Question

Two particles each of mass \(m\) are separated by a distance \(d\). If the mass of one of the particles is doubled without changing the distance between the two particles, then the shift in the position of the center of mass is

• A. \( \frac{d}{6} \)
• B. \( \frac{d}{2} \)
• C. \( \frac{d}{4} \)
• D. \( \frac{d}{5} \)

Hint:

In problems involving the center of mass, remember that the position of the center of mass shifts towards the heavier object when one mass is increased, while the distance between them remains constant.

Answer 1

The Correct Option is A

The formula for the position of the center of mass of two particles is given by: \[ x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \] Let the positions of the two particles be \(x_1 = 0\) and \(x_2 = d\). Initially, both particles have the same mass \(m\), so the center of mass is located at: \[ x_{\text{cm}} = \frac{m(0) + m(d)}{m + m} = \frac{md}{2m} = \frac{d}{2} \] Now, if the mass of one particle is doubled, then the new mass is \(2m\), and the new center of mass is: \[ x_{\text{cm}} = \frac{m(0) + 2m(d)}{m + 2m} = \frac{2md}{3m} = \frac{2d}{3} \] The shift in the position of the center of mass is: \[ \text{Shift} = \frac{2d}{3} - \frac{d}{2} = \frac{4d}{6} - \frac{3d}{6} = \frac{d}{6} \] Thus, the shift in the position of the center of mass is \( \frac{d}{6} \).