Particles of masses 2M, m and 3M are respectively at points A, B and C with $\mathrm{AB} = \frac{1}{4}\left(\mathrm{BC}\right)$. m is much-much smaller than M and at time t = 0, they are all at rest. At subsequent times before any collision takes place
 

 

Force on mass at B due to mass 2M at A is
${\mathrm{F}}_1 = \frac{\mathrm{GM}\times 2\mathrm{M}}{{\left(\mathrm{AB}\right)}^2}\mathrm{along} \mathrm{BA}$

Force on mass m at B due to mass M at C is
${\mathrm{F}}_2 = \frac{\mathrm{Gm}\times 3\mathrm{M}}{{\left(\mathrm{BC}\right)}^2} \mathrm{along} \mathrm{BC}$

($\therefore$ Resultant force on mass rn at B due to masses at A and C is
${\mathrm{F}}_{\mathrm{R}} = {\mathrm{F}}_1-{\mathrm{F}}_2$

($\because$ ${\mathrm{F}}_1 \mathrm{and} {\mathrm{F}}_2$ are acting in opposite directions)
   = $\frac{2\mathrm{GmM}}{{\left(\mathrm{AB}\right)}^2}-\frac{\mathrm{Gm}\times 3\mathrm{M}}{{\left(\mathrm{BC}\right)}^2}$

$\because \mathrm{AB} = \frac{1}{4}\mathrm{BC}$

$\because$ ${\mathrm{F}}_{\mathrm{R}} = \frac{2\mathrm{GmM}}{{\left({\displaystyle \frac{1}{4}}\mathrm{BC}\right)}^2}-\frac{3\mathrm{GmM}}{{\left(\mathrm{BC}\right)}^2} \mathrm{along} \mathrm{BA}$

         $=\frac{32\mathrm{GmM}}{{\left(\mathrm{BC}\right)}^2}-\frac{3\mathrm{GmM}}{{\left(\mathrm{BC}\right)}^2}\mathrm{along} \mathrm{BA}$

        = $\frac{29\mathrm{GmM}}{{\left(\mathrm{BC}\right)}^2}\mathrm{along} \mathrm{BA}$

Therefore, m will move towards 2M.