Question

In a two- particle system with particle masses m1 and m2, the first particle is pushed towards the centre of mass through a distance d, the distance through which second particle must be moved to keep the centre of mass at the same position is-
(A)$\dfrac{{{m_2}d}}{{{m_1}}}$
(B) $d$
(C) $\dfrac{{{m_1}d}}{{({m_1} + {m_2})}}$
(D) $\dfrac{{({m_1} + {m_2})d}}{{{m_1}}}$

Answer 1

As we know that the centre of mass is the distribution of mass in space.
Sometimes it is known as the balance point where the relative position of the distributed mass is added to zero.

Complete step by step answer:
Consider P1 and P2 are the position of masses m1 and m2.
The position of centre of mass is given by Pcm=$\dfrac{{{m_1}{P_1} + {m_2}{P_2}}}{{{m_1} + {m_2}}}$
Let us consider the position is shifted by little difference, If P1 changes by$\Delta {P_1}$ and P2 changes by $\Delta {P_2}$ then the change in the centre of mass will be-
$\Delta {P_{cm}} = \dfrac{{{m_1}\Delta {P_1} + {m_2}\Delta {P_2}}}{{{m_1} + {m_2}}}$ --- (1)
Now, according to the question- $\Delta {P_{cm}} = 0$ and $\Delta {P_1} = d$.
Now substitute the given values in equation (1), we get-
${m_1}d + {m_2}\Delta {P_2} = 0$
For finding the change in position, we can rearrange the terms,
$\Delta {P_2} = - \dfrac{{{m_1}d}}{{{m_2}}}$
So, the distance moved by m2 is $= - \dfrac{{{m_1}d}}{{{m_2}}}$. We can also write the resulted equation as-
$= \dfrac{{{m_2}d}}{{{m_1}}}$.

So, the correct answer is “Option A”.

Note:
The centre of mass of an object is the point at which the object can be balanced. Mathematically, it is the point at which the torques from the mass elements of an object sum to zero. The centre of mass is useful because problems can often be simplified by treating a collection of masses as one mass at their common centre of mass.