Question

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

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

Answer 1

The correct option is A $7.5R$

Assuming velocities of masses M and 5M are $v_1$ and $v_2$ respectively

Using conservation of momentum principle

$M{v}_1+5M{v}_2=0\Rightarrow v_2=-\frac{v_1}{5}$...(i)

Using energy conservation principle

$\Delta P.E.=\Delta K.E.$

The P.E. is given as $\frac{G{m}_1{m}_2}{r}$

$-\frac{G5M^2}{12R}-(-\frac{G5M^2}{3R})=\frac{1}{2}M{v}_1^2+\frac{1}{2}5M{v}_2^2-0$

from equation (i) substituting value of $v_2$

$\frac{15G{M}^2}{12R}=\frac{1}{2}M{v}_1^2+\frac{1}{2}5M(\frac{v_1}{5}{)}^2$

$\frac{G5M^2}{4R}=\frac{1}{2}M{v}_1^2+\frac{1}{2}M\frac{v_1^2}{5}$

$\frac{1}{2}M{v}_1^2=$K.E. of mass M just before collision $=\frac{25G{M}^2}{24R}$

Using work energy theorem:

work done by gravitational force =$\Delta K.E.$

${\int }_{12R}^{r}F.dr=\frac{1}{2}m{v}^1$

${\int }_{12R}^{r^{\prime }}\frac{GM5M}{r^2}dr=\frac{1}{2}m{v}^1$

$[\frac{GM5M}{r}{]}_{12R}^{r^{\prime }}=\frac{1}{2}m{v}^1$

$GM5M(\frac{1}{r}-\frac{1}{12R})=\frac{25G{M}^2}{24R}$

$\Rightarrow r=7.5R$

hence correct answer is option A