1. Interpret the figure

   This is the standard setup of a block of mass $m$ sliding on a smooth wedge of mass $M$, with the wedge free to move on a smooth horizontal surface.

   From the answer choices, the incline angle is the usual $45^\circ$ case (as implied by the figure in the original problem).

2. Required quantity

   We need the acceleration of the block with respect to the wedge along the incline.

3. Use the standard result

   For a smooth wedge of angle $\theta$ free to move horizontally, the acceleration of the block relative to the wedge is

   $a_{\text{rel}}=\frac{(M+m)g\sin\theta}{M+m\sin^2\theta}$

   This is along the incline downward.

4. Substitute $\theta=45^\circ$

   Since $\sin 45^\circ=\frac{1}{\sqrt{2}}, \qquad \sin^2 45^\circ=\frac{1}{2}$

   we get

   $a_{\text{rel}}=\frac{(M+m)g\cdot \frac{1}{\sqrt{2}}}{M+\frac{m}{2}}$

5. Substitute $m=8\text{ kg},\; M=16\text{ kg}$

   =\frac{24g}{20\sqrt{2}} =\frac{6g}{5\sqrt{2}}$$ This is not directly among the options, so let us derive the correct expression carefully in the direction along the plane using Newton's laws.

6. Derivation from Newton's laws

   Let the wedge accelerate rightward with acceleration $A$.

   Let the block slide down the incline with acceleration relative to wedge $a$.

   Taking ground frame, block coordinates for a $45^\circ$ incline:

   $x_b = x_w - \frac{s}{\sqrt{2}}, \qquad y_b = \text{const} - \frac{s}{\sqrt{2}}$

   Hence accelerations are

   $a_{bx}=A-\frac{a}{\sqrt{2}}, \qquad a_{by}=-\frac{a}{\sqrt{2}}$

7. Forces on the block

   Forces are:

   • weight $mg$ downward
   • normal reaction $N$ perpendicular to plane

   For a $45^\circ$ incline, components of $N$ are

   $N_x=\frac{N}{\sqrt{2}}, \qquad N_y=\frac{N}{\sqrt{2}}$

   with $N_x$ to the right and $N_y$ upward.

   Applying Newton's second law:

   In $x$-direction: $\frac{N}{\sqrt{2}} = m\left(A-\frac{a}{\sqrt{2}}\right) \quad ...(1)$

   In $y$-direction: $\frac{N}{\sqrt{2}} - mg = -m\frac{a}{\sqrt{2}} \quad ...(2)$

8. Equation for the wedge

   Horizontal force on wedge is due to reaction from block:

   $M A = \frac{N}{\sqrt{2}} \quad ...(3)$

9. Solve equations

   From (1) and (3):

   $M A = m\left(A-\frac{a}{\sqrt{2}}\right)$

   $MA = mA - \frac{ma}{\sqrt{2}}$

   $\frac{ma}{\sqrt{2}} = (m-M)A$

   This sign issue indicates we must be careful with wedge direction. Since wedge moves backward (left), let its acceleration magnitude be $A$ to the left. Then block acceleration components become

   $a_{bx}=-A-\frac{a}{\sqrt{2}}, \qquad a_{by}=-\frac{a}{\sqrt{2}}$

   and the block's $x$-equation becomes

   $\frac{N}{\sqrt{2}} = m\left(-A-\frac{a}{\sqrt{2}}\right)$

   if right is positive this is inconsistent with force to right, so instead take left positive. Then a simpler standard approach is preferable.

10. Use constrained-coordinate method

Let $X$ be wedge displacement horizontally, and $s$ be block displacement down the incline relative to wedge.

Then for $\theta=45^\circ$, block velocity components are

$\dot x = \dot X - \frac{\dot s}{\sqrt{2}}, \qquad \dot y = -\frac{\dot s}{\sqrt{2}}$

Kinetic energy:

$T=\frac12 M\dot X^2 + \frac12 m\left[\left(\dot X-\frac{\dot s}{\sqrt{2}}\right)^2+\left(\frac{\dot s}{\sqrt{2}}\right)^2\right]$

$T=\frac12(M+m)\dot X^2 - \frac{m}{\sqrt{2}}\dot X\dot s + \frac12 m\dot s^2$

Potential energy:

$V = -\frac{mg}{\sqrt{2}}s$

Using Lagrange equation for $X$:

$\frac{d}{dt}\left((M+m)\dot X - \frac{m}{\sqrt{2}}\dot s\right)=0$

Since initially at rest,

$(M+m)\dot X - \frac{m}{\sqrt{2}}\dot s = 0$

Differentiating:

$(M+m)\ddot X - \frac{m}{\sqrt{2}}a = 0$

$\ddot X = \frac{m}{\sqrt{2}(M+m)}a \quad ...(4)$

For $s$:

$m a - \frac{m}{\sqrt{2}}\ddot X = \frac{mg}{\sqrt{2}}$

$a - \frac{1}{\sqrt{2}}\ddot X = \frac{g}{\sqrt{2}}$

Substitute (4):

$a - \frac{1}{\sqrt{2}}\cdot \frac{m}{\sqrt{2}(M+m)}a = \frac{g}{\sqrt{2}}$

$a\left(1-\frac{m}{2(M+m)}\right)=\frac{g}{\sqrt{2}}$

$a\left(\frac{2M+m}{2(M+m)}\right)=\frac{g}{\sqrt{2}}$

$a=\frac{2(M+m)}{2M+m}\cdot \frac{g}{\sqrt{2}}$

11. Substitute $M=16, m=8$