Question

Please help me out in that question. Step by step explanation required​

Answer 1

By applying law of conservation of momentum,

$\sf{{m_1}{v_2}-{m_2}{v_2}=0\implies{m_1}{v_1}={m_2}{v_2}....(i)}$

Where v₁ and v₂ are the velocities of masses m₁ and m₂ at a distance r from each other.

By conservation of energy,

Change in P.E = Change in K.E.

$\sf{\dfrac{Gm_1m_2}{r}=\dfrac{1}{2}m_1 {v_1}^{2}}+\dfrac{1}{2}+m_2{v_2}^{2}...(ii)$

Solving equation (i) and (ii) we get,

$\sf{v_1= \sqrt{ \dfrac{2{GM_2}^{2} }{r(m_1 + m_2)}}} \: and \: v_2 = \sqrt{ \dfrac{2{GM_1}^{2}}{r(m_1 + m_2)}}$

Relative velocity of approach,$\rm{\:v_R}$

$\sf{= |v_1| + |v_2| = \sqrt{ \dfrac{2G}{r}(m_1 + m_2)} }$