Question

Consider a relativistic particle of rest mass 2m moving with a speed v along the x direction. It collides with another relativistic particle of rest mass m moving with the same speed but in the opposite direction. These two particles coalesce to form one particle whose rest mass M is
(\(\beta = \frac{v}{c}\), where c is the speed of light)

• A. \(m \sqrt{\frac{9 - \beta^2}{1 - \beta^2}}\)
• B. \(2m \sqrt{\frac{3 - \beta^2}{1 - \beta^2}}\)
• C. \(\frac{m}{2} \sqrt{\frac{9 - \beta^2}{2 - \beta^2}}\)
• D. \(\frac{m}{4} \sqrt{\frac{1 - \beta^2}{2 - \beta^2}}\)

Hint:

In relativistic collision problems, using the invariant quantity \(E^2 - (pc)^2 = (m_0c^2)^2\) for the entire system before and after the collision often simplifies the calculation. This avoids the need to explicitly find the final velocity of the coalesced particle.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem involves an inelastic collision between two relativistic particles. In any collision, both total energy and total momentum are conserved. We can use the conservation laws and the relativistic energy-momentum relation to find the rest mass of the final particle.
Step 2: Key Formula or Approach:
The key principles are the conservation of relativistic energy and momentum.
Relativistic energy: \(E = \gamma m_0 c^2 = \frac{m_0 c^2}{\sqrt{1 - v^2/c^2}}\)
Relativistic momentum: \(p = \gamma m_0 v = \frac{m_0 v}{\sqrt{1 - v^2/c^2}}\)
Conservation of Energy: \(E_{\text{initial}} = E_{\text{final}}\)
Conservation of Momentum: \(p_{\text{initial}} = p_{\text{final}}\)
A very useful tool is the energy-momentum invariant relation for a system: \(E_{\text{total}}^2 - (p_{\text{total}}c)^2 = (M_{\text{rest}}c^2)^2\), where \(M_{\text{rest}}\) is the total rest mass of the system. For a single particle, this is its rest mass M.
Step 3: Detailed Explanation:
Let's denote the Lorentz factor as \(\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - \beta^2}}\).
Initial State (before collision):
Particle 1: rest mass \(m_1 = 2m\), velocity \(v_1 = +v\).
Particle 2: rest mass \(m_2 = m\), velocity \(v_2 = -v\).
Total initial energy (\(E_i\)):
\[ E_i = E_1 + E_2 = \gamma (2m) c^2 + \gamma (m) c^2 = 3 \gamma m c^2 = \frac{3mc^2}{\sqrt{1-\beta^2}} \] Total initial momentum (\(p_i\)):
\[ p_i = p_1 + p_2 = \gamma (2m) v + \gamma (m) (-v) = \gamma m v = \frac{mv}{\sqrt{1-\beta^2}} \] Final State (after collision):
The two particles coalesce into a single particle of rest mass M. Let its velocity be V.
Final energy \(E_f\) and momentum \(p_f\) of this single particle are related by \(E_f^2 - (p_f c)^2 = (M c^2)^2\).
Applying Conservation Laws:
By the principle of conservation, the total energy and momentum of the system are conserved.
\(E_f = E_i = \frac{3mc^2}{\sqrt{1-\beta^2}}\)
\(p_f = p_i = \frac{mv}{\sqrt{1-\beta^2}}\)
Now we use the energy-momentum invariant for the final particle:
\[ (Mc^2)^2 = E_f^2 - (p_f c)^2 \] Substituting the values of \(E_f\) and \(p_f\):
\[ (Mc^2)^2 = \left(\frac{3mc^2}{\sqrt{1-\beta^2}}\right)^2 - \left(\frac{mv}{\sqrt{1-\beta^2}} c\right)^2 \] \[ M^2c^4 = \frac{9m^2c^4}{1-\beta^2} - \frac{m^2v^2c^2}{1-\beta^2} \] Since \(\beta = v/c\), we have \(v = \beta c\). Substituting this:
\[ M^2c^4 = \frac{9m^2c^4}{1-\beta^2} - \frac{m^2(\beta c)^2c^2}{1-\beta^2} = \frac{9m^2c^4}{1-\beta^2} - \frac{m^2\beta^2c^4}{1-\beta^2} \] \[ M^2c^4 = \frac{(9 - \beta^2)m^2c^4}{1-\beta^2} \] Cancelling \(c^4\) from both sides:
\[ M^2 = \frac{(9 - \beta^2)m^2}{1-\beta^2} \] \[ M = \sqrt{\frac{(9 - \beta^2)m^2}{1-\beta^2}} = m \sqrt{\frac{9 - \beta^2}{1 - \beta^2}} \] Step 4: Final Answer:
The rest mass M of the resulting particle is \(m \sqrt{\frac{9 - \beta^2}{1 - \beta^2}}\). This corresponds to option (A).