A big particle of mass (3+m) kg blasts into 3 pieces , such that a particle of mass 1 kg moves along x - axis , with velocity `2ms^(-1)` and a particle of mass 2 kg moves with velocity `1ms^(-1)` perpendicular to direction of 1 kg particle . If the third particle moves with velocity `sqrt(2) ms ^(-1)`, then m is

Explanation

To solve the problem step by step, we will apply the principle of conservation of momentum.

### Step 1: Understand the problem and define the variables
We have a big particle of mass \( (3 + m) \) kg that splits into three pieces. The masses and velocities of the pieces are given as follows:
- Mass of the first piece \( m_1 = 1 \) kg, moving along the x-axis with velocity \( v_1 = 2 \) m/s.
- Mass of the second piece \( m_2 = 2 \) kg, moving perpendicular to the first piece with velocity \( v_2 = 1 \) m/s.
- Mass of the third piece \( m_3 = m \) kg, moving with velocity \( v_3 = \sqrt{2} \) m/s.

### Step 2: Calculate the momentum of each piece
The momentum of each piece can be calculated using the formula:
\[ \text{Momentum} = \text{mass} \times \text{velocity} \]
- For the first piece:
\[ P_1 = m_1 \cdot v_1 = 1 \cdot 2 = 2 \, \text{kg m/s} \quad \text{(along x-axis)} \]
- For the second piece:
\[ P_2 = m_2 \cdot v_2 = 2 \cdot 1 = 2 \, \text{kg m/s} \quad \text{(along y-axis)} \]
- For the third piece:
\[ P_3 = m_3 \cdot v_3 = m \cdot \sqrt{2} \quad \text{(direction unknown)} \]

### Step 3: Apply conservation of momentum
Since the initial momentum of the system is zero (the big particle was at rest), the total final momentum must also be zero:
\[ P_{\text{total}} = P_1 + P_2 + P_3 = 0 \]
This can be expressed as:
\[ P_3 = - (P_1 + P_2) \]
Substituting the values:
\[ P_3 = - (2 \hat{i} + 2 \hat{j}) \]
Thus,
\[ P_3 = -2 \hat{i} - 2 \hat{j} \]

### Step 4: Find the magnitude of \( P_3 \)
The magnitude of \( P_3 \) can be calculated as:
\[ |P_3| = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]

### Step 5: Relate \( P_3 \) to mass and velocity
From the momentum of the third piece, we have:
\[ |P_3| = m \cdot v_3 \]
Substituting the known values:
\[ 2\sqrt{2} = m \cdot \sqrt{2} \]

### Step 6: Solve for \( m \)
Dividing both sides by \( \sqrt{2} \):
\[ m = \frac{2\sqrt{2}}{\sqrt{2}} = 2 \, \text{kg} \]

### Conclusion
Thus, the value of \( m \) is \( 2 \) kg.

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