Question

A particle with rest mass \(m\) is at rest and decays into two particles of equal rest mass \(\tfrac{3}{10}m\), which move along the z-axis. Their velocities are given by:

• A. \(v_1 = v_2 = 0.8c\,\hat{z}\) axis
• B. \(v_1 = -v_2 = 0.8c\,\hat{z}\) axis
• C. \(v_1 = -v_2 = 0.8c\,\hat{z}\)
• D. \(v_1 = 0.6c\,\hat{z},\ -v_2 = 0.8c\,\hat{z}\)

Hint:

Parent at rest gives zero total momentum, so daughters go opposite ways with equal speed. Set \(mc^2 = 2\gamma(\tfrac{3}{10}m)c^2\) to get \(\gamma = 5/3\).

Answer 1

The Correct Option is B

Step 1: The parent particle is at rest, so total momentum is zero. By conservation of momentum the two daughters must carry equal and opposite momenta, hence equal speeds in opposite directions along the z-axis: \(\vec{v}_1 = -\vec{v}_2\).

Step 2: Conserve total (relativistic) energy. The initial energy is the rest energy \(mc^2\). Each daughter has energy \(\gamma\left(\tfrac{3}{10}m\right)c^2\), so \[2\,\gamma\left(\tfrac{3}{10}m\right)c^2 = mc^2.\]

Step 3: Solve for the Lorentz factor: \[\gamma = \frac{mc^2}{2\cdot\tfrac{3}{10}mc^2} = \frac{10}{6} = \frac{5}{3}.\]

Step 4: Recover the speed from \(\gamma = \dfrac{1}{\sqrt{1-v^2/c^2}}\): \[1-\frac{v^2}{c^2} = \frac{1}{\gamma^2} = \frac{9}{25} \;\Rightarrow\; \frac{v^2}{c^2} = \frac{16}{25} \;\Rightarrow\; v = 0.8c.\]

Step 5: The daughters move oppositely along z, so \(v_1 = -v_2 = 0.8c\,\hat{z}\).\[\boxed{v_1 = -v_2 = 0.8c\,\hat{z}}\]