Two beads A and B of masses m_1 and m_2 respe | vedclass

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(A) When bead $A$ slides down from the top to the bottom to collide with $B$,its potential energy is converted into kinetic energy.Using the law of co

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$1 : \sqrt{2}$

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(A) When bead $A$ slides down from the top to the bottom to collide with $B$,its potential energy is converted into kinetic energy.Using the law of conservation of energy: $\frac{1}{2} m_1 v_1^2 = m_1 g (2R)$$	herefore v_1^2 = 4gR \implies v_1 = \sqrt{4gR} \quad ... (1)$Since bead $A$ comes to rest after the collision,all its momentum is transferred to bead $B$. By the law of conservation of momentum: $m_1 v_1 = m_2 v_2$$	herefore v_2 = \frac{m_1}{m_2} v_1 = \frac{m_1}{m_2} \sqrt{4gR} \quad ... (2)$After the collision,bead $B$ reaches the height of the center $O$,which is a vertical displacement of $R$. Its kinetic energy is converted into potential energy:$\frac{1}{2} m_2 v_2^2 = m_2 g R$$	herefore v_2^2 = 2gR \implies v_2 = \sqrt{2gR}$Substituting $v_2$ into equation $(2)$: $\sqrt{2gR} = \frac{m_1}{m_2} \sqrt{4gR}$$	herefore \frac{m_1}{m_2} = \frac{\sqrt{2gR}}{\sqrt{4gR}} = \frac{1}{\sqrt{2}}$

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