A 10 g bullet, moving with a velocity of 500 m/s. enters a stationary piece of ice of mass 10 kg and stops. If the piece of ice is lying on a frictionless plane, then its velocity will be

Options

A. 5 cm/s
B. 5 m/s
C. 0·5 m/s
D. 0·5 cm/s

Answer

To solve the problem, we will use the principle of conservation of momentum. The momentum before the bullet hits the ice block must equal the momentum after the bullet has stopped inside the block.

### Step-by-Step Solution:

1. **Identify the masses and velocities:**
- Mass of the bullet, \( m_b = 10 \, \text{g} = 10 \times 10^{-3} \, \text{kg} = 0.01 \, \text{kg} \)
- Velocity of the bullet, \( v_b = 500 \, \text{m/s} \)
- Mass of the ice block, \( m_i = 10 \, \text{kg} \)
- Initial velocity of the ice block, \( v_i = 0 \, \text{m/s} \)

2. **Calculate the initial momentum:**
The initial momentum of the system (bullet + ice block) is given by:
\[ p_{\text{initial}} = m_b \cdot v_b + m_i \cdot v_i \]
Since the ice block is stationary, its initial momentum is zero:
\[ p_{\text{initial}} = (0.01 \, \text{kg} \cdot 500 \, \text{m/s}) + (10 \, \text{kg} \cdot 0) = 5 \, \text{kg m/s} \]

3. **Determine the final momentum:**
After the bullet embeds itself in the ice block, they move together with a common velocity \( v \). The total mass of the system after the collision is:
\[ m_{\text{total}} = m_b + m_i = 0.01 \, \text{kg} + 10 \, \text{kg} = 10.01 \, \text{kg} \]
The final momentum of the system is:
\[ p_{\text{final}} = m_{\text{total}} \cdot v = 10.01 \, \text{kg} \cdot v \]

4. **Apply conservation of momentum:**
According to the conservation of momentum:
\[ p_{\text{initial}} = p_{\text{final}} \]
Therefore:
\[ 5 \, \text{kg m/s} = 10.01 \, \text{kg} \cdot v \]

5. **Solve for \( v \):**
Rearranging the equation to solve for \( v \):
\[ v = \frac{5 \, \text{kg m/s}}{10.01 \, \text{kg}} \approx 0.4995 \, \text{m/s} \]
Rounding this gives us:
\[ v \approx 0.5 \, \text{m/s} \]

### Final Answer:
The velocity of the piece of ice after the bullet stops inside it is approximately \( 0.5 \, \text{m/s} \).

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