Question

Two spherical bodies of masses $M$ and $9\, M$ and radii $R$ and $3\, R$ respectively are released in free space with initial separation between their centres equal $9\, R$. 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

Now, $Mx _{1}=9\, M x _{2}$
$\Rightarrow x_{2}=\frac{x_{1}}{9}$
As $x_{1}+\frac{x_{1}}{9}=5\, R$
$\Rightarrow 10 x_{1}=45\, R$
$\Rightarrow x_{1}=4.5\, R$
Alternate method
Since no external force is acting on the system of two masses $M$ and $9\, M$.
The centre of mass of the system will remain unaltered. Considering mass $M$ at the origin,
Centre of mass $X =\frac{(0 \times M +9 M \times 9 R )}{( M +9 M )}$
$\therefore X =\frac{81 R }{10}$
$\therefore X =8.1\, R$
$\therefore $ The $CM$ is at a distance of $8.1\, R$ which remains unchanged before and after collision. At collision the centre of spherical masses will be $4\, R$ apart from each then.
Let collision take place at a distance $x$ from origin,
$\therefore \frac{ Mx +9 M ( x +4 R )}{10 M }=8.1\, R$
$\therefore x+9(x+4 R)=81\, R$
$\therefore x+9 x=81 R-36\, R$
$\therefore 10 x=45\, R$
$\therefore x=4.5\, R$