If a bullet of mass 5gm moving with velocity `100m/sec, penertates the wooden block upto 6cm. Then the average force imposed by the bullet on the block is

Options

A. 8300 N
B. 417 N
C. 830 N
D. zero

Answer

To solve the problem of finding the average force imposed by a bullet on a wooden block, we can follow these steps:

### Step 1: Convert Units
Convert the mass of the bullet from grams to kilograms and the penetration distance from centimeters to meters.
- Mass of the bullet:
\[ m = 5 \text{ grams} = 5 \times 10^{-3} \text{ kg} \]
- Penetration distance:
\[ s = 6 \text{ cm} = 0.06 \text{ m} \]

### Step 2: Identify Initial and Final Velocities
Identify the initial velocity (u) and final velocity (v) of the bullet.
- Initial velocity:
\[ u = 100 \text{ m/s} \]
- Final velocity (when the bullet comes to rest):
\[ v = 0 \text{ m/s} \]

### Step 3: Use the Third Equation of Motion
Use the third equation of motion to find the acceleration (a) of the bullet as it penetrates the block:
\[ v^2 = u^2 + 2as \]
Substituting the known values:
\[ 0 = (100)^2 + 2a(0.06) \]
This simplifies to:
\[ 0 = 10000 + 0.12a \]

### Step 4: Solve for Acceleration
Rearranging the equation to solve for acceleration (a):
\[ 0.12a = -10000 \]
\[ a = \frac{-10000}{0.12} = -83333.33 \text{ m/s}^2 \]

### Step 5: Calculate the Average Force
Using Newton's second law, calculate the average force (F) exerted by the bullet on the block:
\[ F = m \cdot a \]
Substituting the values:
\[ F = (5 \times 10^{-3} \text{ kg}) \cdot (-83333.33 \text{ m/s}^2) \]
\[ F = -416.67 \text{ N} \]
Since force is a vector, we can take the magnitude:
\[ F \approx 417 \text{ N} \]

### Final Answer
The average force imposed by the bullet on the block is approximately **417 N**.

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