Question

Particles of masses 2M, m and M are respectively 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.
At subsequent times before any collision takes place [NCERT Exemplar]

Moderate

Solution

Force on B due to $A = F_{B A} = \frac{G (2 M m)}{(A B)^{2}} towards B A$
Force on B due to $C = F_{B C} = \frac{G M m}{(B C)^{2}} towards B C$
As, $(B C) = 2 A B$
$\Rightarrow F_{B C} = \frac{G M m}{(2 A B)^{2}} = \frac{G M m}{4 (A B)^{2}} < F_{B A}$
Hence, m will move towards 2M.

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