Consider a two particle system with particles having masses m_1 and m_2 . If the first particle is pushed towards the center of mass through a distance d, by what distance should the second particle is moved, so as to keep the center of mass at the same position?

Center of Mass PYQ — JEE Main 2006 · Shift 0 · Q130 with Solution

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Correct answer: C

For a two-particle system, if the center of mass is to stay at the same position, the net change in the numerator of

x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}

must be zero.

So, for small displacements $\Delta x_1$ and $\Delta x_2$ ,

m_1 \Delta x_1 + m_2 \Delta x_2 = 0.

The first particle is moved towards the center of mass by a distance $d$ . So its displacement is of magnitude $d$ .

Let us take this displacement as positive toward the center of mass:

\Delta x_1 = d.

Then the second particle must move in the opposite direction so that the center of mass remains fixed.

Using

m_1 \Delta x_1 + m_2 \Delta x_2 = 0,

we get

m_1 d + m_2 \Delta x_2 = 0.

Thus,

\Delta x_2 = -\frac{m_1}{m_2} d.

The negative sign shows that the second particle must move in the opposite direction to the first particle. Hence the required distance is

\left|\Delta x_2\right| = \frac{m_1}{m_2} d.

This corresponds to:

\boxed{\frac{m_1}{m_2}d}

which is Option C.