A bullet when fixed at a target with a velocity of `100 ms^(-1)`, penetrates one metre into it. If the bullet is fired with the same velocity as a similar target with a thickness `0.5` metre, then it will emerge from it with a velocity of

Options

A. `50sqrt2m//s`
B. `(50)/(sqrt2)m//s`
C. `50 m//s`
D. `10 m//s`

Answer

To solve the problem, we need to analyze the motion of the bullet as it penetrates the target. We will use the equations of motion to find the final velocity of the bullet after penetrating a target of thickness 0.5 meters.

### Step-by-step Solution:

1. **Identify the given data for the first case:**
- Initial velocity of the bullet, \( u = 100 \, \text{m/s} \)
- Final velocity after penetrating 1 meter, \( v = 0 \, \text{m/s} \)
- Distance penetrated, \( s = 1 \, \text{m} \)

2. **Use the equation of motion to find the retardation (deceleration):**
\[ v^2 = u^2 + 2as \]
Substituting the known values:
\[ 0 = (100)^2 + 2a(1) \]
\[ 0 = 10000 + 2a \]
Rearranging gives:
\[ 2a = -10000 \implies a = -5000 \, \text{m/s}^2 \]

3. **Identify the given data for the second case:**
- The thickness of the target, \( s = 0.5 \, \text{m} \)
- Initial velocity, \( u = 100 \, \text{m/s} \)
- Retardation, \( a = -5000 \, \text{m/s}^2 \)

4. **Use the equation of motion to find the final velocity after penetrating 0.5 meters:**
\[ v^2 = u^2 + 2as \]
Substituting the known values:
\[ v^2 = (100)^2 + 2(-5000)(0.5) \]
\[ v^2 = 10000 - 5000 \]
\[ v^2 = 5000 \]
Taking the square root gives:
\[ v = \sqrt{5000} = 10\sqrt{50} \, \text{m/s} \]

5. **Simplify the final velocity:**
\[ v = 10\sqrt{50} = 10 \cdot 5\sqrt{2} = 50\sqrt{2} \, \text{m/s} \]

### Final Answer:
The bullet will emerge from the target with a velocity of \( 50\sqrt{2} \, \text{m/s} \).

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