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(B) Let the velocities of these masses at distance $r$ be $v_1$ and $v_2$ respectively. Since the system is initially at rest,by the conservation of l

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$[2G(m_1 + m_2)/r]^{1/2}$

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(B) Let the velocities of these masses at distance $r$ be $v_1$ and $v_2$ respectively. Since the system is initially at rest,by the conservation of linear momentum,$m_1 v_1 = m_2 v_2$,which implies $v_1 = (m_2/m_1)v_2$. By the law of conservation of energy,the loss in potential energy equals the gain in kinetic energy: $G m_1 m_2 / r = (1/2) m_1 v_1^2 + (1/2) m_2 v_2^2$. Substituting $v_1 = (m_2/m_1)v_2$ into the energy equation: $G m_1 m_2 / r = (1/2) m_1 (m_2^2/m_1^2) v_2^2 + (1/2) m_2 v_2^2 = (1/2) (m_2^2/m_1 + m_2) v_2^2 = (1/2) [m_2(m_1 + m_2)/m_1] v_2^2$. Solving for $v_2$,we get $v_2 = \sqrt{2 G m_1^2 / (r(m_1 + m_2))}$. Similarly,$v_1 = \sqrt{2 G m_2^2 / (r(m_1 + m_2))}$. The relative velocity of approach is $v_{	ext{rel}} = v_1 + v_2 = \sqrt{2 G / (r(m_1 + m_2))} (m_2 + m_1) = \sqrt{2 G (m_1 + m_2) / r}$.

Two bodies of masses $m_1$ and $m_2$ are initially at rest at an infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach at a separation distance $r$ between them is

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