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QPage 458: Q4: HC Verma: Concepts of Physics - Volume 2 The Special Theory of Relativity Page 458: Q4 Solution: A person standing on a platform finds that a train moving with…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently invert the length contraction formula, calculating the rest length as $L_0 = L\sqrt{1 - v^2/c^2}$, which incorrectly implies the moving object is longer than its rest length. Remember that the observed length $L$ is always shorter than the rest length $L_0$, so you must divide by the Lorentz factor (or multiply by $1/\gamma$) to find $L_0$.

A person standing on a platform finds that a train moving with velocity 0.6c takes one second to pass by him. Find a) the length of the train as seen by the person and b) the rest length of the train.

Step-by-Step Verified Solution

**Answer:** (a) Length of train = 1.8×1081.8 \times 10^8 m (b) Rest length = 2.25×1082.25 \times 10^8 m. **Step-by-Step Explanation:** (a) The person on the platform measures the moving train. Time to pass = 1 s, speed = 0.6c=1.8×1080.6c = 1.8 \times 10^8 m/s. Contracted length L=v×t=1.8×108×1=1.8×108L = v \times t = 1.8 \times 10^8 \times 1 = 1.8 \times 10^8 m. (b) L=L01−v2/c2=L01−0.36=0.8L0L = L_0\sqrt{1 - v^2/c^2} = L_0\sqrt{1 - 0.36} = 0.8L_0. L0=L/0.8=1.8×108/0.8=2.25×108L_0 = L/0.8 = 1.8 \times 10^8 / 0.8 = 2.25 \times 10^8 m.

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