Subtopic: Relativistic Momentum, Energy & 4-Vectors
Question Statement:
Two identical particles of rest mass m_0 approach each other with equal and
opposite velocity v = 0.5c, where c is the speed of light. The total energy of one
particle as measured in the rest frame of the other is E = \alpham_0c2. The value of \alpha
is _______________ (Round off to two decimal places)
Official Answer Key:
1.65 to 1.70
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.61)**
**1. Special Relativity (Relativistic Velocity Addition & Energy)**:
- Two particles of rest mass $m_0$ approach each other with velocities $u = +0.5c$ and $v = -0.5c$.
- Relative velocity $v_{\text{rel}}$ of one particle in the rest frame of the other:
$v_{\text{rel}} = \frac{u - v}{1 - \frac{uv}{c^2}}$
**2. Detailed Step-by-Step Derivation**:
- Substitute the velocity values:
$v_{\text{rel}} = \frac{0.5c - (-0.5c)}{1 - \frac{(0.5c)(-0.5c)}{c^2}} = \frac{1.0c}{1 + 0.25} = \frac{1.0c}{1.25} = 0.8c$
- Compute the relativistic Lorentz factor $\gamma_{\text{rel}}$:
$\gamma_{\text{rel}} = \frac{1}{\sqrt{1 - \left(\frac{v_{\text{rel}}}{c}\right)^2}} = \frac{1}{\sqrt{1 - (0.8)^2}} = \frac{1}{\sqrt{1 - 0.64}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.6} = \frac{5}{3} \approx 1.6667$
- Total relativistic energy:
$E = \gamma_{\text{rel}} m_0 c^2 = \frac{5}{3} m_0 c^2 \approx 1.67 m_0 c^2$
- Therefore, $\alpha = 1.67$.
**3. Official Verification**:
The verified official IIT Master Answer range is **1.65 to 1.70**.
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