Question

Two masses $m$ and $2m$ are at rest initially. Find the velocity of mass $2m$, when their separation reduces to $\frac{d}{2}$ due to gravitational force of attraction.

• A. $\sqrt{\frac{Gm}{2d}}$
• B. $\sqrt{\frac{Gm}{d}}$
• C. $\sqrt{\frac{2Gm}{3d}}$
• D. $\sqrt{\frac{Gm}{3d}}$

Answer 1

The correct option is C $\sqrt{\frac{2Gm}{3d}}$


Let $v_1$ and $v_2$ are the velocities of $m$ & $2m$ respectively when the separation between both decreases to $\frac{d}{2}$.


Since, there is no external force on the given system, therefore momentum will be conserved

$p_i=p_f$

$\Rightarrow 0=m{v}_1-\left(2m\right)v_2$

$\Rightarrow v_1=2v_2..........\left(1\right)$


Initial potential energy of the system

$U_i=\frac{-Gm\left(2m\right)}{d}=-\frac{2G{m}^2}{d}$

Final potential energy of the system

$U_f=\frac{-Gm\left(2m\right)}{\frac{d}{2}}=-\frac{4G{m}^2}{d}$

Initial and final kinetic energy of the system will be

$K_i=0;$

$K_f=\frac{1}{2}m{v}_1^2+\frac{1}{2}\left(2m\right)v_2^2$

From eq. $\left(1\right)$,

$K_f=\frac{1}{2}m(2v_2{)}^2+\frac{1}{2}(2m)v_2^2=3m{v}_2^2$

Energy of the system is also conserved because only gravitation force is present, which is conservative force.

$E_i=E_f$

$U_i+K_i=U_f+K_f$

$\Rightarrow -\frac{2G{m}^2}{d}+0=-\frac{4G{m}^2}{d}+3m{v}_2^2$

$\Rightarrow \frac{2G{m}^2}{d}=3m{v}_2^2$

$\Rightarrow v_2=\sqrt{\frac{2Gm}{3d}}$

Hence, option $\left(c\right)$ is correct.