A hemispherical wedge of mass $M$ is placed on horizontal floor. A man of mass m starts moving on the wedge from position $A$ with constant speed $v$ relative to wedge. Initially, both man and wedge were at rest. Neglect friction between wedge and horizontal floor and man doesn't slip on wedge. While man moves from position $A$ to $C$ relative to wedge. Choose the correct statement.

$\Rightarrow$As man moves from $A$ to $B$ , $\theta \downarrow$ , $v\cos \theta \uparrow$, So, velocity of wedge also $\uparrow$ . This       is because $F_{ext}$ in horizontal direction is $0$ . So, $V_{cm}=$ constant $=0$ (as it was         $0$ initially).

      So $v_{wedge}$ increases upto $B$ then decreases.

$\Rightarrow$(Man $+$ wedge) system

      $W_{friction}+W_{normal}+W_{gravity}=∆K$

      In horizontal direction ( $A\to C$ ), $W_{gravity}=0$

      Friction is static, ( $W_{friction}$ ) $system=0$

      $∆K=k_f-k_i=0-0=0$ ( As velocity of COM of the system is always zero )

$\therefore$  ${\left(W_{normal}\right)}_{A\to C}=0$

       $$\begin{gathered} \overrightarrow{S_{COM}}=0 \\ m\left(2R-x\right)-Mx=0 \\ x=\frac{2mR}{m+M} \end{gathered}$$

      During the motion of the man from A to B ,horizontal component of frictional force (leftwards)acting on the wedge gradually increases and horizontal component of normal reaction (rightwards) on the wedge gradually decreases. As a result the wedge accelerates during the motion of the man from A to B. Hence velocity of the wedge is maximum at point B .