A small block of mass m is placed on a wedge | vedclass

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(D) Consider the system of the block $(m)$ and the wedge $(M)$. Since all surfaces are frictionless and there are no external horizontal forces acting

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(D) Consider the system of the block $(m)$ and the wedge $(M)$. Since all surfaces are frictionless and there are no external horizontal forces acting on the system,the horizontal momentum of the system is conserved.Let $v$ be the velocity of the block relative to the ground and $V$ be the velocity of the wedge relative to the ground at any instant.By conservation of horizontal momentum: $mv_x + MV = 0$,where $v_x$ is the horizontal component of the block's velocity.Also,the work-energy theorem states that the loss in potential energy of the block is equal to the gain in kinetic energy of the system (since the spring is not compressed at the start and we look for the velocity of the wedge as the block reaches the bottom,assuming the spring is not yet significantly involved in the energy balance or is negligible for the peak velocity of the wedge).Using the conservation of energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}MV^2$.From momentum conservation,$V = -\frac{mv_x}{M}$.For the wedge to attain maximum velocity,we analyze the motion. The correct expression for the velocity of the wedge $M$ when the block $m$ reaches the bottom is $V = \sqrt{\frac{2m^2gh}{M(M+m)}}$.Comparing this with the given options,none of the expressions match the derived result. Thus,the correct option is $D$.

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