Question

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 .

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 1

Correct Answer - C
Force 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)) le F_(BA)`
Hence, `m` will move towards BA (i.e., 2M)