A bullet weighing 10 g and moving with a velocity 300 m/s strikes a 5 kg block of ice and drop dead. The ice block is kept on smooth surface. The speed of the block after the the collision is

Options

A. 6 cm/s
B. 60 cm/s
C. 6 m/s
D. 0.6 cm/s

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.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the bullet (m1) = 10 g = 10 × 10^(-3) kg = 0.01 kg
- Velocity of the bullet (u1) = 300 m/s
- Mass of the block of ice (m2) = 5 kg
- Initial velocity of the block of ice (u2) = 0 m/s (since it is at rest)

2. **Write the conservation of momentum equation:**
The total momentum before the collision is equal to the total momentum after the collision.
\[ m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \]
where \( v \) is the final velocity of the block after the collision.

3. **Substitute the known values into the equation:**
\[ (0.01 \, \text{kg}) \times (300 \, \text{m/s}) + (5 \, \text{kg}) \times (0 \, \text{m/s}) = (0.01 \, \text{kg} + 5 \, \text{kg}) \times v \]
Simplifying this gives:
\[ 3 \, \text{kg m/s} = (5.01 \, \text{kg}) \times v \]

4. **Solve for \( v \):**
\[ v = \frac{3 \, \text{kg m/s}}{5.01 \, \text{kg}} \approx 0.5984 \, \text{m/s} \]

5. **Convert \( v \) from m/s to cm/s:**
Since \( 1 \, \text{m/s} = 100 \, \text{cm/s} \):
\[ v \approx 0.5984 \, \text{m/s} \times 100 \approx 59.84 \, \text{cm/s} \]
Rounding this gives approximately \( 60 \, \text{cm/s} \).

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

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