A bullet of mass `10^-3kg` strikes an obstacle and moves at `60^@` to its original direction. If its speed also changes from `20m//s` to `10m//s`. Find the magnitude of impulse acting on the bullet.

Explanation

To find the magnitude of the impulse acting on the bullet, we will follow these steps:

### Step 1: Identify the given data
- Mass of the bullet, \( m = 10^{-3} \, \text{kg} \)
- Initial speed, \( v_1 = 20 \, \text{m/s} \)
- Final speed, \( v_2 = 10 \, \text{m/s} \)
- Angle of deflection, \( \theta = 60^\circ \)

### Step 2: Calculate the initial momentum
The initial momentum of the bullet can be calculated as:
\[ p_1 = m \cdot v_1 = 10^{-3} \, \text{kg} \cdot 20 \, \text{m/s} = 2 \times 10^{-2} \, \text{kg m/s} \]

### Step 3: Resolve the final velocity into components
The final velocity \( v_2 \) can be resolved into two components:
1. Along the original direction (x-direction):
\[ v_{2x} = v_2 \cdot \cos(60^\circ) = 10 \cdot \frac{1}{2} = 5 \, \text{m/s} \]
2. Perpendicular to the original direction (y-direction):
\[ v_{2y} = v_2 \cdot \sin(60^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \]

### Step 4: Calculate the final momentum components
1. Final momentum in the x-direction:
\[ p_{2x} = m \cdot v_{2x} = 10^{-3} \cdot 5 = 5 \times 10^{-3} \, \text{kg m/s} \]
2. Final momentum in the y-direction:
\[ p_{2y} = m \cdot v_{2y} = 10^{-3} \cdot 5\sqrt{3} = 5\sqrt{3} \times 10^{-3} \, \text{kg m/s} \]

### Step 5: Calculate the change in momentum
1. Change in momentum in the x-direction:
\[ \Delta p_x = p_{2x} - p_1 = 5 \times 10^{-3} - 2 \times 10^{-2} = -1.5 \times 10^{-2} \, \text{kg m/s} \]
2. Change in momentum in the y-direction:
\[ \Delta p_y = p_{2y} - 0 = 5\sqrt{3} \times 10^{-3} \, \text{kg m/s} \]

### Step 6: Calculate the impulse components
1. Impulse in the x-direction:
\[ J_x = \Delta p_x = -1.5 \times 10^{-2} \, \text{N s} \]
2. Impulse in the y-direction:
\[ J_y = \Delta p_y = 5\sqrt{3} \times 10^{-3} \, \text{N s} \]

### Step 7: Calculate the magnitude of the net impulse
The magnitude of the net impulse can be calculated using the Pythagorean theorem:
\[ J = \sqrt{J_x^2 + J_y^2} \]
Substituting the values:
\[ J = \sqrt{(-1.5 \times 10^{-2})^2 + (5\sqrt{3} \times 10^{-3})^2} \]
Calculating:
\[ J = \sqrt{(2.25 \times 10^{-4}) + (75 \times 10^{-6})} = \sqrt{(2.25 \times 10^{-4}) + (7.5 \times 10^{-5})} = \sqrt{3 \times 10^{-4}} = \sqrt{3} \times 10^{-2} \, \text{N s} \]

### Final Answer
The magnitude of the impulse acting on the bullet is:
\[ J = \sqrt{3} \times 10^{-2} \, \text{N s} \]