Question

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

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

Answer 1

The correct option is C $7.5R$

From Newton's 3rd Law, the force of attraction will be the same on both bodies, say, F.
Thus, the acceleration of the smaller body will be $a=\frac{F}{M}$ and $A=\frac{F}{5M}$of the bigger, respectively.
Until the collision, the smaller body travels $s=\frac{1}{2}a{t}^2$ and the bigger one $S=\frac{1}{2}A{t}^2$
The criterion of the collision is
$s+S+R+2R=12R$
By taking the ratio of $S$ and $s$, we get,
$\frac{S}{s}=\frac{\frac{F}{5M}}{\frac{F}{M}}=\frac{1}{5}$
$S=\frac{s}{5}$
By substituting $S=\frac{s}{5}$ into $s+S+R+2R=12R$, we get,
$s+\frac{s}{5}+R+2R=12R$
$\Rightarrow \frac{6}{5}s=(12-1-2)R=9R$
$\Rightarrow s=9R\times \frac{5}{6}=\frac{15}{2}R=7.5R$