A bullet of mass 20 g has an initial speed of `1 ms^(-1)` just before it starts penetrating a mud wall of thickness 20 cm. If the wall offers a mean resistance of `2.5 xx 10^(-2)N` the speed of the bullet after emerging from the other side of the wall is close to:

Options

A. `0.7 ms^(-1)`
B. `0.4 ms^(-1)`
C. `0.1 ms^(-1)`
D. `0.3 ms^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert the mass of the bullet to kilograms
The mass of the bullet is given as 20 grams. To convert this to kilograms:
\[ m = 20 \, \text{g} = 20 \times 10^{-3} \, \text{kg} = 0.02 \, \text{kg} \]

**Hint:** Remember that 1 gram = \(10^{-3}\) kg.

### Step 2: Identify the initial speed of the bullet
The initial speed \(u\) of the bullet is given as:
\[ u = 1 \, \text{m/s} \]

**Hint:** The initial speed is the speed just before the bullet starts penetrating the wall.

### Step 3: Convert the thickness of the wall to meters
The thickness of the wall is given as 20 cm. To convert this to meters:
\[ S = 20 \, \text{cm} = 20 \times 10^{-2} \, \text{m} = 0.2 \, \text{m} \]

**Hint:** Remember that 1 centimeter = \(10^{-2}\) meters.

### Step 4: Identify the mean resistance force
The mean resistance force \(F\) offered by the wall is given as:
\[ F = 2.5 \times 10^{-2} \, \text{N} \]

**Hint:** This force acts in the opposite direction to the bullet's motion.

### Step 5: Calculate the acceleration of the bullet
Using Newton's second law, the acceleration \(a\) can be calculated as:
\[ F = ma \implies a = \frac{F}{m} \]
Substituting the values:
\[ a = \frac{-2.5 \times 10^{-2}}{0.02} = -1.25 \, \text{m/s}^2 \]
(Note: The acceleration is negative because it is acting opposite to the direction of motion.)

**Hint:** The negative sign indicates deceleration.

### Step 6: Use the kinematic equation to find the final speed
We will use the kinematic equation:
\[ v^2 = u^2 + 2aS \]
Substituting the known values:
\[ v^2 = (1)^2 + 2 \times (-1.25) \times 0.2 \]
Calculating this:
\[ v^2 = 1 - 0.5 = 0.5 \]
Taking the square root to find \(v\):
\[ v = \sqrt{0.5} \approx 0.707 \, \text{m/s} \]

**Hint:** The final speed is the speed of the bullet after emerging from the wall.

### Conclusion
The speed of the bullet after emerging from the other side of the wall is approximately:
\[ \boxed{0.707 \, \text{m/s}} \]