A bullet of mass `10g` travelling horizontally with a velocity of `150 ms^(-1)` strikes a stationary wooden block and come to rest in `0.03 s`. Calculate the distance of penetration of the bullet into the block. Also, Calculate the magnitude of the force exerted by the wooden block on the bullet,

Explanation

To solve the problem step by step, we need to calculate two things: the distance of penetration of the bullet into the block and the magnitude of the force exerted by the wooden block on the bullet.

### Step 1: Identify the given data
- Mass of the bullet (m) = 10 g = 10 × 10^(-3) kg = 0.01 kg
- Initial velocity of the bullet (u) = 150 m/s
- Final velocity of the bullet (v) = 0 m/s (since it comes to rest)
- Time taken to come to rest (t) = 0.03 s

### Step 2: Calculate the acceleration of the bullet
Using the equation of motion:
\[ v = u + at \]
Substituting the known values:
\[ 0 = 150 + a \times 0.03 \]

Rearranging to solve for acceleration (a):
\[ a = \frac{0 - 150}{0.03} \]
\[ a = \frac{-150}{0.03} \]
\[ a = -5000 \, \text{m/s}^2 \]

### Step 3: Calculate the distance of penetration (x)
Using the equation:
\[ v^2 = u^2 + 2ax \]
Substituting the known values:
\[ 0^2 = 150^2 + 2 \times (-5000) \times x \]

This simplifies to:
\[ 0 = 22500 - 10000x \]
Rearranging gives:
\[ 10000x = 22500 \]
\[ x = \frac{22500}{10000} \]
\[ x = 2.25 \, \text{m} \]

### Step 4: Calculate the magnitude of the force exerted by the wooden block on the bullet
Using Newton's second law:
\[ F = ma \]
Substituting the mass and acceleration:
\[ F = 0.01 \, \text{kg} \times (-5000 \, \text{m/s}^2) \]
\[ F = -50 \, \text{N} \]

The negative sign indicates that the force is in the opposite direction of the bullet's motion (resistive force). Therefore, the magnitude of the force is:
\[ |F| = 50 \, \text{N} \]

### Final Answers:
- Distance of penetration of the bullet into the block: **2.25 m**
- Magnitude of the force exerted by the wooden block on the bullet: **50 N**