A bullet of mass 40g moving with a speed of `90ms^(-1)` enters a heavy wooden block and is stopped after a direction of 60cm. The average resistive force exered by the block on the bullet is

Options

A. 1. 180 N
B. 2. 220 N
C. 3. 270 N
D. 4. 320 N

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- Mass of the bullet, \( m = 40 \, \text{g} = 0.04 \, \text{kg} \) (since \( 1 \, \text{g} = 0.001 \, \text{kg} \))
- Initial speed of the bullet, \( u = 90 \, \text{m/s} \)
- Final speed of the bullet, \( v = 0 \, \text{m/s} \) (since the bullet comes to a stop)
- Distance traveled while stopping, \( s = 60 \, \text{cm} = 0.6 \, \text{m} \) (since \( 1 \, \text{cm} = 0.01 \, \text{m} \))

### Step 2: Use the third equation of motion
We will use the third equation of motion which states:
\[ v^2 = u^2 + 2as \]
Rearranging it to solve for acceleration \( a \):
\[ a = \frac{v^2 - u^2}{2s} \]

### Step 3: Substitute the values into the equation
Substituting the known values into the equation:
\[ a = \frac{0^2 - (90)^2}{2 \times 0.6} \]
\[ a = \frac{0 - 8100}{1.2} \]
\[ a = \frac{-8100}{1.2} = -6750 \, \text{m/s}^2 \]

### Step 4: Calculate the average resistive force
Using Newton's second law, the force \( F \) can be calculated as:
\[ F = m \cdot a \]
Substituting the values:
\[ F = 0.04 \, \text{kg} \times (-6750 \, \text{m/s}^2) \]
\[ F = -270 \, \text{N} \]

### Step 5: Interpret the result
The negative sign indicates that the force is a resistive force acting in the opposite direction to the motion of the bullet. Thus, the magnitude of the average resistive force is:
\[ |F| = 270 \, \text{N} \]

### Final Answer
The average resistive force exerted by the block on the bullet is \( 270 \, \text{N} \).

---