Two bodies with masses $M_{1}$ and $M_{2}$ are initially at the rest and a distance $R$ apart. Then they move directly towards one another under the influence of their mutual gravitational attraction. What is the ratio of the distances traveled by $M_{1}$ to the distance traveled by $M_{2}$ ?

\frac{M_{1}}{M_{2}}

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\frac{M_{2}}{M_{1}}

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1

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\frac{1}{2}

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Solution

The correct option is A $\frac{M_{2}}{M_{1}}$
The forces of attraction between the particles are the internal forces. Therefore, the center of mass of the system will have no acceleration. The particles move, but the center of mass will continue to be at the same place. At the time of collision, the two particles are at one place and the center of mass will also be at that place. As the center of mass does not move, the collision will take place at the center of mass.
Distance of center of mass from $M_{1} = \frac{M_{1} (0) + M_{2} (R)}{M_{1} + M_{2}} = \frac{M_{2} R}{M_{1} + M_{2}}$
Distance of center of mass form $M_{2} = R - \frac{M_{2} R}{M_{1} + M_{2}} = \frac{M_{1} R}{M_{1} + M_{2}}$
Ratio of the distances $= \frac{M_{2}}{M_{1}}$

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Two bodies with masses M1 and M2 are initially at the rest and a distance R apart. Then they move directly towards one another under the influence of their mutual gravitational attraction. What is the ratio of the distances traveled by M1 to the distance traveled by M2?

Two bodies with masses $M_{1}$ and $M_{2}$ are initially at the rest and a distance $R$ apart. Then they move directly towards one another under the influence of their mutual gravitational attraction. What is the ratio of the distances traveled by $M_{1}$ to the distance traveled by $M_{2}$ ?