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JEE Main 2026 (2 Apr, Shift 2) · Motion of Centre of Mass

Motion of Centre of Mass question from JEE Main 2026 (2 Apr, Shift 2)

Q3JEE Main 2026 · 2 APR · SHIFT 2Medium

Motion of Centre of Mass

Two blocks of masses 2 kg and 1 kg respectively, are tied to the ends of a string which passes over a light frictionless pulley as shown in the figure below. The masses are held at rest at the same horizontal level and then released. The distance traversed by the centre of mass in 2 s is _______ m. (Take g=10 m/s2g = 10\text{ m/s}^2)

Figure

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Answer: (c)

  1. Block acceleration: a=(2−1)g2+1=103a = \frac{(2-1)g}{2+1} = \frac{10}{3} m/s2^2 (2 kg down, 1 kg up).
  2. acm=2(a)−1(a)3=a3=109a_{cm} = \frac{2(a) - 1(a)}{3} = \frac{a}{3} = \frac{10}{9} m/s2^2 downward.
  3. Starting from rest: s=12acmt2=12⋅109⋅4=209s = \frac{1}{2}a_{cm}t^2 = \frac{1}{2}\cdot\frac{10}{9}\cdot 4 = \frac{20}{9}
  4. So the distance is 2.222.22 m, answer (c).
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