Question

Two spherical bodies of mass $M$ and $5$ $M$ and radius $R$ and $2$ $R$ 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 before collision is

• A. $1.5$ $R$
• B. $2.5$ $R$
• C. $4.5$ $R$
• D. $7.5$ $R$

Answer 1

The correct option is D $7.5$ $R$

As the distance between two sphere is $9R$. Let before collision small sphere moves $x$ distance and the bigger sphere move $(9R-x)$ distance,

Force of attraction between two sphare due to gravity is given by,
$f=\frac{G{M}^{\prime }m}{r^2}$
As given, $M^{\prime }=M$, $m=5M$, $r=12R$
So, $F=\frac{5G{M}^2}{(12R{)}^2}$
Now, acceleration of small sphere will be
$a_{small}=\frac{F}{M}=\frac{5GM}{(12R{)}^2}$
Acceleration of bigger sphere will be
$a_{big}=\frac{F}{5M}=\frac{GM}{(12R{)}^2}$
By using Newton's law of motion
$x=\frac{1}{2}{a}_{small}{t}^2$
$x=\frac{1}{2}\frac{5GM}{(12R{)}^2}{t}^2$
$\begin{array}{}\frac{x}{5}=\frac{1}{2}\frac{GM}{(12R{)}^2}{t}^2& \text{(1)}\end{array}$
$(9R-x)=\frac{1}{2}{a}_{big}{t}^2$
$\begin{array}{}(9R-x)=\frac{1}{2}\frac{GM}{(12R{)}^2}{t}^2& \text{(2)}\end{array}$
From eq. (1) and eq. (2)
$\frac{x}{5}=(9R-x)$
$x=45R-5x$
$x=7.5R$