Three particles of masses `20g`, `30g` and `40g` are initially moving along the positive direction of the three coordinate axes respectively with the same velocity of `20cm//s`. When due to their mutual interaction, the first particle comes to rest, the second acquires a velocity `(10hati+20hatk)cm//s`. What is then the velocity of the third particle?

Explanation

To solve the problem, we will use the principle of conservation of momentum. Here are the steps to find the velocity of the third particle:

### Step 1: Identify the masses and initial velocities of the particles
- Mass of particle 1, \( m_1 = 20 \, \text{g} \)
- Mass of particle 2, \( m_2 = 30 \, \text{g} \)
- Mass of particle 3, \( m_3 = 40 \, \text{g} \)

The initial velocities for each particle are:
- \( \vec{u_1} = 20 \hat{i} \, \text{cm/s} \) (along x-axis)
- \( \vec{u_2} = 20 \hat{j} \, \text{cm/s} \) (along y-axis)
- \( \vec{u_3} = 20 \hat{k} \, \text{cm/s} \) (along z-axis)

### Step 2: Write down the final velocities after interaction
- After the interaction:
- The first particle comes to rest: \( \vec{v_1} = 0 \)
- The second particle acquires a new velocity: \( \vec{v_2} = 10 \hat{i} + 20 \hat{k} \, \text{cm/s} \)
- The third particle's velocity is unknown: \( \vec{v_3} \)

### Step 3: Apply the conservation of momentum
According to the conservation of momentum:
\[ \text{Initial momentum} = \text{Final momentum} \]

The initial momentum \( \vec{P_i} \) is given by:
\[ \vec{P_i} = m_1 \vec{u_1} + m_2 \vec{u_2} + m_3 \vec{u_3} \]
Substituting the values:
\[ \vec{P_i} = 20 \hat{i} \cdot 20 + 30 \hat{j} \cdot 20 + 40 \hat{k} \cdot 20 \]
Calculating each term:
\[ \vec{P_i} = 400 \hat{i} + 600 \hat{j} + 800 \hat{k} \]

The final momentum \( \vec{P_f} \) is given by:
\[ \vec{P_f} = m_1 \vec{v_1} + m_2 \vec{v_2} + m_3 \vec{v_3} \]
Substituting the known values:
\[ \vec{P_f} = 20 \cdot 0 + 30 (10 \hat{i} + 20 \hat{k}) + 40 \vec{v_3} \]
Calculating:
\[ \vec{P_f} = 0 + 300 \hat{i} + 600 \hat{k} + 40 \vec{v_3} \]

### Step 4: Set initial momentum equal to final momentum
Setting \( \vec{P_i} = \vec{P_f} \):
\[ 400 \hat{i} + 600 \hat{j} + 800 \hat{k} = 300 \hat{i} + 600 \hat{k} + 40 \vec{v_3} \]

### Step 5: Solve for \( \vec{v_3} \)
Rearranging gives:
\[ 400 \hat{i} + 600 \hat{j} + 800 \hat{k} = 300 \hat{i} + 600 \hat{k} + 40 \vec{v_3} \]
This can be separated into components:
1. For \( \hat{i} \):
\[ 400 = 300 + 40 v_{3x} \implies 100 = 40 v_{3x} \implies v_{3x} = \frac{100}{40} = 2.5 \, \text{cm/s} \]
2. For \( \hat{j} \):
\[ 600 = 0 + 0 + 40 v_{3y} \implies 600 = 40 v_{3y} \implies v_{3y} = \frac{600}{40} = 15 \, \text{cm/s} \]
3. For \( \hat{k} \):
\[ 800 = 0 + 600 + 40 v_{3z} \implies 200 = 40 v_{3z} \implies v_{3z} = \frac{200}{40} = 5 \, \text{cm/s} \]

### Final Step: Combine the components to find \( \vec{v_3} \)
Thus, the velocity of the third particle is:
\[ \vec{v_3} = 2.5 \hat{i} + 15 \hat{j} + 5 \hat{k} \, \text{cm/s} \]

### Conclusion
The velocity of the third particle is:
\[ \vec{v_3} = 2.5 \hat{i} + 15 \hat{j} + 5 \hat{k} \, \text{cm/s} \]