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 .

Options

A. m will remain at rest
B. m will move towards 3M
C. m will move towards 2M
D. m will have oscillatory motion

Answer

C

Explanation

Force on mass at B due to mass 2M at A is
`F_(1)=(GMxx3M)/((AB)^(2)) along BA`
Force on mass at B due to mass M at C is
`F_(2)=(GMxx3M)/((BC)^(2)) along BC`
`:.` Resultant force on mass m 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 directions)
`=(2GmM)/((AB)^(2))-(Gmxx3M)/((BC)^(2))`
`:. AB=1/2BC`
`:' F_(R)=(GmM)/((1/4BC)^(2))-(3GmM)/((BC)^(2))` along BA
`=(32GmM)/((BC)^(2))-(3GmM)/((BC)^(2)) along BA`
`=(29GmM)/((BC)^(2)) along BA`
Therefore, m will move towards 2M