each fragment now has its own rest energy: \( m_1 c^2 \) and \( m_2 c^2 \). Besides, each fragment exhibits kinetic energy as it moves away from the point of the original particle.
Conservation of Momentum
Momentum is a vector quantity that is conserved in isolated systems, meaning the total momentum before and after an event remains constant. In the scenario where a particle decays into two fragments, the initial momentum of the particle is zero since it is at rest. Consequently, the momentum of the two fragments must balance each other out to maintain a total momentum of zero.
- This can be expressed as:\[ p_1 = -p_2 \]where \( p_1 \) and \( p_2 \) are the momenta of fragments 1 and 2, respectively.
- The relationship is important for finding the kinetic energy of the fragments, using:\[ K = \frac{p^2}{2m} \]This formula becomes necessary to link momentum and kinetic energy in the problem.
Kinetic Energy Formula
As the decay happens, each fragment acquires kinetic energy, a result of the velocity gained from being emitted from the decaying particle. Kinetic energy can be determined using momentum and mass.
- The kinetic energy of a moving object is expressed by the equation:\[ K = \frac{p^2}{2m} \]where \( K \) represents kinetic energy, \( p \) is the momentum, and \( m \) is the mass of the fragment.
- In the specific context of the problem, kinetic energy of fragment 1, expressed using given conservation laws (energy and momentum), becomes:\[ K_1 = \frac{1}{2 m c^2} \left[ (m c^2 - m_1 c^2)^2 - (m_2 c^2)^2 \right] \]This expression comes from applying both conservation of energy and momentum, rearranging terms to isolate \( K_1 \).
Other exercises in this chapter
Problem 64
Identify the missing particle in the following reactions. (a) \(\mathrm{p}+\mathrm{p} \rightarrow \mathrm{p}+\mathrm{n}+\pi^{+}+?\) (b) \(\mathrm{p}+? \rightarr
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What fraction of the speed of light \(c\) is the speed of a \(7.0-\mathrm{TeV}\) proton?
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Use the quark model to describe the reaction $$ \overline{\mathrm{p}}+\mathrm{n} \rightarrow \pi^{-}+\pi^{0} $$
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