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

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze the motion of the two balls and how their velocities affect the center of mass (CM) of the system.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- We have two balls of the same mass (let's denote the mass as \( m \)).
- One ball is released from height \( h \) and the other from height \( 2h \).
- Both balls are released simultaneously.

2. **Calculating the Time of Fall**:
- For the ball falling from height \( h \):
\[ h = \frac{1}{2} g t_1^2 \implies t_1 = \sqrt{\frac{2h}{g}} \]
- For the ball falling from height \( 2h \):
\[ 2h = \frac{1}{2} g t_2^2 \implies t_2 = \sqrt{\frac{4h}{g}} = 2\sqrt{\frac{h}{g}} \]

3. **Velocity of Each Ball Just Before Collision**:
- The velocity of the ball falling from height \( h \) just before it hits the ground:
\[ v_1 = g t_1 = g \sqrt{\frac{2h}{g}} = \sqrt{2gh} \]
- The velocity of the ball falling from height \( 2h \) just before it hits the ground:
\[ v_2 = g t_2 = g (2\sqrt{\frac{h}{g}}) = 2\sqrt{gh} \]

4. **Velocity of Center of Mass Before Collision**:
- The velocity of the center of mass (CM) before the collision can be calculated using:
\[ V_{CM} = \frac{m v_1 + m v_2}{m + m} = \frac{v_1 + v_2}{2} = \frac{\sqrt{2gh} + 2\sqrt{gh}}{2} = \frac{(1 + \sqrt{2})\sqrt{gh}}{2} \]

5. **Collision with the Ground**:
- When the first ball (from height \( h \)) hits the ground at time \( t_1 \), it sticks to the ground, and its velocity becomes \( 0 \).
- The second ball continues to fall and hits the ground at time \( t_2 \).

6. **Velocity of Center of Mass After Collision**:
- After the first ball sticks to the ground, the velocity of the center of mass becomes:
\[ V_{CM} = \frac{m \cdot 0 + m \cdot 2\sqrt{gh}}{2m} = \frac{2\sqrt{gh}}{2} = \sqrt{gh} \]

7. **Behavior of the Center of Mass After Collision**:
- After the second ball hits the ground, both balls are at rest, so the velocity of the center of mass becomes \( 0 \).

8. **Constructing the Velocity-Time Graph**:
- From \( t = 0 \) to \( t = t_1 \), the velocity of the center of mass increases linearly.
- At \( t_1 \), the velocity drops to \( \sqrt{gh} \) and remains constant until \( t_2 \).
- After \( t_2 \), the velocity drops to \( 0 \) as both balls are at rest.

### Conclusion:
The velocity-time graph of the center of mass will show a linear increase until \( t_1 \), a constant value until \( t_2 \), and then a drop to \( 0 \). The graph that best represents this behavior is option **B**.