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Q.
Particle $A$ makes a perfectly elastic collision with another particle $B$ at rest. They fly apart in opposite directions with equal speeds. The ratio of their masses $m_A/m_B$ is
According to law of conservation of linear momentum, we get
$m_1 u_1 + m_2 \times 0 = m_1 v_1 + m_2 (-v_1)$
$m_1 u_1 = (m_1 - m_2) v_1$ .....(i)
$\therefore \:\: \frac{u_1}{v_1} = \frac{m_1 - m_2}{m_1}$ ....(ii)
According to law of conservation of kinetic energy, we get
$\frac{1}{2} m_1 u_1^2 = \frac{1}{2} (m_1 + m_2)v_1^2$ ...(iii)
Divide (iii) by (i), we get
$u_{1}=\frac{\left(m_{1}+m_{2}\right)v_{1}}{m_{1}-m_{2}} $ or $\frac{u_{1}}{v_{1}}=\frac{m_{1}+m_{2}}{m_{1}-m_{2}}$ ...(iv)
From (ii) and (iv), we get $\frac{m_{1}-m_{2}}{m_{1}}=\frac{m_{1}+m_{2}}{m_{1}-m_{2}}$
On solving, we get $\frac{m_1}{m_2} = \frac{1}{3}$ or $\frac{m_A}{m_B} = \frac{1}{3}$