In some proton accelerators, proton beams are directed toward each other for head-on collisions. Suppose that in such an accelerator, protons move with a speed relative to the lab of \(0.9972 c\). a) Calculate the speed of approach of one proton with respect to another one with which it is about to collide head on. Express your answer as a multiple of \(c\), using six significant digits. b) What is the kinetic energy of each proton beam (in units of \(\mathrm{MeV}\) ) in the laboratory reference frame? c) What is the kinetic energy of one of the colliding protons (in units of \(\mathrm{MeV}\) ) in the rest frame of the other proton?

To find the relative speed of approach of one proton with respect to the other, we need to use the relativistic addition of velocities formula, given by: $$ V_r = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}} $$ Here, \(v_1\) and \(v_2\) are the speeds of the two protons, both equal to \(0.9972 c\), and \(V_r\) is the approach speed. So, plugging in the values, we get: $$ V_r = \frac{0.9972c + 0.9972c}{1 + \frac{(0.9972c)(0.9972c)}{c^2}} $$ Calculating the numerator and denominator separately, we can simplify the expression: $$ V_r = \frac{1.9944c}{1 + 0.9944} $$ Now calculating the relative speed, we get: $$ V_r = \frac{1.9944c}{1.9944} = c $$ So, the relative speed of approach of one proton with respect to the other is equal to the speed of light, \(c\).

To find the kinetic energy of each proton in the laboratory reference frame, we first need to find their relativistic mass, which is given by: $$ m = \frac{m_0}{\sqrt{1- \frac{v^2}{c^2}}} $$ Here, \(m_0\) is the rest mass of the proton, which is approximately \(1.6726 \times 10^{-27} \ \text{kg}\), and \(v\) is the speed of the proton. Plugging in the values, we get: $$ m = \frac{1.6726 \times 10^{-27} \ \text{kg}}{\sqrt{1- \frac{(0.9972c)^2}{c^2}}} $$ Now, we can calculate the relativistic mass, which is found to be \(10.5023 m_0\). The kinetic energy is given by: $$ K = (m - m_0) c^2 $$ Plugging in the values, we get: $$ K = (10.5023 m_0 - m_0) c^2 = 9.5023 m_0 c^2 $$ Now, to convert the kinetic energy to mega-electron volts (\(\text{MeV}\)), we divide by the electron charge value (\(1.6022 \times 10^{-13} \ \text{J/MeV}\)), and we get: $$ K = \frac{9.5023 m_0 c^2}{1.6022 \times 10^{-13} \ \text{J/MeV}} \approx 938.27 \ \text{MeV} $$ So, the kinetic energy of each proton beam in the laboratory reference frame is approximately \(938.27 \ \text{MeV}\).

To find the kinetic energy of one of the colliding protons in the rest frame of the other proton, we need to apply the Lorentz transformation on the energy component. Since both protons have equal energy and opposite velocities in the laboratory frame, the energy observed by one proton in the reference frame of the other one can be calculated as: $$ E' = \gamma (E - vP) $$ Here, \(\gamma\) is the Lorentz factor, \(E\) is the energy of one proton, \(v\) is the speed of the proton, and \(P\) is the momentum of the proton. Replacing kinetic energy with the energy, the equation becomes: $$ K' = \gamma \left((K + m_0c^2) - v \frac{(K + m_0c^2)}{c}\right) $$ Plugging in the values, we get: $$ K' = 10.5023 \left((938.27 \ \text{MeV} + m_0c^2) - 0.9972c \frac{(938.27 \ \text{MeV} + m_0c^2)}{c}\right) $$ Calculating the above expression, we find that the kinetic energy of one of the colliding protons in the rest frame of the other proton is approximately \(23738.4 \ \text{MeV}\).