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 $2M$.