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 .

Answer

C

Explanation

( c)
Force on mass m at B due to mass M at C is
`F_(1)=(Gmxx2M)/((AB)^(2))` along BA
Force on mass m at B due to mass M at C is
`F_(2)=(GmM)/((BC)^(2))` along BC
`:.` Resultant force on mass ma at B due to masses at A and C is
`F_(R)=F_(1)-F_(2)" " `( `:.F_(1)` and `F_(2)` are acting in opposite direction )
`=(2GmM)/((AB)^(2))-(GmM)/((BC)^(2))`
`because AB=(1)/(2)BC`
`:. F_(R)=(2GmM)/((1)/(2)BC)^(2))-(GmM)/((BC)^(2))` along BA
`=(8GmM)/((BC)^(2))-(GmM)/((BC)^(2))` along BA
`=(7GmM)/((BC)^(2))` along BA
Therefore, m will towards 2M.