Question

The maximum separation between their centers after their first collision
A. $4R$
B. $6R$
C. $8R$
D. $12R$

Answer 1

we will use the formula of gravitational potential energy and the kinetic energy between the two particles. We will equate both the formulas of gravitational potential energy and the kinetic energy between the two particles. Also, we will use the law of conservation of linear momentum in the solution.

Formula used:
The formula of gravitational potential energy between the two particles is given below
$F = \dfrac{{G{m_1}{m_2}}}{d}$
Here, $F$ is the gravitational force, $G$ is the gravitational constant, ${m_1}$ and ${m_2}$ are the masses of the two bodies and $d$ is the radius between these two bodies.
Also, the formula of the kinetic energies of the two particles is given below
$K.E. = \dfrac{1}{2}{m_1}v_1^2 + \dfrac{1}{2}{m_2}v_2^2$
Here, ${m_1}$ and ${m_2}$ are the masses of the two bodies and ${v_1}$ and ${v_2}$ are the velocities attained by the two bodies.

Complete step by step answer:
Here, we will equate both the gravitational potential energy and the kinetic energy as shown below,
$\dfrac{{G{m_1}{m_2}}}{d} = \dfrac{1}{2}{m_1}v_1^2 + \dfrac{1}{2}{m_2}v_2^2$
Now, according to law of conservation of linear momentum, the momentum of both the masses are equal as given below
${m_1}{v_1} = {m_2}{v_2}$
$\Rightarrow {v_2} = \dfrac{{{m_1}{v_1}}}{{{m_2}}}$
Now, putting this value in the above equation, we get
$\dfrac{{G{m_1}{m_2}}}{d} = \dfrac{1}{2}{m_1}v_1^2 + \dfrac{1}{2}{m_2}{\left( {\dfrac{{{m_1}{v_1}}}{{{m_2}}}} \right)^2}$
$\Rightarrow \,\dfrac{{G{m_1}{m_2}}}{d} = \dfrac{1}{2}{m_1}v_1^2 + \dfrac{1}{2}\dfrac{{m_1^2v_1^2}}{{{m_2}}}$
$\Rightarrow \,\dfrac{{G{m_1}{m_2}}}{d} = \dfrac{{{m_1}{m_2}v_1^2 + {m_1}^2v_1^2}}{{2{m_2}}}$
$\Rightarrow \,\dfrac{{G{m_1}{m_2}}}{d} = \dfrac{{{m_1}v_1^2\left( {{m_2} + {m_1}} \right)}}{{2{m_2}}}$
$\Rightarrow \,\dfrac{{G{m_2}}}{d} = \dfrac{{v_1^2\left( {{m_2} + {m_1}} \right)}}{{2{m_2}}}$
$\Rightarrow \,v_1^2 = \dfrac{{2Gm_2^2}}{{d\left( {{m_1} + {m_2}} \right)}}$
$\Rightarrow \,{v_1} = \sqrt {\dfrac{{2Gm_2^2}}{{d\left( {{m_1} + {m_2}} \right)}}}$
$\Rightarrow \,{v_1} = {m_2}\sqrt {\dfrac{{2G}}{{d\left( {{m_1} + {m_2}} \right)}}}$
Similarly, we can calculate the value of ${v_2}$ and is given below
$\Rightarrow \,{v_2} = \sqrt {\dfrac{{2Gm_1^2}}{{d\left( {{m_1} + {m_2}} \right)}}}$
$\Rightarrow \,{v_2} = {m_1}\sqrt {\dfrac{{2G}}{{d\left( {{m_1} + {m_2}} \right)}}}$
Therefore, the relative velocity of the particles is given below
${v_1} - \left( {{v_2}} \right) = {v_1} + {v_2} = {m_2}\sqrt {\dfrac{{2G}}{{d\left( {{m_1} + {m_2}} \right)}}} + {m_1}\sqrt {\dfrac{{2G}}{{d\left( {{m_1} + {m_2}} \right)}}}$
$\Rightarrow \,{v_1} + {v_2} = \left( {{m_1} + {m_2}} \right)\sqrt {\dfrac{{2G}}{{d\left( {{m_1} + {m_2}} \right)}}}$
$\Rightarrow \,{v_1} + {v_2} = \sqrt {\dfrac{{2G{{\left( {{m_1} + {m_2}} \right)}^2}}}{{d\left( {{m_1} + {m_2}} \right)}}}$
$\Rightarrow \,{v_1} + {v_2} = \sqrt {\dfrac{{2G\left( {{m_1} + {m_2}} \right)}}{d}}$
This will be the relative velocity before the collision of the particles. Now, the relative velocity after the collision of the particles will be $\dfrac{{{v_1} + {v_2}}}{2}$. This is because the coefficient of restitution after collision is $\dfrac{1}{2}$. Now, we can say that the relative velocity after the collision is half the relative velocity before collision. Therefore, the kinetic energy will reduce by the factor of $4$.

Now, if we equate this kinetic energy with gravitational potential energy, we will get that this potential energy will decrease by $4$. This means that the distance between the particles will increase by a factor of $4$. Now, we know that the separation of the particles before collision is $\left( {R + 2R} \right) = 3R$. Also, the separation between the particles after collision increases by $4$. Therefore, the final separation between the particles will be $\left( {3R \times 4} \right) = 12R$. Therefore, the maximum separation between their centers after their first collision is $12R$.

Hence, option D is the correct option.

Note: Here, the relative velocities of both the particles before the collisions is opposite in direction. That is why, we have taken the velocity ${v_2}$ as negative. Here, we have considered that the coefficient of restitution after collision is $\dfrac{1}{2}$, that is why the separation of the particles is $\left( {R + 2R} \right) = 3R$.