Question

Two spherical bodies of masses $M$ and $9M$ and radii $R$ and $3R$ respectively are released in free space with initial separation between their centres equal $9R$. If they attract each other due to gravitational force only, then distance covered by the smaller body just before collision is $NR$. The value of $N$ is_______.

Answer 1

Answer: 4.50

Now, $M{x}_1=9M{x}_2$
$\Rightarrow x_2=\frac{x_1}{9}$
As $x_1+\frac{x_1}{9}=5R$
$\Rightarrow 10x_1=45R$
$\Rightarrow x_1=4.5R$
Alternate method
Since no external force is acting on the system of two masses $M$ and $9M$.
The centre of mass of the system will remain unaltered. Considering mass $M$ at the origin,
Centre of mass $X=\frac{(0\times M+9M\times 9R)}{(M+9M)}$
$\therefore X=\frac{81R}{10}$
$\therefore X=8.1R$
$\therefore$ The $CM$ is at a distance of $8.1R$ which remains unchanged before and after collision. At collision the centre of spherical masses will be $4R$ apart from each then.
Let collision take place at a distance $x$ from origin,
$\therefore \frac{Mx+9M(x+4R)}{10M}=8.1R$
$\therefore x+9(x+4R)=81R$
$\therefore x+9x=81R-36R$
$\therefore 10x=45R$
$\therefore x=4.5R$