Question

Two balls, A and B of masses 0.4 kg and 1.2 kg respectively, are placed at a distance of 12 cm from each other. Another ball of mass 0.1 kg is placed in between them such that there is no net force acting on it. Find the distance of the third ball from ball A.

• A. $-6-6\sqrt{3}\text{cm}$
• B. $-6+6\sqrt{3}\text{cm}$
• C. $12-6\sqrt{3}\text{cm}$
• D. $-12+6\sqrt{3}\text{cm}$

Answer 1

The correct option is B $-6+6\sqrt{3}\text{cm}$

Given:
Mass of the ball(A), ${\text{m}}_A=0.4\text{kg}$
Mass of the ball(B), ${\text{m}}_B=1.2\text{kg}$
Mass of the another ball(C), ${\text{m}}_C=0.1\text{kg}$
Let the third mass is placed at a distance $\text{x}\text{cm}$ from ${\text{m}}_A$ such that there is no net force acting on it.

If ${\text{r}}_{AC}$ is the distance between ${\text{m}}_A$ and ${\text{m}}_C$, and ${\text{r}}_{BC}$ is the distance between ${\text{m}}_B$ and ${\text{m}}_C$, then for the net force on ${\text{m}}_C$ to be zero:
${\text{F}}_{AC}={\text{F}}_{BC}$
$\frac{\text{G}{\text{m}}_A{\text{m}}_C}{{\text{r}}_{AC}^2}=\frac{\text{G}{\text{m}}_B{\text{m}}_C}{{\text{r}}_{BC}^2}$
$\frac{{\text{m}}_A}{{\text{r}}_{AC}^2}=\frac{{\text{m}}_B}{{\text{r}}_{BC}^2}$
$\frac{{\text{m}}_A}{{\text{m}}_B}=\frac{{\text{r}}_{AC}^2}{{\text{r}}_{BC}^2}$
$⟹\frac{4}{12}=\frac{{\text{x}}^2}{(12-\text{x}{)}^2}$
$3{\text{x}}^2={\text{x}}^2-24\text{x}+144$
$2{\text{x}}^2+24\text{x}-144=0$
${\text{x}}^2+12\text{x}-72=0$
$⟹\text{x}=-6\pm 6\sqrt{3}$
But, distance can't be negative,
So,
$\text{x}=-6+6\sqrt{3}\text{cm}$