STATEMENT-1: A ball of mass m is released from the top of a smooth inclined plane. At the same time another ball of same mass allowed to fall freely from same height. When both balls reaches the bottom of the plane, momentum will not be equal
because
STATEMENT-2: Momentum depends on the direction of velocity as well as its magnitude.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Answer

To solve the given question, we need to analyze the two statements provided and understand the concepts of momentum and its dependence on velocity.

### Step-by-Step Solution:

1. **Understanding the Scenario**:
- We have two balls of the same mass \( m \).
- One ball is released from the top of a smooth inclined plane, and the other ball is dropped freely from the same height \( h \).

2. **Momentum Definition**:
- Momentum (\( p \)) is defined as the product of mass and velocity:
\[ p = m \cdot v \]
- Since momentum is a vector quantity, it has both magnitude and direction.

3. **Velocity of the Free-Falling Ball**:
- For the ball that is dropped freely, we can use the equation of motion to find its velocity just before it hits the ground:
\[ v_{\text{free}} = \sqrt{2gh} \]
- Here, \( g \) is the acceleration due to gravity, and \( h \) is the height from which it is dropped.

4. **Velocity of the Ball on the Inclined Plane**:
- For the ball sliding down the inclined plane, we can also find its velocity using energy conservation:
\[ mgh = \frac{1}{2} mv^2 \]
- Simplifying this, we find:
\[ v_{\text{inclined}} = \sqrt{2gh} \]
- However, the direction of the velocity will be along the incline, which is different from the vertical direction of the free-falling ball.

5. **Comparison of Momentum**:
- The momentum of the free-falling ball is:
\[ p_{\text{free}} = m \cdot v_{\text{free}} = m \cdot \sqrt{2gh} \]
- The momentum of the ball on the inclined plane is:
\[ p_{\text{inclined}} = m \cdot v_{\text{inclined}} = m \cdot \sqrt{2gh} \]
- Although the magnitudes of the momenta are equal, the directions are different (vertical vs. along the incline).

6. **Conclusion**:
- Since momentum is a vector quantity, the two momenta cannot be considered equal due to their different directions. Therefore, Statement 1 is true.
- Statement 2 correctly explains that momentum depends on both the magnitude and direction of velocity, which is also true.

### Final Answer:
- **Statement 1 is true**: The momentum will not be equal.
- **Statement 2 is true**: Momentum depends on the direction of velocity as well as its magnitude.
- **Conclusion**: Both statements are true, and Statement 2 is the correct explanation for Statement 1.