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

Options

A. 2 kg
B. 1 kg
C. `2sqrt(2) ` kg
D. none of these

Answer

To solve the problem, we will apply the principle of conservation of momentum. The initial momentum of the system is zero since the big particle is at rest before it blasts. After the explosion, the total momentum of the three pieces must also be zero.

### Step 1: Identify the momentum of each piece
1. **First piece**: Mass = 1 kg, Velocity = 2 m/s (along the x-axis)
- Momentum \( p_1 = m_1 \cdot v_1 = 1 \cdot 2 = 2 \, \text{kg m/s} \) (to the right)

2. **Second piece**: Mass = 2 kg, Velocity = 1 m/s (along the y-axis)
- Momentum \( p_2 = m_2 \cdot v_2 = 2 \cdot 1 = 2 \, \text{kg m/s} \) (upward)

3. **Third piece**: Mass = \( m \) kg, Velocity = \( \sqrt{2} \) m/s (unknown direction)
- Let’s denote the velocity of the third piece as \( \vec{v_3} \).

### Step 2: Set up the momentum equations
Since the initial momentum is zero, the final momentum must also be zero. We will consider the momentum in the x and y directions separately.

#### In the x-direction:
- Total momentum in x-direction:
\[ p_{x, \text{final}} = p_1 + p_3 = 2 - m \cdot v_{3x} = 0 \]
Where \( v_{3x} \) is the x-component of the velocity of the third piece.

#### In the y-direction:
- Total momentum in y-direction:
\[ p_{y, \text{final}} = p_2 + p_3 = 2 - m \cdot v_{3y} = 0 \]
Where \( v_{3y} \) is the y-component of the velocity of the third piece.

### Step 3: Solve for the components of the third piece's velocity
From the x-direction equation:
\[ 2 = m \cdot v_{3x} \implies v_{3x} = \frac{2}{m} \]

From the y-direction equation:
\[ 2 = m \cdot v_{3y} \implies v_{3y} = \frac{2}{m} \]

### Step 4: Use the velocity magnitude of the third piece
The magnitude of the velocity of the third piece is given as \( \sqrt{2} \) m/s. Therefore:
\[ \sqrt{v_{3x}^2 + v_{3y}^2} = \sqrt{2} \]

Substituting the expressions for \( v_{3x} \) and \( v_{3y} \):
\[ \sqrt{\left(\frac{2}{m}\right)^2 + \left(\frac{2}{m}\right)^2} = \sqrt{2} \]

### Step 5: Simplify the equation
\[ \sqrt{2 \cdot \left(\frac{2}{m}\right)^2} = \sqrt{2} \]
Squaring both sides:
\[ 2 \cdot \left(\frac{4}{m^2}\right) = 2 \]
\[ \frac{8}{m^2} = 2 \]
\[ m^2 = 4 \implies m = 2 \]

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

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