A bullet of mass 50 g enters a block of thickness t with speed of 500 `ms^-1`. It emerges with 4% of its initial kinetic energy. The emergent speed is

Options

A. `100 ms^-1`
B. `200ms^-1`
C. `300ms^-1`
D. `500ms^-1`

Answer

To solve the problem, we need to find the emergent speed of a bullet after it passes through a block. The bullet has an initial speed and kinetic energy, and it emerges with a certain percentage of its initial kinetic energy.

### Step-by-Step Solution:

1. **Identify the given data:**
- Mass of the bullet, \( m = 50 \, \text{g} = 0.05 \, \text{kg} \) (convert grams to kilograms)
- Initial speed of the bullet, \( u = 500 \, \text{m/s} \)
- The bullet emerges with \( 4\% \) of its initial kinetic energy.

2. **Calculate the initial kinetic energy (KE_initial):**
\[ KE_{\text{initial}} = \frac{1}{2} m u^2 \]
Substituting the values:
\[ KE_{\text{initial}} = \frac{1}{2} \times 0.05 \, \text{kg} \times (500 \, \text{m/s})^2 \]
\[ KE_{\text{initial}} = \frac{1}{2} \times 0.05 \times 250000 \]
\[ KE_{\text{initial}} = 0.025 \times 250000 = 6250 \, \text{J} \]

3. **Calculate the final kinetic energy (KE_final):**
Since the bullet emerges with \( 4\% \) of its initial kinetic energy:
\[ KE_{\text{final}} = 0.04 \times KE_{\text{initial}} = 0.04 \times 6250 \, \text{J} \]
\[ KE_{\text{final}} = 250 \, \text{J} \]

4. **Relate the final kinetic energy to the emergent speed (v):**
\[ KE_{\text{final}} = \frac{1}{2} m v^2 \]
Substituting the values:
\[ 250 = \frac{1}{2} \times 0.05 \times v^2 \]
\[ 250 = 0.025 v^2 \]

5. **Solve for \( v^2 \):**
\[ v^2 = \frac{250}{0.025} = 10000 \]

6. **Calculate the emergent speed \( v \):**
\[ v = \sqrt{10000} = 100 \, \text{m/s} \]

### Final Answer:
The emergent speed of the bullet is \( 100 \, \text{m/s} \).

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