particles of masses 2M m and M are resectively at points A , B and C with ` AB = (1)/(2) (BC)` m is much - much smaller than M and at time ` t=0` they are all at rest as given in figure . As subsequent times before any collision takes palce .

Options

A. m will remain at rest
B. m will move towards M.
C. m will move towards 2M.
D. m will have oscillatory motion

Answer

C

Explanation

Gravitational force on mass m at B due to mass 2M at A
`F_(BA) = (Gm2M)/((AB)^(2))`
Similarly, `F_(BC) = (GmM)/((BC)^(2)) = (GmM)/(4(AB)^(2)) [because BC 2AB]`
So `F_(BA) gt F_(BC)`
Hence, m will move towards 2M.
The correct option is (c ).