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QPage 458: Q28: HC Verma: Concepts of Physics - Volume 2 The Special Theory of Relativity Page 458: Q28 Solution: Find the speed at which the kinetic energy of a particle will…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often mistakenly set the leading term $\frac{1}{2}v^2$ equal to the non-relativistic expression, which yields $v=0$ or a contradiction. One must expand the relativistic kinetic energy to the next order, $K \approx \frac{1}{2}m_0v^2 + \frac{3}{8}m_0v^4/c^2$, and equate the higher-order correction term to the specified percentage difference.

Find the speed at which the kinetic energy of a particle will differ by 1% from its nonrelativistic value 12m0v2\frac{1}{2}m_0v^2.

Step-by-Step Verified Solution

**Answer:** v=0.115cv = 0.115c. **Step-by-Step Explanation:** Relativistic KE: K=(γ−1)m0c2K = (\gamma - 1)m_0c^2. Non-relativistic KE: Knr=12m0v2K_{nr} = \frac{1}{2}m_0v^2. The difference is 1%: K=1.01KnrK = 1.01 K_{nr}. (γ−1)c2=1.01×12v2(\gamma - 1)c^2 = 1.01 \times \frac{1}{2}v^2. For small v/cv/c, γ−1≈12v2/c2+38v4/c4\gamma - 1 \approx \frac{1}{2}v^2/c^2 + \frac{3}{8}v^4/c^4. 12v2+38v4/c2=1.01×12v2\frac{1}{2}v^2 + \frac{3}{8}v^4/c^2 = 1.01 \times \frac{1}{2}v^2. 38v2/c2=0.01×12⇒v2/c2=0.0133⇒v=0.115c\frac{3}{8}v^2/c^2 = 0.01 \times \frac{1}{2} \Rightarrow v^2/c^2 = 0.0133 \Rightarrow v = 0.115c .

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