A stationary shell breaks into three fragments The momentum of two of the fragments is `P` each and move at `60^(@)` to each other.The momentum of the third fragment is

Options

A. P
B. 2P
C. `P/sqrt3`
D. `sqrt3P`

Explanation

To solve the problem of the stationary shell breaking into three fragments, we can follow these steps:

### Step 1: Understand the Initial Conditions
The shell is initially stationary, which means its total momentum is zero. When it breaks into fragments, the momentum must be conserved.

### Step 2: Define the Momentum of the Fragments
Let’s denote the momentum of the two fragments as \( P_1 \) and \( P_2 \), where both have a magnitude of \( P \) and are moving at an angle of \( 60^\circ \) to each other.

### Step 3: Calculate the Resultant Momentum of the First Two Fragments
To find the resultant momentum \( P_{12} \) of the first two fragments, we can use the law of cosines. The formula for the magnitude of the resultant \( R \) of two vectors \( A \) and \( B \) at an angle \( \theta \) is given by:

\[ R = \sqrt{A^2 + B^2 + 2AB \cos(\theta)} \]

In our case, both \( A \) and \( B \) are equal to \( P \), and \( \theta = 60^\circ \):

\[ P_{12} = \sqrt{P^2 + P^2 + 2P \cdot P \cdot \cos(60^\circ)} \]

Since \( \cos(60^\circ) = \frac{1}{2} \):

\[ P_{12} = \sqrt{P^2 + P^2 + 2P^2 \cdot \frac{1}{2}} = \sqrt{P^2 + P^2 + P^2} = \sqrt{3P^2} = P\sqrt{3} \]

### Step 4: Determine the Momentum of the Third Fragment
Since the total momentum of the system must remain zero (as it was initially), the momentum of the third fragment \( P_3 \) must be equal in magnitude but opposite in direction to the resultant momentum of the first two fragments:

\[ P_3 = -P_{12} = -P\sqrt{3} \]

### Final Answer
Thus, the momentum of the third fragment is \( P\sqrt{3} \) in the opposite direction to the resultant momentum of the first two fragments.

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