Question

Two particles $A$ and $B$ of mass $1\text{kg}$ and $2\text{kg}$ respectively are kept $1\text{m}$ apart and are released to move under mutual attraction. When speed of $B$ is $3.6\text{cm/hour}$, what is the separation (in cm) between the particles. Take $G=\frac{20}{3}\times {10}^{-11}\frac{{\text{Nm}}^2}{\text{kg}}$

Answer 1

Given:

When particle started to move, let the separation is $r$ and at that instant speed of particle $A$ is $V_A$.
Since only internal forces are actimg, Linear momentum of the system is conserved.
By the conservation of linear momentum, $P_i=P_f$
$\Rightarrow 0=1\times V_A-2\times 3.6$
$\Rightarrow V_A=7.2\text{cm/hr}=\frac{7.2\times {10}^{-2}}{3600}\text{m/s}$
$\Rightarrow V_A=2\times {10}^{-5}\text{m/s}$
$V_B=3.6\text{cm/hr}=\frac{3.6\times {10}^{-2}}{3600}\text{m/s}={10}^{-5}\text{m/s}$

Now by the conservation of energy-
${\left(PE\right)}_i+{\left(KE\right)}_i={\left(PE\right)}_f+{\left(KE\right)}_f$

Using, $PE=\frac{-G{m}_1{m}_2}{r}$ and $KE=\frac{1}{2}m{v}^2$

$\Rightarrow -\frac{G\times 2\times 1}{1}+0=-\frac{G\times 2\times 1}{r}+\frac{1}{2}\times 1{(2\times {10}^{-5})}^2+\frac{1}{2}\times 2({10}^{-5}{)}^2$

$\Rightarrow -2G=-\frac{2G}{r}+3\times {10}^{-10}$

$\Rightarrow -2\times \frac{20}{3}\times {10}^{-11}=-\frac{2G}{r}+3\times {10}^{-10}$

$\Rightarrow -\frac{13}{3}\times {10}^{-10}=-\frac{2}{r}\times \frac{2\times {10}^{-10}}{3}$

$\Rightarrow r=\frac{4}{13}=0.307\simeq 0.31\text{m}$
$\Rightarrow r=31\text{cm}$