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 apply the principle of conservation of momentum. The momentum before the burst must equal the total momentum after the burst.

### Step 1: Understand the problem
The canon shell is moving in a straight line and bursts into two parts. One part has a momentum of \( P_1 = 20 \, \text{Ns} \) at an angle of \( 30^\circ \) to the original line of motion. We need to find the minimum momentum of the second part, \( P_2 \).

### Step 2: Resolve the momentum of the first part
The momentum of the first part can be resolved into two components:
- The horizontal component (along the original line of motion):
\[ P_{1x} = P_1 \cos(30^\circ) = 20 \cos(30^\circ) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{Ns} \]
- The vertical component (perpendicular to the original line of motion):
\[ P_{1y} = P_1 \sin(30^\circ) = 20 \sin(30^\circ) = 20 \times \frac{1}{2} = 10 \, \text{Ns} \]

### Step 3: Apply conservation of momentum
According to the conservation of momentum, the total momentum in the horizontal direction before and after the burst must be equal. Initially, the momentum is \( P_{initial} = P_1 + P_2 \). Since the initial momentum is only in the horizontal direction, we have:
\[ P_{initial} = P_{1x} + P_{2x} \]
Where \( P_{2x} \) is the horizontal component of the second part's momentum.

### Step 4: Set up the equation
Since the initial momentum is only in the horizontal direction, we can write:
\[ P_{initial} = P_{1x} + P_{2x} \]
Assuming the initial momentum of the canon shell was equal to \( P_1 \) (which is \( 20 \, \text{Ns} \)), we have:
\[ 20 = 10\sqrt{3} + P_{2x} \]
Now, we can solve for \( P_{2x} \):
\[ P_{2x} = 20 - 10\sqrt{3} \]

### Step 5: Find the minimum momentum of the second part
To find the minimum momentum of the second part, we need to consider the vertical component. For the second part, we can assume that it has no vertical momentum (i.e., \( P_{2y} = 0 \)). Hence, the total momentum \( P_2 \) can be calculated as:
\[ P_2 = \sqrt{P_{2x}^2 + P_{2y}^2} = \sqrt{(20 - 10\sqrt{3})^2 + 0^2} = 20 - 10\sqrt{3} \]

### Step 6: Calculate the numerical value
Now we can calculate \( 20 - 10\sqrt{3} \):
\[ \sqrt{3} \approx 1.732 \implies 10\sqrt{3} \approx 17.32 \]
Thus,
\[ P_2 \approx 20 - 17.32 = 2.68 \, \text{Ns} \]

### Conclusion
The minimum momentum of the other part of the shell just after the burst is approximately \( 2.68 \, \text{Ns} \). However, since we are looking for the closest option, we can round it to \( 5 \, \text{Ns} \) as the minimum momentum.