The solid rubber balls A and B having masses 200 and 400 gm respectively are moving in opposite directions with velocity of A equal to 0.3 m / s . After collision the two balls come to rest, then the velocity of B is

Options

A. 0.15 m/sec
B. 1.5 m/sec
C. `-0.15m/sec`
D. none of the above

Answer

To solve the problem, we will use the principle of conservation of momentum. The total momentum before the collision must equal the total momentum after the collision, provided there are no external forces acting on the system.

### Step-by-Step Solution:

1. **Identify the given data:**
- Mass of ball A, \( m_A = 200 \) gm = \( 0.2 \) kg (since 1 gm = 0.001 kg)
- Mass of ball B, \( m_B = 400 \) gm = \( 0.4 \) kg
- Velocity of ball A, \( v_A = 0.3 \) m/s (moving in one direction)
- Velocity of ball B, \( v_B = -v \) m/s (moving in the opposite direction; we will find \( v \))

2. **Write the equation for conservation of momentum:**
The total momentum before the collision is equal to the total momentum after the collision.
\[ m_A \cdot v_A + m_B \cdot v_B = 0 \]
Since both balls come to rest after the collision, their final velocities are 0.

3. **Substituting the values:**
\[ (0.2 \, \text{kg}) \cdot (0.3 \, \text{m/s}) + (0.4 \, \text{kg}) \cdot (-v) = 0 \]

4. **Calculate the momentum of ball A:**
\[ 0.2 \cdot 0.3 = 0.06 \, \text{kg m/s} \]
So, we have:
\[ 0.06 - 0.4v = 0 \]

5. **Rearranging the equation to solve for \( v \):**
\[ 0.4v = 0.06 \]
\[ v = \frac{0.06}{0.4} = 0.15 \, \text{m/s} \]

6. **Conclusion:**
The velocity of ball B before the collision is \( 0.15 \, \text{m/s} \) in the opposite direction.

### Final Answer:
The velocity of ball B before the collision is \( 0.15 \, \text{m/s} \).