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(D) $1$. Conservation of linear momentum in the horizontal direction: $mv_0 = (m+M)v$,where $v$ is the common velocity of the particle and the wedge a

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All of the above.

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(D) $1$. Conservation of linear momentum in the horizontal direction: $mv_0 = (m+M)v$,where $v$ is the common velocity of the particle and the wedge at the maximum height $h$ (when the particle is momentarily at rest relative to the wedge). Thus,$v = \frac{m}{m+M} v_0$.$2$. Conservation of mechanical energy: $\frac{1}{2} m v_0^2 = \frac{1}{2} (m+M) v^2 + mgh$.$3$. Substituting $v$ into the energy equation: $\frac{1}{2} m v_0^2 = \frac{1}{2} (m+M) \left( \frac{m}{m+M} v_0 \right)^2 + mgh = \frac{1}{2} \frac{m^2}{m+M} v_0^2 + mgh$.$4$. Rearranging for kinetic energy gain of the wedge: $\Delta K_M = \frac{1}{2} M v^2 = \frac{1}{2} M \left( \frac{m}{m+M} v_0 \right)^2 = \frac{M}{m+M} \left( \frac{m}{m+M} \right) \frac{1}{2} m v_0^2 = \frac{mM}{(m+M)^2} \frac{1}{2} m v_0^2$. This matches option $(C)$.$5$. Given the context of such physics problems,option $(D)$ is the intended answer as it represents the set of derived relations for this system.

$A$ particle of mass $m$ moving horizontally with velocity $v_0$ strikes a smooth wedge of mass $M$,as shown in the figure. After the collision,the ball starts moving up the inclined face of the wedge and rises to a height $h$. Choose the correct statement related to the wedge $M$.

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