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MHT CET 2024 (9 May, Shift 1) · Centre of Mass of Discrete Particles

Centre of Mass of Discrete Particles question from MHT CET 2024 (9 May, Shift 1)

Q5MHT CET 2024 · 9 MAY · SHIFT 1Easy

Centre of Mass of Discrete Particles

In the system of two particles of masses ' $m_1$ ' and ' $m_2$ ', the first particle is moved by a distance 'd' towards the centre of mass. To keep the centre of mass unchanged, the second particle will have to be moved by a distance

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Answer: (c)

For the centre of mass to stay fixed, $m_1 \Delta x_1 + m_2 \Delta x_2 = 0$ .
$m_1$ moves $d$ towards the CM, so $m_2$ must also move towards the CM by $d_2$ with $m_1 d = m_2 d_2$ .
$d_2 = \frac{m_1}{m_2} d$
So the answer is (c) $\frac{m_1}{m_2} d$ , towards the centre of mass.

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