Two masses $m_{1}$ and $m_{2} (m_{1} < m_{2})$ are released from rest from a finite distance. They start to move under their mutual gravitational attraction. Identity the incorrect statement.

Acceleration of mass

m_{1}

is more than that of

m_{2}

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Force experience by the both masses follow Newton's third law.

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Centre of mass of system will remain at rest in all the reference frame

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Total energy of system remain constant.

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Solution

The correct option is C Centre of mass of system will remain at rest in all the reference frame
Since gravitational force is equal and opposite in direction acting on both masses $m_{1}$ & $m_{2}$

Acceleration of masses will be

$a_{1} = \frac{F}{m_{1}}$ and $a_{2} = \frac{F}{m_{2}}$

Thus, $a_{2} < a_{1} (∵ m_{2} > m_{1})$

Here, both masses experiencing gravitational force only, and we know that gravitational force is an example of action- reaction pair. So, force experience by both masses follow Newton's third law.

$\Rightarrow$ Centre of mass will remain at rest only in the frame of reference which is at rest, because initially centre of mass is at rest and $F_{e x t} = 0$ on the system.

In absence of external force (remember mutual gravitational force is an internal force for the system) total energy remains constant.

Hence, option $(c)$ is correct choice.

$\begin{matrix} Why this question? Caution:Velocity of centre of mass is dependent on the frame of reference. \end{matrix}$

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