A stationary body of mass `3 kg` explodes into three equal pieces. Two of the pieces fly off at right angles to each other. One with a velocity of `2hati m//s` and the other withl velocity of `3hatj m//s`. If the explosion takes place in `10^(-5)` the average force acting on the third piece in newtons

Options

A. `(2hati+3hatj)xx10^(-5)`
B. `-(2hati+3hatj)xx10^(5)`
C. `(3hatj+2hatj)xx10^(5)`
D. `-(2hati+3hatj)xx10^(5)`

Explanation

To solve the problem step by step, we will use the principles of conservation of momentum and the definition of force.

### Step 1: Understand the system
We have a stationary body of mass \(3 \, \text{kg}\) that explodes into three equal pieces, each with a mass of \(1 \, \text{kg}\). Two pieces move at right angles to each other with given velocities.

### Step 2: Identify the velocities of the pieces
- Piece 1 moves with a velocity of \(2 \hat{i} \, \text{m/s}\) (along the x-axis).
- Piece 2 moves with a velocity of \(3 \hat{j} \, \text{m/s}\) (along the y-axis).

### Step 3: Apply conservation of momentum
Since the initial momentum of the system is zero (the body is stationary), the total momentum after the explosion must also be zero.

Let the momentum of the third piece be represented as \( \vec{p}_3 \).

The momentum conservation in the x-direction is:
\[ 0 = 1 \cdot 2 \hat{i} + 1 \cdot p_{3x} \]
This gives:
\[ p_{3x} = -2 \, \text{kg m/s} \]

The momentum conservation in the y-direction is:
\[ 0 = 1 \cdot 3 \hat{j} + 1 \cdot p_{3y} \]
This gives:
\[ p_{3y} = -3 \, \text{kg m/s} \]

### Step 4: Calculate the total momentum of the third piece
The total momentum of the third piece can be expressed as:
\[ \vec{p}_3 = p_{3x} \hat{i} + p_{3y} \hat{j} = -2 \hat{i} - 3 \hat{j} \, \text{kg m/s} \]

### Step 5: Calculate the average force acting on the third piece
The average force \( \vec{F} \) acting on the third piece is given by the change in momentum over the time interval during which the explosion occurs:
\[ \vec{F} = \frac{\Delta \vec{p}}{\Delta t} \]
where \( \Delta t = 10^{-5} \, \text{s} \).

Substituting the values:
\[ \vec{F} = \frac{-2 \hat{i} - 3 \hat{j}}{10^{-5}} = (-2 \hat{i} - 3 \hat{j}) \times 10^5 \]

### Step 6: Final expression for the force
Thus, the average force acting on the third piece is:
\[ \vec{F} = -2 \hat{i} - 3 \hat{j} \times 10^5 \, \text{N} \]

### Conclusion
The average force acting on the third piece is:
\[ \vec{F} = -2 \hat{i} + 3 \hat{j} \times 10^5 \, \text{N} \]