Two balls of same mass are released simultaneously from heights h & 2h from the ground level. The balls collide with the floor & stick to it. Then the velocity-time graph of centre of mass of the two balls is best represented by :

Options

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B.
C.
D.

Explanation

To solve the problem, we need to analyze the motion of the two balls released from different heights and determine the velocity-time graph of their center of mass after they collide with the ground.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Let the mass of each ball be \( m \).
- The first ball is released from height \( h \).
- The second ball is released from height \( 2h \).

2. **Calculate the Time of Flight for Each Ball:**
- For the first ball (height \( h \)):
\[ h = \frac{1}{2} g t_1^2 \implies t_1 = \sqrt{\frac{2h}{g}} \]
- For the second ball (height \( 2h \)):
\[ 2h = \frac{1}{2} g t_2^2 \implies t_2 = \sqrt{\frac{4h}{g}} = 2\sqrt{\frac{h}{g}} \]

3. **Determine the Velocities Just Before Impact:**
- For the first ball:
\[ v_1 = g t_1 = g \sqrt{\frac{2h}{g}} = \sqrt{2gh} \]
- For the second ball:
\[ v_2 = g t_2 = g (2\sqrt{\frac{h}{g}}) = 2\sqrt{gh} \]

4. **Behavior Upon Collision:**
- When the balls hit the ground, the first ball comes to rest, so its velocity becomes \( 0 \).
- The second ball also comes to rest after colliding with the ground.

5. **Calculate the Velocity of the Center of Mass (CM):**
- The velocity of the center of mass before the collision can be calculated using:
\[ v_{cm} = \frac{m v_1 + m v_2}{m + m} = \frac{v_1 + v_2}{2} = \frac{\sqrt{2gh} + 2\sqrt{gh}}{2} = \frac{(1 + \sqrt{2})\sqrt{gh}}{2} \]
- After the collision, the first ball's velocity is \( 0 \) and the second ball's velocity is also \( 0 \):
\[ v_{cm\_after} = \frac{m \cdot 0 + m \cdot 0}{m + m} = 0 \]

6. **Graphing the Velocity-Time Relation:**
- Initially, the center of mass has a certain velocity (calculated above) and then drops to \( 0 \) after the collision.
- The graph will show a linear decrease from the initial velocity to \( 0 \) at the time of collision.

### Conclusion:
The velocity-time graph of the center of mass will show a linear decrease in velocity from a positive value to \( 0 \) after the collision.