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

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$\left( {\frac{M}{{m + M}}} \right)\frac{{v_0^2}}{{2g}}$

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(A) $1$. Since the wedge is smooth and there are no external horizontal forces acting on the system (particle + wedge),the horizontal momentum of the system is conserved.$2$. Let $v$ be the common horizontal velocity of the particle and the wedge when the particle reaches the maximum height $h$ relative to the wedge.$3$. By conservation of linear momentum in the horizontal direction: $m v_0 = (m + M) v$,which gives $v = \frac{m v_0}{m + M}$.$4$. By conservation of mechanical energy: The initial kinetic energy of the particle equals the sum of the final kinetic energy of the system and the potential energy of the particle at height $h$.$5$. $\frac{1}{2} m v_0^2 = \frac{1}{2} (m + M) v^2 + mgh$.$6$. Substituting $v = \frac{m v_0}{m + M}$ into the energy equation: $\frac{1}{2} m v_0^2 = \frac{1}{2} (m + M) \left( \frac{m v_0}{m + M} \right)^2 + mgh$.$7$. $\frac{1}{2} m v_0^2 = \frac{1}{2} \frac{m^2 v_0^2}{m + M} + mgh$.$8$. $mgh = \frac{1}{2} m v_0^2 \left( 1 - \frac{m}{m + M} \right) = \frac{1}{2} m v_0^2 \left( \frac{M}{m + M} \right)$.$9$. Solving for $h$,we get $h = \left( \frac{M}{m + M} \right) \frac{v_0^2}{2g}$.

$A$ 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 particle starts moving up the inclined face of the wedge and rises to a height $h$. The maximum height $h$ attained by the particle is:

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