Three interacting particles of masses `100g, 200g` and `400g` each hace a velocity of `20 m//s` magnitude along the positive direction of x-axis, y-axis and z-axis. Due to force of interaction the third particle stops moving. The velocity of the second particle is `(10hat(j) + 5hat(k))`. What is the velocity of the first particle ?

Options

A. `20hat(i) + 20hat(j) + 70hat(k)`
B. `10hat(i) + 20hat(j) + 8hat(k)`
C. `30hat(i) + 10hat(j) + 7hat(k)`
D. `15hat(i) + 5hat(j) + 60hat(k)`

Explanation

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

### Step-by-Step Solution:

**Step 1: Write down the given data.**
- Mass of the first particle, \( m_1 = 100 \, \text{g} = 0.1 \, \text{kg} \)
- Mass of the second particle, \( m_2 = 200 \, \text{g} = 0.2 \, \text{kg} \)
- Mass of the third particle, \( m_3 = 400 \, \text{g} = 0.4 \, \text{kg} \)
- Initial velocity of the first particle, \( \mathbf{V_1} = 20 \hat{i} \, \text{m/s} \)
- Initial velocity of the second particle, \( \mathbf{V_2} = 20 \hat{j} \, \text{m/s} \)
- Initial velocity of the third particle, \( \mathbf{V_3} = 20 \hat{k} \, \text{m/s} \)
- Final velocity of the second particle, \( \mathbf{V_2'} = 10 \hat{j} + 5 \hat{k} \)
- Final velocity of the third particle, \( \mathbf{V_3'} = 0 \hat{k} \) (stops moving)

**Step 2: Calculate initial momentum.**
Using the formula for momentum \( \mathbf{p} = m \mathbf{v} \):
\[ \text{Initial momentum} = m_1 \mathbf{V_1} + m_2 \mathbf{V_2} + m_3 \mathbf{V_3} \]
Calculating each term:
- \( m_1 \mathbf{V_1} = 0.1 \times 20 \hat{i} = 2 \hat{i} \)
- \( m_2 \mathbf{V_2} = 0.2 \times 20 \hat{j} = 4 \hat{j} \)
- \( m_3 \mathbf{V_3} = 0.4 \times 20 \hat{k} = 8 \hat{k} \)

Thus, the total initial momentum is:
\[ \mathbf{P_{initial}} = 2 \hat{i} + 4 \hat{j} + 8 \hat{k} \]

**Step 3: Calculate final momentum.**
The final momentum after the interaction is:
\[ \text{Final momentum} = m_1 \mathbf{V_1'} + m_2 \mathbf{V_2'} + m_3 \mathbf{V_3'} \]
Where \( \mathbf{V_1'} \) is the final velocity of the first particle (unknown), \( \mathbf{V_2'} = 10 \hat{j} + 5 \hat{k} \), and \( \mathbf{V_3'} = 0 \).

Calculating the final momentum:
\[ \mathbf{P_{final}} = 0.1 \mathbf{V_1'} + 0.2 (10 \hat{j} + 5 \hat{k}) + 0.4 (0) \]
This simplifies to:
\[ \mathbf{P_{final}} = 0.1 \mathbf{V_1'} + 2 \hat{j} + 1 \hat{k} \]

**Step 4: Set initial momentum equal to final momentum.**
According to the conservation of momentum:
\[ \mathbf{P_{initial}} = \mathbf{P_{final}} \]
Thus:
\[ 2 \hat{i} + 4 \hat{j} + 8 \hat{k} = 0.1 \mathbf{V_1'} + 2 \hat{j} + 1 \hat{k} \]

**Step 5: Solve for \( \mathbf{V_1'} \).**
Rearranging the equation:
\[ 2 \hat{i} + 4 \hat{j} + 8 \hat{k} - 2 \hat{j} - 1 \hat{k} = 0.1 \mathbf{V_1'} \]
This simplifies to:
\[ 2 \hat{i} + (4 - 2) \hat{j} + (8 - 1) \hat{k} = 0.1 \mathbf{V_1'} \]
\[ 2 \hat{i} + 2 \hat{j} + 7 \hat{k} = 0.1 \mathbf{V_1'} \]

**Step 6: Isolate \( \mathbf{V_1'} \).**
To find \( \mathbf{V_1'} \):
\[ \mathbf{V_1'} = \frac{1}{0.1} (2 \hat{i} + 2 \hat{j} + 7 \hat{k}) = 20 \hat{i} + 20 \hat{j} + 70 \hat{k} \]

### Final Answer:
The velocity of the first particle is:
\[ \mathbf{V_1'} = 20 \hat{i} + 20 \hat{j} + 70 \hat{k} \, \text{m/s} \]

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