A ring of mass m can freely slide along the m | vedclass

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(B) Let the mass of the ring be $m$ and the mass of the block be $M$. The system is released from rest,so the initial total momentum is zero.Since all

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$v < \sqrt {2gH} $

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(B) Let the mass of the ring be $m$ and the mass of the block be $M$. The system is released from rest,so the initial total momentum is zero.Since all surfaces are smooth and there are no external horizontal forces,the horizontal momentum of the system is conserved.As the ring slides down the curved rod,it exerts a horizontal force on the rod,causing the block $M$ to move in the opposite direction.Let $v_m$ be the horizontal velocity of the ring and $V_M$ be the velocity of the block $M$ at the lowermost point.By conservation of horizontal momentum: $m v_m - M V_M = 0$,which implies $v_m = (M/m) V_M$.By conservation of mechanical energy: $mgH = \frac{1}{2} m v_m^2 + \frac{1}{2} M V_M^2$.Substituting $V_M = (m/M) v_m$ into the energy equation: $mgH = \frac{1}{2} m v_m^2 + \frac{1}{2} M (m/M)^2 v_m^2 = \frac{1}{2} m v_m^2 (1 + m/M)$.Solving for $v_m$: $v_m = \sqrt{\frac{2gH}{1 + m/M}}$.Since $(1 + m/M) > 1$,it follows that $v_m < \sqrt{2gH}$.

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