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particles of masses 2M m and M are resec...

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

Text Solution

Verified by Experts

The correct Answer is:
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)
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