We are given two particles:
- Particle 1: Rest mass \( m_0 \) and velocity \( u \)
- Particle 2: Rest mass \( \frac{m_0}{3} \) and velocity \( -u \)
After the collision, the particles stick together, forming a single system. We need to determine the rest mass \( M_0 \) of this combined system.
In special relativity, the total momentum before and after the collision must be conserved. The relativistic momentum \( p \) is given by:
\[ p = \gamma m v \]
where \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \).
For Particle 1:
\[ p_1 = \gamma_1 m_0 u \]
where \( \gamma_1 = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \).
For Particle 2:
\[ p_2 = \gamma_2 \left(\frac{m_0}{3}\right) (-u) \]
where \( \gamma_2 = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \).
Since \( \gamma_1 = \gamma_2 \) (both particles have the same speed \( u \)):
\[ p_2 = -\gamma_1 \left(\frac{m_0}{3}\right) u \]
The total momentum before the collision is:
\[ p_{\text{total}} = p_1 + p_2 = \gamma_1 m_0 u - \gamma_1 \left(\frac{m_0}{3}\right) u \]
\[ p_{\text{total}} = \gamma_1 m_0 u \left(1 - \frac{1}{3}\right) \]
\[ p_{\text{total}} = \gamma_1 m_0 u \left(\frac{2}{3}\right) \]
The total energy before and after the collision must also be conserved. The relativistic energy \( E \) is given by:
\[ E = \gamma m c^2 \]
For Particle 1:
\[ E_1 = \gamma_1 m_0 c^2 \]
For Particle 2:
\[ E_2 = \gamma_1 \left(\frac{m_0}{3}\right) c^2 \]
The total energy before the collision is:
\[ E_{\text{total}} = E_1 + E_2 = \gamma_1 m_0 c^2 + \gamma_1 \left(\frac{m_0}{3}\right) c^2 \]
\[ E_{\text{total}} = \gamma_1 m_0 c^2 \left(1 + \frac{1}{3}\right) \]
\[ E_{\text{total}} = \gamma_1 m_0 c^2 \left(\frac{4}{3}\right) \]
After the collision, the combined system has rest mass \( M_0 \) and is at rest (since the momenta cancel out). The total energy of the system is:
\[ E_{\text{total}} = M_0 c^2 \]
Equating the total energy before and after the collision:
\[ \gamma_1 m_0 c^2 \left(\frac{4}{3}\right) = M_0 c^2 \]
\[ M_0 = \gamma_1 m_0 \left(\frac{4}{3}\right) \]
Since \( \gamma_1 = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \), we have:
\[ M_0 = \frac{m_0 \left(\frac{4}{3}\right)}{\sqrt{1 - \frac{u^2}{c^2}}} \]
The rest mass \( M_0 \) of the system after the collision is:
\[
\boxed{M_0 = \frac{4 m_0}{3 \sqrt{1 - \frac{u^2}{c^2}}}}
\]
Answered
