A bullet weighing 10 gm and moving at 300 m/s strikes a 5 kg ice and drops dead. The ice block is placed on a frictionless level surface. The speed of the block after the collision, is

Options

A. 6 cm/sec
B. 6 m/sec
C. 60 cm/sec
D. 60 m/sec

Answer

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

### Step-by-Step Solution:

1. **Identify the masses and initial velocities:**
- Mass of the bullet, \( m_b = 10 \text{ gm} = 0.010 \text{ kg} \)
- Velocity of the bullet before collision, \( u_b = 300 \text{ m/s} \)
- Mass of the ice block, \( m_i = 5 \text{ kg} \)
- Velocity of the ice block before collision, \( u_i = 0 \text{ m/s} \) (since it is at rest)

2. **Write the momentum conservation equation:**
The total momentum before the collision is equal to the total momentum after the collision.
\[ \text{Total momentum before} = \text{Total momentum after} \]
\[ m_b \cdot u_b + m_i \cdot u_i = (m_b + m_i) \cdot v \]
where \( v \) is the velocity of the combined system (bullet + ice block) after the collision.

3. **Substitute the known values into the equation:**
\[ (0.010 \text{ kg} \cdot 300 \text{ m/s}) + (5 \text{ kg} \cdot 0 \text{ m/s}) = (0.010 \text{ kg} + 5 \text{ kg}) \cdot v \]
\[ 3 \text{ kg m/s} = 5.010 \text{ kg} \cdot v \]

4. **Solve for \( v \):**
\[ v = \frac{3 \text{ kg m/s}}{5.010 \text{ kg}} \approx 0.598 \text{ m/s} \]

5. **Convert the velocity to centimeters per second:**
\[ v \approx 0.598 \text{ m/s} \times 100 \text{ cm/m} \approx 59.8 \text{ cm/s} \]
Rounding this, we can say:
\[ v \approx 60 \text{ cm/s} \]

### Final Answer:
The speed of the block after the collision is approximately **60 cm/s**.