An unstable particle of mass \(M=m_{1}+m_{2}\) decays into two particles of masses \(m_{1}\) and \(m_{2}\), releasing an amount of energy \(Q .\) Determine the kinetic energies of the two particles in the CM frame. Given that \(m_{1} / m_{2}=4, Q=1 \mathrm{MeV}\), and that the unstable particle is moving in the Lab with kinetic energy \(2.25 \mathrm{MeV}\), find the maximum and minimum Lab kinetic energies of the particle of mass \(m_{1}\).

To determine the kinetic energies of the two particles in the CM frame, we will use energy conservation as well as conservation of momentum. The energy conservation equation is: $$E_{i} = E_{f}$$ where \(E_{i}\) is the initial energy of the unstable particle and \(E_{f}\) is the final energy of the two particles. Considering that \(M=m_{1}+m_{2}\), we can set up the following conservation equations: $$Q + \frac{1}{2} M V^2 = K_{1} + K_{2} + K_{rel}$$ $$M V = m_{1} v_{1} + m_{2} v_{2}$$ where \(V\) is the initial velocity of the unstable particle, \(K_{1}\) and \(K_{2}\) are the kinetic energies of particles of masses \(m_{1}\) and \(m_{2}\), and \(K_{rel}\) is the released energy due to the decay.

Divide both sides of the momentum conservation equation by \(M\): $$V = \frac {m_{1}}{M} v_{1} + \frac {m_{2}}{M} v_{2}$$ Now, we know \(Q = K_{1}+K_{2}\). Using the given ratio \(m_{1}/m_{2}=4\), we can solve for \(Q\) in terms of \(K_{1}\) and \(K_{2}\): $$Q = K_{1} + K_{2} = \frac{1}{2} m_{1} v_{1}^2 + \frac{1}{2} m_{2} v_{2}^2$$ Substitute \(m_{1}=4m_{2}\) and solve for \(K_{1}\) and \(K_{2}\): $$1\; \text{MeV} = \frac{1}{2} (4m_{2}) v_{1}^2 + \frac{1}{2} m_{2} v_{2}^2$$ Since in the CM frame, the momentum of the two particles should cancel each other: $$m_{1} v_{1} = m_{2} v_{2} \Rightarrow 4m_{2} v_{1} = m_{2} v_{2}$$ Solve the above equation for \(v_{1}^2\): $$v_{1}^2 = \frac{1}{4} v_{2}^2$$ Substitute this back into the energy conservation equation: $$1\; \text{MeV} = \frac{1}{2} (4m_{2}) \frac{1}{4} v_{2}^2 + \frac{1}{2} m_{2} v_{2}^2$$ $$\Rightarrow 1\; \text{MeV} = m_{2} v_{2}^2$$ Now we can find the kinetic energies of the two particles in the CM frame: $$K_{1} = \frac{1}{2} m_{1} v_{1}^2 = \frac{1}{2}(4m_{2})\frac{1}{4}v_{2}^2 = \frac{1}{2}m_{2} v_{2}^2$$ $$K_{2} = \frac{1}{2} m_{2} v_{2}^2$$ Since we know the total energy released is \(1 \text{MeV}\), we have: $$K_{1} = 0.5 \; \text{MeV}, \; K_{2} = 0.5 \; \text{MeV}$$

To find the maximum and minimum Lab kinetic energies of the particle of mass \(m_{1}\), we will use the Lorentz transformation. The initial total energy of the system in the Lab frame is \(E_{i}^{'}=\gamma(Mc^2+K_{i}^{'})\), where \(\gamma = \frac{1}{\sqrt{1-\beta^{2}}}\) and \(\beta = \frac{V}{c}\) with \(V\) being the initial velocity of the unstable particle. Similarly, for the final energy, $$E_{f}^{'}=\gamma \left[(m_{1} c^2 + K_{1}^{'}) + (m_{2} c^2 + K_{2}^{'})\right]$$ where \(K_{i}^{'}\) and \(K_{f}^{'}\) are the initial and final kinetic energies, respectively. Using the total energy conservation equation: $$\gamma(Mc^2+K_{i}^{'}) = \gamma \left[(m_{1} c^2 + K_{1}^{'}) + (m_{2} c^2 + K_{2}^{'})\right]$$ Substitute the given initial kinetic energy of the unstable particle \(2.25\; \text{MeV}\) and solving for \(K_{1}^{'}\) and \(K_{2}^{'}\), we get: $$\gamma(Mc^2 + 2.25\; \text{MeV}) = \gamma \left[(m_{1} c^2 + K_{1}^{'}) + (m_{2} c^2 + K_{2}^{'})\right]$$ Using the previous result \(K_1 = 0.5\;\text{MeV}\) as a starting point, we can find Maximum of \(K_1^{'}\) by adding all released energy to it, and Minimum of \(K_1^{'}\) by conservation of energy: $$\max(K_{1}^{'}) = 0.5\; \text{MeV} + 1\; \text{MeV} = 1.5\; \text{MeV}$$ $$\min(K_{1}^{'}) = 0.5\; \text{MeV}$$ The maximum and minimum Lab kinetic energies of the particle of mass \(m_{1}\) are \(1.5\; \text{MeV}\) and \(0.5\; \text{MeV}\), respectively.