A man goes at the top of a smooth inclined plane. He releases a bag to fall freely and himself slides down on inclined plane to reach the bottom. If `u_1 and u_2` are the respective velocities of the man and bag at the bottom of inclined plane, then

Options

A. `u_1 gt u_2`
B. `u_1 lt u_2`
C. `u_1 = u_2`
D. `u_1 and u_2` cannot be compared

Answer

To solve the problem, we need to analyze the motion of both the man sliding down the inclined plane and the bag falling freely. We will use the principle of conservation of energy for both cases.

### Step-by-step Solution:

1. **Identify the System**:
- We have a man sliding down a smooth inclined plane and a bag falling freely from the same height \( H \).

2. **Define Variables**:
- Let \( m_1 \) be the mass of the man and \( m_2 \) be the mass of the bag.
- Let \( u_1 \) be the final velocity of the man at the bottom of the incline.
- Let \( u_2 \) be the final velocity of the bag when it reaches the ground.

3. **Energy Conservation for the Man**:
- At the top of the incline, the man has potential energy and no kinetic energy (since he starts from rest).
- Initial potential energy of the man: \( PE_{initial} = m_1 g H \)
- Final potential energy at the bottom: \( PE_{final} = 0 \)
- Final kinetic energy of the man: \( KE_{final} = \frac{1}{2} m_1 u_1^2 \)
- By conservation of energy:
\[ m_1 g H = \frac{1}{2} m_1 u_1^2 \]
- Cancel \( m_1 \) from both sides (assuming \( m_1 \neq 0 \)):
\[ g H = \frac{1}{2} u_1^2 \]
- Rearranging gives:
\[ u_1^2 = 2 g H \quad \Rightarrow \quad u_1 = \sqrt{2 g H} \]

4. **Energy Conservation for the Bag**:
- The bag also starts from rest and falls freely.
- Initial potential energy of the bag: \( PE_{initial} = m_2 g H \)
- Final potential energy at the ground: \( PE_{final} = 0 \)
- Final kinetic energy of the bag: \( KE_{final} = \frac{1}{2} m_2 u_2^2 \)
- By conservation of energy:
\[ m_2 g H = \frac{1}{2} m_2 u_2^2 \]
- Cancel \( m_2 \) from both sides (assuming \( m_2 \neq 0 \)):
\[ g H = \frac{1}{2} u_2^2 \]
- Rearranging gives:
\[ u_2^2 = 2 g H \quad \Rightarrow \quad u_2 = \sqrt{2 g H} \]

5. **Comparing Velocities**:
- From the equations derived:
\[ u_1 = \sqrt{2 g H} \quad \text{and} \quad u_2 = \sqrt{2 g H} \]
- Therefore, we conclude:
\[ u_1 = u_2 \]

### Final Relation:
The relation between the velocities of the man and the bag at the bottom of the inclined plane is:
\[ u_1 = u_2 \]