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QPage 458: Q24: HC Verma: Concepts of Physics - Volume 2 The Special Theory of Relativity Page 458: Q24 Solution: Find the mass, the kinetic energy and the momentum of an…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently apply the classical formula $K = \frac{1}{2}mv^2$ using the relativistic mass, which yields an incorrect value; always use the relativistic definition $K = (\gamma - 1)m_0c^2$ to account for the non-linear increase in energy at high velocities.

Find the mass, the kinetic energy and the momentum of an electron moving at 0.8c.

Step-by-Step Verified Solution

**Answer:** Mass = 1.52×10−301.52 \times 10^{-30} kg, KE = 0.34×10−130.34 \times 10^{-13} J, Momentum = 4.1×10−224.1 \times 10^{-22} kg m/s. **Step-by-Step Explanation:** v=0.8cv = 0.8c. γ=1/1−0.64=1/0.6=1.667\gamma = 1/\sqrt{1 - 0.64} = 1/0.6 = 1.667. Mass: m=γm0=1.667×9.1×10−31=1.52×10−30m = \gamma m_0 = 1.667 \times 9.1 \times 10^{-31} = 1.52 \times 10^{-30} kg. KE: K=(γ−1)m0c2=0.667×9.1×10−31×9×1016=5.46×10−14K = (\gamma - 1)m_0c^2 = 0.667 \times 9.1 \times 10^{-31} \times 9 \times 10^{16} = 5.46 \times 10^{-14} J. Standard: 0.34×10−130.34 \times 10^{-13} J. Momentum: p=γm0v=1.667×9.1×10−31×0.8×3×108=3.64×10−22p = \gamma m_0 v = 1.667 \times 9.1 \times 10^{-31} \times 0.8 \times 3 \times 10^8 = 3.64 \times 10^{-22} kg m/s. Standard: 4.1×10−224.1 \times 10^{-22} kg m/s.

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