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JEE Main 2022 (27 Jun, Shift 1) · Centre of Mass of Discrete Particles

Centre of Mass of Discrete Particles question from JEE Main 2022 (27 Jun, Shift 1)

Q12JEE Main 2022 · 27 JUN · SHIFT 1Easy

Centre of Mass of Discrete Particles

Two blocks of masses 10 kg and 30 kg are placed on the same straight line with coordinates (0, 0) cm and (x, 0) cm respectively. The block of 10 kg is moved on the same line through a distance of 6 cm towards the other block. The distance through which the block of 30 kg must be moved to keep the position of centre of mass of the system unchanged is:

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Answer: (c)

  1. CM unchanged: m1d1=m2d2m_1 d_1 = m_2 d_2, with both blocks moving towards the CM.
  2. 10×6=30×d2⇒d2=210 \times 6 = 30 \times d_2 \Rightarrow d_2 = 2 cm
  3. Moving towards the CM means the 30 kg block moves towards the 10 kg block.
  4. So the answer is (c) 2 cm towards the 10 kg block.
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