TS EAMCET · Physics · Center of Mass Momentum and Collision
A system consists of two particles of masses \(m_1\) and \(m_2\). If the particle of mass \(m_1\) is moved towards the centre of mass through a distance \(d\), then the distance the second particle should be moved, so as to keep the centre of mass at the same position is
- A \(-\frac{m_2}{m_1} d\)
- B \(\frac{m_2}{m_1+m_2} d\)
- C \(-\frac{m_1}{m_2} d\)
- D \(\frac{m_1}{m_2} d\)
Answer & Solution
Correct Answer
(D) \(\frac{m_1}{m_2} d\)
Step-by-step Solution
Detailed explanation
Let \(x\) and \(y\) be the distance of \(m_1\) and \(m_2\) from the centre of mass. then \(\mathrm{m}_1 \mathrm{x}=\mathrm{m}_2 \mathrm{y}\) Given \(m_1\) is moves by a distance \(d\). Let \(\mathrm{m}_2\) is moves by a distance \(\mathrm{s}\).…
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