Two balls of different masses are thrown in air with different velocities. While they are in air acceleration of centre of mass of the system. (neglect air resistance)?

Options

A. Depends on the direction of the motion of two balls
B. Depends on the masses of the two balls
C. Depends on the magnitude of velocities of the two balls
D. Is equal to g

Answer

To find the acceleration of the center of mass of a system of two balls thrown in the air with different velocities and masses, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the System**: We have two balls with masses \( m_1 \) and \( m_2 \) thrown in the air with initial velocities \( v_1 \) and \( v_2 \) at angles \( \alpha \) and \( \beta \) respectively.

2. **Understand the Forces Acting on the Balls**: Since we are neglecting air resistance, the only force acting on both balls is the force of gravity. This force causes a downward acceleration of \( g \) (approximately \( 9.81 \, \text{m/s}^2 \)) for both balls.

3. **Determine the Accelerations of Each Ball**:
- The acceleration of ball 1, \( a_1 \), is \( g \) downward.
- The acceleration of ball 2, \( a_2 \), is also \( g \) downward.

4. **Use the Formula for Acceleration of the Center of Mass**: The acceleration of the center of mass \( a_{cm} \) of a system of particles is given by:
\[ a_{cm} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2} \]

5. **Substitute the Values**:
- Substitute \( a_1 = g \) and \( a_2 = g \) into the formula:
\[ a_{cm} = \frac{m_1 g + m_2 g}{m_1 + m_2} \]

6. **Factor Out \( g \)**:
\[ a_{cm} = \frac{g(m_1 + m_2)}{m_1 + m_2} \]

7. **Simplify the Expression**: The \( m_1 + m_2 \) terms cancel out:
\[ a_{cm} = g \]

8. **Conclusion**: The acceleration of the center of mass of the system is \( g \) downward.

### Final Answer:
The acceleration of the center of mass of the system is \( g \) (downward).

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