Question

Particles of masses $2M$, $m$ and $M$ are respectively at points $A$, $B$ and $C$ with $AB = 1/2(BC)$. $m$ is muchmuch smaller than $M$ and at time $t=0$, they are all at rest. At subsequent times before any collision takes place

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

Answer 1

Answer: (C)

Force on mass $m$ at $B$ due to mass $2M$ at $A$ is
$F_{1} = \frac{Gm \times 2M}{\left(AB\right)^{2}}$ along $BA$
Force on mass $m$ at $B$ due to mass $M$ at $C$ is
$F_{2} = \frac{Gm \times 2M}{\left(BC\right)^{2}}$ along $BC$
$\therefore $ Resultant force on mass $m$ at $B$ due to masses at $A$ and $C$ is
$F_{R} =F_{1}- F_{2}$
($\because F_{1}$ and $F_{2}$ are acting in opposite directions)
$= \frac{2GmM}{\left(AB\right)^{2}}-\frac{GmM}{\left(BC\right)^{2}}$
$\because AB = \frac{1}{2}BC$
$\therefore F_{R} = \frac{2GmM}{\left(\frac{1}{2}BC\right)^{2}}-\frac{GmM}{\left(BC\right)^{2}}$ along $BA$
$= \frac{8GmM}{\left(BC\right)^{2}}-\frac{GmM}{\left(BC\right)^{2}}$ along $BA$
$= \frac{7GmM}{\left(BC\right)^{2}}$ along $BA$
Therefore, $m$ will move towards $2M$.