A person slides freely down a frictionless inclined plane while his bag falls down vertically from the same height. The final speed of man `(v_(m))` and the bag `(v_(B))` should be such that

Options

A. `v_(M) lt v_(B)`
B. `v_(M) = v_(B)`
C. `v_(M) gt v_(B)`
D. they depen on masses

Answer

To solve the problem of comparing the final speeds of a person sliding down a frictionless inclined plane and a bag falling vertically from the same height, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Define the scenario
- A person slides down a frictionless inclined plane from height \( h \).
- A bag falls vertically from the same height \( h \).

### Step 2: Analyze the motion of the person
- For the person sliding down the inclined plane, we apply the conservation of mechanical energy:
\[ \text{Initial Potential Energy} = \text{Final Kinetic Energy} \]
- The initial potential energy (PE) of the person at height \( h \) is given by:
\[ PE_{initial} = mgh \]
(where \( m \) is the mass of the person and \( g \) is the acceleration due to gravity).

- The initial kinetic energy (KE) is zero because the person starts from rest:
\[ KE_{initial} = 0 \]

- The final kinetic energy when the person reaches the bottom of the incline is:
\[ KE_{final} = \frac{1}{2} m v_m^2 \]

- Setting the initial potential energy equal to the final kinetic energy:
\[ mgh = \frac{1}{2} m v_m^2 \]

### Step 3: Simplify the equation for the person
- Cancel the mass \( m \) from both sides (since it appears in every term):
\[ gh = \frac{1}{2} v_m^2 \]

- Rearranging gives:
\[ v_m^2 = 2gh \]

- Taking the square root:
\[ v_m = \sqrt{2gh} \]

### Step 4: Analyze the motion of the bag
- For the bag falling vertically, we again apply the conservation of mechanical energy:
\[ \text{Initial Potential Energy} = \text{Final Kinetic Energy} \]
- The initial potential energy of the bag is also:
\[ PE_{initial} = m_bgh \]
(where \( m_b \) is the mass of the bag).

- The initial kinetic energy is zero:
\[ KE_{initial} = 0 \]

- The final kinetic energy when the bag reaches the ground is:
\[ KE_{final} = \frac{1}{2} m_b v_b^2 \]

- Setting the initial potential energy equal to the final kinetic energy:
\[ m_bgh = \frac{1}{2} m_b v_b^2 \]

### Step 5: Simplify the equation for the bag
- Cancel the mass \( m_b \) from both sides:
\[ gh = \frac{1}{2} v_b^2 \]

- Rearranging gives:
\[ v_b^2 = 2gh \]

- Taking the square root:
\[ v_b = \sqrt{2gh} \]

### Step 6: Compare the speeds
- From the calculations, we have:
\[ v_m = \sqrt{2gh} \]
\[ v_b = \sqrt{2gh} \]

- Therefore, we conclude that:
\[ v_m = v_b \]

### Final Answer
The final speeds of the man and the bag are equal:
\[ v_m = v_b \]

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