Two spherical bodies of the mass M and 5M and radii R and 2R respectively are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is

Let the collision takes place at a point O. Suppose the smaller sphere moves to a distance x. Now the bigger mass moves to a distance (9R - x). Therefore
$\mathrm{F}=\mathrm{G}\frac{\mathrm{M}\times 5\mathrm{M}}{(12\mathrm{R}-\mathrm{x}{)}^2}$
Now ${\mathrm{a}}_{\text{small }}=\frac{\mathrm{F}}{\mathrm{M}}=\mathrm{G}\times \frac{5\mathrm{M}}{(12\mathrm{R}-\mathrm{x}{)}^2}$
${\mathrm{a}}_{\mathrm{big}}=\frac{\mathrm{F}}{5\mathrm{M}}=\mathrm{G}\times \frac{\mathrm{M}}{(12\mathrm{R}-\mathrm{x}{)}^2}$
$\therefore \mathrm{x}=\frac{1}{2}\cdot {\mathrm{a}}_{\text{small }}\cdot {\mathrm{t}}^2=\frac{1}{2}\cdot \mathrm{G}\cdot \frac{\mathrm{M}}{(12\mathrm{R}-\mathrm{x}{)}^2}\cdot {\mathrm{t}}^2$………….(1)
and $(9\mathrm{R}-\mathrm{x})=\frac{1}{2}\cdot {\mathrm{a}}_{\text{big }}\cdot {\mathrm{t}}^2=\frac{1}{2}\cdot \mathrm{G}\cdot \frac{\mathrm{M}}{(12\mathrm{R}-\mathrm{x}{)}^2}\cdot {\mathrm{t}}^2$…….(2)
Dividing eq. (1) by eq. (2), we get
$\frac{\mathrm{X}}{(9\mathrm{R}-\mathrm{x})}=5\text{ or }\mathrm{X}=45\mathrm{R}-5\mathrm{X}$
Solving we get, $\mathrm{x}=7\cdot 5\mathrm{R}$