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(C) $1$. Since the surface is smooth and there are no external horizontal forces acting on the system (particle + wedge),the linear momentum of the sy

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When the particle reaches the horizontal surface,its velocity relative to the wedge is $v_0$

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(C) $1$. Since the surface is smooth and there are no external horizontal forces acting on the system (particle + wedge),the linear momentum of the system in the horizontal direction is conserved.$2$. The centre of mass of the system moves with a constant velocity. It is not stationary.$3$. When the particle reaches the maximum height $h$,it has the same horizontal velocity as the wedge. Let this velocity be $V$. By conservation of momentum: $mv_0 = (m+M)V$,so $V = \frac{mv_0}{m+M}$.$4$. When the particle starts moving downward,it will eventually leave the wedge. Due to the conservation of energy and momentum in the horizontal direction,when the particle returns to the horizontal surface,its velocity relative to the wedge will be equal in magnitude to its initial velocity $v_0$ but in the opposite direction.$5$. Thus,the velocity of the particle relative to the wedge is $v_0$ when it returns to the horizontal surface.

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