Analysis of the Physical System

1. System: A point mass m and a block M. The total mass of the system is m + M.
2. Forces:
   • External forces: Gravity on m and M, normal force from the table on M.
   • Internal forces: Normal force between m and M.
3. Conservation Laws:
   • Horizontal Momentum: The horizontal surface is frictionless, and there are no external horizontal forces acting on the system (m + M). Therefore, the total horizontal momentum of the system is conserved.
   • Mechanical Energy: Gravity is a conservative force. The normal forces do no work (the normal force from the table is on a block that moves horizontally, so W=0; the normal force between m and M is always perpendicular to the velocity of m relative to the surface of M). Thus, the total mechanical energy of the system is conserved.
   • Center of Mass: Since the net external horizontal force is zero, the x-coordinate of the center of mass of the system remains at a constant velocity. Since the system starts from rest, the x-coordinate of the center of mass does not move.

Coordinate System

Let's use the coordinate system fixed to the table, with the origin at the initial position of the right edge of the block M. The block is on the x < 0 side. The circular cut is a quarter-circle of radius R.

• Initial position of the right edge of M: $x_{edge}(0) = 0$.
• Initial position of mass m (at the top of the cut): $x_m(0) = -R$, $y_m(0) = R$.
• Initial velocities: $v_m(0) = 0$, $v_M(0) = 0$.

When the mass m reaches the bottom of the cut, it loses contact. Let's analyze the state at this instant.

• The block M has moved to the left. Let its velocity be $\vec{V}_M = (-V, 0)$ (where V is its speed).
• The mass m is moving horizontally to the right. Let its velocity be $\vec{v}_m = (v, 0)$ (where v is its speed).

Step 1: Analyze Velocities (Options B and D)

1. Conservation of Horizontal Momentum: The initial horizontal momentum is $P_{x,i} = 0$. The final horizontal momentum is $P_{x,f} = m v + M (-V) = mv - MV$. Equating initial and final momentum: $0 = mv - MV \implies MV = mv \quad (1)$

2. Conservation of Mechanical Energy: The initial energy is purely potential energy (taking the final height of m as h=0): $E_i = PE_i + KE_i = mgR + 0 = mgR$ The final energy is purely kinetic: $E_f = PE_f + KE_f = 0 + \frac{1}{2}mv^2 + \frac{1}{2}MV^2$ Equating initial and final energy: $mgR = \frac{1}{2}mv^2 + \frac{1}{2}MV^2 \quad (2)$

3. Solving for v and V: From equation (1), we get $V = \frac{m}{M}v$. Substitute this into equation (2): $mgR = \frac{1}{2}mv^2 + \frac{1}{2}M\left(\frac{m}{M}v\right)^2$ $mgR = \frac{1}{2}mv^2 + \frac{1}{2}M\frac{m^2}{M^2}v^2 = \frac{1}{2}mv^2 + \frac{1}{2}\frac{m^2}{M}v^2$ $mgR = \frac{1}{2}v^2 \left(m + \frac{m^2}{M}\right) = \frac{1}{2}v^2 m\left(1 + \frac{m}{M}\right)$ $v^2 = \frac{2gR}{1 + \frac{m}{M}}$ $v = \sqrt{\frac{2gR}{1 + \frac{m}{M}}}$ This matches Option B, which is therefore correct.

4. Check Option D: The velocity of the block M has magnitude $V = \frac{m}{M}v = \frac{m}{M} \sqrt{\frac{2gR}{1 + \frac{m}{M}}}$. The velocity vector is directed to the left. Option D states the velocity is $V = - \frac{m}{M}\sqrt{2gR}$. The magnitude in this option is incorrect as it misses the $\sqrt{1 + m/M}$ factor in the denominator. Therefore, Option D is incorrect.

Step 2: Analyze Displacements (Options A and C)

1. Conservation of Center of Mass Position: Since the x-coordinate of the center of mass of the system does not change, its total displacement is zero: $\Delta X_{cm} = 0$. $\Delta X_{cm} = \frac{m \Delta x_m + M \Delta X_M}{m+M} = 0$ This implies $m \Delta x_m + M \Delta X_M = 0$, where $\Delta x_m$ is the displacement of mass m and $\Delta X_M$ is the displacement of the center of mass of block M.

2. Relating Displacements: The mass m moves from x=-R to the bottom of the cut at x=0 relative to the block. So, its horizontal displacement relative to the block is +R. The absolute displacement of m is the displacement of the block plus the displacement of m relative to the block: $\Delta x_m = \Delta X_M + R$

3. Solving for $\Delta X_M$: Substitute the relation for $\Delta x_m$ into the center of mass equation: $m (\Delta X_M + R) + M \Delta X_M = 0$ $m \Delta X_M + mR + M \Delta X_M = 0$ $(m+M) \Delta X_M = -mR$ $\Delta X_M = - \frac{mR}{M+m}$ This is the x-component of the displacement of the center of mass of block M. This matches Option C, which is therefore correct.

4. Solving for the final position of m (x): The final position of mass m is x. The mass m is at the right edge of the block when it loses contact. The initial position of the right edge was x=0. The block (and its right edge) has displaced by $\Delta X_M$. So, the final position of the right edge, and thus of mass m, is: $x = 0 + \Delta X_M = - \frac{mR}{M+m}$ Option A gives the position as $x = - \sqrt{2} \frac{mR}{M+m}$. This is different by a factor of $\sqrt{2}$. Therefore, Option A is incorrect.

Conclusion

• Option A: Incorrect. The final position of m is x = -mR/(M+m).
• Option B: Correct. The velocity of m is $v = \sqrt{2gR / (1 + m/M)}$.
• Option C: Correct. The displacement of the CM of block M is -mR/(M+m).
• Option D: Incorrect. The velocity of M has a different magnitude.

Hence, the correct options are B and C.