More significant than the kinematic features of the special theory of relativity are the dynamical processes that it describes that Newtonian dynamics does not. Suppose a hypothetical particle with rest mass \(1.000 \mathrm{GeV} / c^{2}\) and \(\mathrm{ki}-\) netic energy \(1.000 \mathrm{GeV}\) collides with an identical particle at rest. Amazingly, the two particles fuse to form a single new particle. Total energy and momentum are both conserved in the collision. a) Find the momentum and speed of the first particle. b) Find the rest mass and speed of the new particle.

We know the rest mass energy (\(E_0\)) and the kinetic energy (KE) of the first particle. The total energy of the first particle can be found by adding these two energies. Total energy (E) = Rest mass energy (E_0) + Kinetic energy (KE) = \(1.000\,\mathrm{GeV}+1.000\,\mathrm{GeV}=2.000\,\mathrm{GeV}\)

Now, we will find the momentum of the first particle using energy-momentum relationship: \(E^2=p^2c^2+E_0^2\) Solving for momentum (p), we get \(p=\dfrac{\sqrt{E^2 - E_0^2}}{c}\) Plug in the values: \(p=\dfrac{\sqrt{(2.000\,\mathrm{GeV})^2 - (1.000\,\mathrm{GeV})^2}}{c}=\dfrac{\sqrt{3}\,\mathrm{GeV}}{c}\)

Now, we will find the speed of the first particle using the momentum and rest mass: \(p=\dfrac{mv}{\sqrt{1 - \dfrac{v^2}{c^2}}}\) Solving for the speed (v), we get \(v=mc\dfrac{\sqrt{1 - \dfrac{v^2}{c^2}}}{p}\) Replacing \(m\) and \(p\) with their values: \(v=c\,\dfrac{\sqrt{1- \dfrac{(1.000\,\mathrm{GeV}/c^2)^2}{(\sqrt{3}\,\mathrm{GeV}/c)^2}}}{\sqrt{3}}\)

Since particle 2 is at rest, its kinetic energy is 0. The total energy and momentum of two particles are the sum of their individual energies and momenta. Total energy = \(E_1 + E_2 = (2.000 + 1.000)\,\mathrm{GeV} = 3.000\,\mathrm{GeV}\) Total momentum = \(p_1 + p_2 = \dfrac{\sqrt{3}\,\mathrm{GeV}}{c}\)

Let's denote the rest mass of the final particle as \(m_f\). We know the total energy and momentum of the two initial particles are conserved. To find \(m_f\), we will use the energy-momentum relation: \(E^2 = p^2c^2 + m_f^2c^4\) Plug in the values of total energy and total momentum: \((3.000\,\mathrm{GeV})^2 = \left(\dfrac{\sqrt{3}\,\mathrm{GeV}}{c}\right)^2c^2 + m_f^2c^4\) Solving for \(m_f\): \(m_f = \dfrac{\sqrt{(3.000\,\mathrm{GeV})^2 - \left(\dfrac{\sqrt{3}\,\mathrm{GeV}}{c}\right)^2c^2}}{c^2}\)

Since the total momentum of the system is conserved, the momentum of the resulting particle, denoted as \(p_f\), is equal to the total momentum before the collision. Knowing \(p_f\) and \(m_f\), we can find the speed of the new particle using the momentum formula: \(p_f=\dfrac{m_fv_f}{\sqrt{1 - \dfrac{v_f^2}{c^2}}}\) Again, solving for speed (\(v_f\)), we get \(v_f=c\,\dfrac{\sqrt{1 - \dfrac{m_f^2c^2}{\left(\dfrac{\sqrt{3}\,\mathrm{GeV}}{c}\right)^2}}}{m_f}\) By obtaining these results, we found: a) The momentum and speed of the first particle: \(p_1 = \dfrac{\sqrt{3}\,\mathrm{GeV}}{c}\) and \(v_1 = c\,\dfrac{\sqrt{1- \dfrac{(1.000\,\mathrm{GeV}/c^2)^2}{(\sqrt{3}\,\mathrm{GeV}/c)^2}}}{\sqrt{3}}\) b) The rest mass and speed of the new particle: \(m_f = \dfrac{\sqrt{(3.000\,\mathrm{GeV})^2 - \left(\dfrac{\sqrt{3}\,\mathrm{GeV}}{c}\right)^2c^2}}{c^2}\) and \(v_f = c\,\dfrac{\sqrt{1 - \dfrac{m_f^2c^2}{\left(\dfrac{\sqrt{3}\,\mathrm{GeV}}{c}\right)^2}}}{m_f}\)