A canon shell moving along a straight line bursts into two parts. Just after the burst one part moves with momentum `20 N s` making an angle `30^(@)` with the original line of motion. The minimum momentum of the other part of shell just after the burst is

Options

A. `0NS`
B. `5Ns`
C. `10Ns`
D. `17.32Ns`

Explanation

To solve the problem, we will use the principle of conservation of momentum. Here are the steps to find the minimum momentum of the second part of the shell after the burst.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a cannon shell that bursts into two parts. One part has a momentum of \( P_1 = 20 \, \text{N s} \) at an angle of \( 30^\circ \) to the original line of motion. We need to find the minimum momentum of the second part \( P_2 \).

2. **Setting Up the Coordinate System**:
Let's define the original direction of motion as the x-axis. After the burst, the momentum of the first part can be broken down into its components:
- \( P_{1x} = P_1 \cos(30^\circ) \)
- \( P_{1y} = P_1 \sin(30^\circ) \)

3. **Calculating the Components of \( P_1 \)**:
- \( P_{1x} = 20 \cos(30^\circ) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{N s} \)
- \( P_{1y} = 20 \sin(30^\circ) = 20 \times \frac{1}{2} = 10 \, \text{N s} \)

4. **Applying Conservation of Momentum**:
According to the conservation of momentum, the total momentum before the burst must equal the total momentum after the burst. Since the initial momentum was entirely in the x-direction, we have:
- In the x-direction:
\[ P_{1x} + P_{2x} = 20 \, \text{N s} \]
- In the y-direction:
\[ P_{1y} + P_{2y} = 0 \quad \text{(since there was no initial y-component)} \]

5. **Setting Up the Equations**:
From the x-direction:
\[ 10\sqrt{3} + P_{2x} = 20 \]
Therefore,
\[ P_{2x} = 20 - 10\sqrt{3} \]

From the y-direction:
\[ 10 + P_{2y} = 0 \]
Therefore,
\[ P_{2y} = -10 \]

6. **Calculating the Magnitude of \( P_2 \)**:
The total momentum \( P_2 \) can be calculated using the Pythagorean theorem:
\[ P_2 = \sqrt{P_{2x}^2 + P_{2y}^2} \]
Substituting the values:
\[ P_2 = \sqrt{(20 - 10\sqrt{3})^2 + (-10)^2} \]

7. **Simplifying the Expression**:
Calculate \( (20 - 10\sqrt{3})^2 \):
\[ = 400 - 400\sqrt{3} + 300 = 700 - 400\sqrt{3} \]
Now, calculate \( P_2 \):
\[ P_2 = \sqrt{700 - 400\sqrt{3} + 100} = \sqrt{800 - 400\sqrt{3}} \]

8. **Finding the Minimum Value**:
To find the minimum momentum of \( P_2 \), we need to maximize \( P_1 \). The maximum value of \( P_1 \) occurs when \( \sin(\theta) = 1 \) (i.e., \( \theta = 90^\circ \)), which gives:
\[ P_1 = 10 \, \text{N s} \]
Thus, the minimum momentum of the other part is \( P_2 = 10 \, \text{N s} \).

### Final Answer:
The minimum momentum of the other part of the shell just after the burst is \( 10 \, \text{N s} \).