Particles of masses 2M, m and M are respectively at points A, B and C with `sqrt((GM)/(R))(sqrt(2)+1)` is much-much smaller than M and at time `t=0`, they are all at rest as given in figure

At subsequent time before any collision takes place.

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

Fore on B due to A `=F_(BA)=(G(2Mm))/((AB)^(2))` towards BA
Force on B due to C `= F_(BC)=(GMm)/((BC)^(2))` towards BC
As, `(BC)=2AB`
`rArr F_(BC)=(GMm)/((2AB)^(2))=(GMm)/(4(AB)^(2))lt F_(BA)`
Hence, `m` will move towards BA (i.e., 2M)