Since the blocks are in contact on a smooth surface, they move together with a common acceleration, $a$. We define the direction to the right as positive.

First, we determine the acceleration of the system by considering both blocks as a single object with total mass $(M+m)$. Applying Newton's second law to the system:

$$F - f = (M+m)a$$

The acceleration of the system is:

$$a = \frac{F-f}{M+m}$$

[Q1] Configuration (a)

To find the contact force $F_{c1}$, we analyze the forces on one of the blocks. Let's consider the free-body diagram for block $M$. The forces acting on it are the applied force $F$ to the right and the contact force from block $m$, $F_{c1}$, to the left. Applying Newton's second law to block $M$:

$$F - F_{c1} = Ma$$

Solving for the contact force $F_{c1}$:

$$F_{c1} = F - Ma$$

Substituting the expression for acceleration $a$:

$$F_{c1} = F - M\left(\frac{F-f}{M+m}\right) = \frac{F(M+m) - M(F-f)}{M+m}$$ $$F_{c1} = \frac{FM + Fm - MF + Mf}{M+m} = \frac{Fm + Mf}{M+m}$$

Now, we substitute the given values: $M = 3$ kg, $m = 2$ kg, $F = 15$ N, and $f = 10$ N.

$$F_{c1} = \frac{(15)(2) + (3)(10)}{3+2} = \frac{30 + 30}{5} = \frac{60}{5} = 12 \text{ N}$$

[Q2] Configuration (b)

The positions of the blocks are swapped. The acceleration of the system remains the same, as the total mass and net external force are unchanged.

$$a = \frac{F-f}{M+m}$$

Let the new contact force be $F_{c2}$. We analyze the forces on block $m$. The forces are the applied force $F$ to the right and the contact force from block $M$, $F_{c2}$, to the left. Applying Newton's second law to block $m$:

$$F - F_{c2} = ma$$

Solving for the contact force $F_{c2}$:

$$F_{c2} = F - ma$$

Substituting the expression for acceleration $a$:

$$F_{c2} = F - m\left(\frac{F-f}{M+m}\right) = \frac{F(M+m) - m(F-f)}{M+m}$$ $$F_{c2} = \frac{FM + Fm - mF + mf}{M+m} = \frac{FM + mf}{M+m}$$

Substituting the given values:

$$F_{c2} = \frac{(15)(3) + (10)(2)}{3+2} = \frac{45 + 20}{5} = \frac{65}{5} = 13 \text{ N}$$