Question

Two lead balls of masses $m$ and $5m$ having radii $R$ and $2R$ are separated by $12R$. If they attract each other by gravitational force, the distance covered by small sphere before they touch each other is:

• A. $10R$
• B. $7.5R$
• C. $9R$
• D. $2.5R$

Answer 1

The correct option is B $7.5R$

Effective distance $=9R$
Distance travelled by smaller mass$=x$
$x=(\frac{m_2}{m_1+m_2}\left)\right(9R)$

The gravitational force between masses $m$ and $5m$ is given by:
$F_G=\frac{G\left(m\right)\left(5m\right)}{(12R-x{)}^2}=\frac{5G{m}^2}{(12R-x{)}^2}$

The acceleration of mass $m$ is given by:
$a_{\left(m\right)}=\frac{F_G}{m}=\frac{5Gm}{(12R-x{)}^2}$

The acceleration of mass $5m$ is given by:
$a_{\left(5m\right)}=\frac{F_G}{5m}=\frac{Gm}{(12R-x{)}^2}$

Distance travelled by mass $m$ is given by:
$x=\frac{1}{2}{a}_{\left(m\right)}{t}^2$
$x=\frac{1}{2}\frac{5Gm}{(12R-x{)}^2}{t}^2$ ................................(1)

The distance covered by mass $5m$ is given by:
$9R-x=\frac{1}{2}{a}_{\left(5m\right)}{t}^2$
$9R-x=\frac{1}{2}\frac{Gm}{(12R-x{)}^2}{t}^2$ .................... (2)

Dividing 1 by 2:
$\frac{x}{9R-x}=\frac{\frac{1}{2}\frac{5Gm}{(12R-x{)}^2}{t}^2}{\frac{1}{2}\frac{Gm}{(12R-x{)}^2}{t}^2}=5$
$\therefore x=45R-5x$
$6x=45R$
$x=7.5R$