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NEET PHYSICSRotational motionMedium

Question

Consider a system of two particles having masses m1m_1 and m2m_2. If the particle of mass m1m_1 is pushed towards the centre of mass of particles through a distance dd, by what distance would the particle of the mass m2m_2 move so as to keep the centre of mass of particles at the original position?

A

m1m1+m2d\frac{m_1}{m_1+m_2}d

B

m1m2d\frac{m_1}{m_2}d

C

dd

D

m2m1d\frac{m_2}{m_1}d

Step-by-Step Solution

The position of the centre of mass for a two-particle system is given by xcm=m1x1+m2x2m1+m2x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}. Since the centre of mass remains at its original position, the shift in the centre of mass is zero, i.e., Δxcm=0\Delta x_{cm} = 0. Therefore, m1Δx1+m2Δx2m1+m2=0\frac{m_1 \Delta x_1 + m_2 \Delta x_2}{m_1 + m_2} = 0, which implies m1Δx1+m2Δx2=0m_1 \Delta x_1 + m_2 \Delta x_2 = 0. Considering only the magnitudes of the displacements (as both must move towards the centre of mass to keep it stationary), we have: m1d=m2dm_1 d = m_2 d' Where dd is the distance moved by m1m_1 and dd' is the distance moved by m2m_2. Thus, the distance moved by the second particle is d=m1m2dd' = \frac{m_1}{m_2}d.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Rotational motion. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRotational motionconsidersystemparticleshavingmasses

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