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(D) Let the initial velocity of particle $A$ be $u_A$ and particle $B$ be $0$. Let the final velocities be $v_A$ and $v_B$. Given that they move in op

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$3 m_A = m_B$

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(D) Let the initial velocity of particle $A$ be $u_A$ and particle $B$ be $0$. Let the final velocities be $v_A$ and $v_B$. Given that they move in opposite directions with equal speeds,let $v_A = -v$ and $v_B = v$.From the conservation of linear momentum:$m_A u_A = m_A (-v) + m_B v$$m_A u_A = (m_B - m_A) v$  --- $(1)$From the conservation of kinetic energy (perfectly elastic collision):$\frac{1}{2} m_A u_A^2 = \frac{1}{2} m_A (-v)^2 + \frac{1}{2} m_B v^2$$m_A u_A^2 = (m_A + m_B) v^2$  --- $(2)$Dividing equation $(2)$ by equation $(1)$:$\frac{m_A u_A^2}{m_A u_A} = \frac{(m_A + m_B) v^2}{(m_B - m_A) v}$$u_A = \frac{(m_A + m_B) v}{(m_B - m_A)}$Substitute $u_A$ back into equation $(1)$:$m_A \left[ \frac{(m_A + m_B) v}{(m_B - m_A)} \right] = (m_B - m_A) v$$m_A (m_A + m_B) = (m_B - m_A)^2$$m_A^2 + m_A m_B = m_B^2 - 2 m_A m_B + m_A^2$$3 m_A m_B = m_B^2$$3 m_A = m_B$

Particle $A$ makes a perfectly elastic collision with another particle $B$ at rest. They fly apart in opposite directions with equal speeds. If their masses are $m_A$ and $m_B$ respectively,then:

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