A small block of mass `m` is placed at rest on the top of a smooth wedge of mass M, which in turn is placed at rest on a smooth horizontal surface as shown in figure. If h be the height of wedge and `theta` is the inclination, then the distance moved by the wedge as the block reaches the foot of the wedge is

Options

A. (a) `(Mhcottheta)/(M+m)`
B. (b) `(mhcottheta)/(M+m)`
C. (c) `(Mhcosectheta)/(M+m)`
D. (d) `(mhcosectheta)/(M+m)`

Explanation

To solve the problem of the small block of mass `m` placed on a smooth wedge of mass `M`, we need to analyze the motion of both the block and the wedge as the block slides down the incline of the wedge.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a block of mass `m` resting on a wedge of mass `M`. The wedge is inclined at an angle `theta` and has a height `h`.
- Both the block and the wedge are on smooth (frictionless) surfaces.

2. **Initial Setup**:
- The block starts from rest at the top of the wedge.
- The wedge can slide horizontally without friction.

3. **Conservation of Momentum**:
- Since there are no external horizontal forces acting on the system, the horizontal momentum of the system is conserved.
- Initially, both the block and the wedge are at rest, so the total momentum is zero.

4. **Analyzing Forces**:
- As the block slides down the wedge, it exerts a normal force on the wedge. By Newton's third law, the wedge exerts an equal and opposite force on the block.
- The normal force can be resolved into horizontal and vertical components. The horizontal component will cause the wedge to move.

5. **Calculating the Distance Moved by the Wedge**:
- When the block slides down the wedge, it will move a vertical distance `h` downwards.
- The horizontal distance moved by the block can be calculated using trigonometry:
\[ d_{block} = h \cdot \tan(\theta) \]
- The wedge will move a distance `d_{wedge}` in the opposite direction due to conservation of momentum. Since the momentum is conserved, the horizontal distance moved by the wedge will be equal to the horizontal component of the block's movement:
\[ d_{wedge} = d_{block} \cdot \frac{m}{M} \]
- Therefore, substituting the expression for `d_{block}`:
\[ d_{wedge} = h \cdot \tan(\theta) \cdot \frac{m}{M} \]

6. **Final Expression**:
- The total distance moved by the wedge as the block reaches the foot of the wedge is:
\[ d_{wedge} = \frac{m}{M} \cdot h \cdot \tan(\theta) \]

### Final Answer:
The distance moved by the wedge as the block reaches the foot of the wedge is \( \frac{m}{M} \cdot h \cdot \tan(\theta) \).