Three particles of masses 1 kg , 2kg and 3 kg are subjected to forces `(3hat(i)-2hat(j)+2hat(k))N , (-hat(i) +2hat(j)-hat(k))N and (hat(i)+hat(j)+hat(k))N ` respectively . The magnitude of acceleration of CM of the system is :

Options

A. `(sqrt(11))/6 "ms"^(-2)`
B. `(sqrt(22))/6 "ms"^(-2)`
C. ` (sqrt(14))/6 "ms"^(-2)`
D. `(sqrt(22))/6 "ms"^(-2)`

Answer

To find the magnitude of acceleration of the center of mass (CM) of the system of three particles, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on Each Particle:**
- For the first particle (mass = 1 kg), the force is \( \mathbf{F_1} = (3\hat{i} - 2\hat{j} + 2\hat{k}) \, \text{N} \).
- For the second particle (mass = 2 kg), the force is \( \mathbf{F_2} = (-\hat{i} + 2\hat{j} - \hat{k}) \, \text{N} \).
- For the third particle (mass = 3 kg), the force is \( \mathbf{F_3} = (\hat{i} + \hat{j} + \hat{k}) \, \text{N} \).

2. **Calculate the Net Force Acting on the System:**
- The net force \( \mathbf{F_{net}} \) is the vector sum of all individual forces:
\[ \mathbf{F_{net}} = \mathbf{F_1} + \mathbf{F_2} + \mathbf{F_3} \]
- Substituting the values:
\[ \mathbf{F_{net}} = (3\hat{i} - 2\hat{j} + 2\hat{k}) + (-\hat{i} + 2\hat{j} - \hat{k}) + (\hat{i} + \hat{j} + \hat{k}) \]
- Combine the components:
- \( \hat{i} \) components: \( 3 - 1 + 1 = 3 \)
- \( \hat{j} \) components: \( -2 + 2 + 1 = 1 \)
- \( \hat{k} \) components: \( 2 - 1 + 1 = 2 \)
- Therefore, the net force is:
\[ \mathbf{F_{net}} = 3\hat{i} + 1\hat{j} + 2\hat{k} \, \text{N} \]

3. **Calculate the Total Mass of the System:**
- The total mass \( M \) of the system is the sum of the individual masses:
\[ M = 1 \, \text{kg} + 2 \, \text{kg} + 3 \, \text{kg} = 6 \, \text{kg} \]

4. **Calculate the Acceleration of the Center of Mass:**
- The acceleration \( \mathbf{a_{CM}} \) of the center of mass is given by Newton's second law:
\[ \mathbf{a_{CM}} = \frac{\mathbf{F_{net}}}{M} \]
- Substituting the values:
\[ \mathbf{a_{CM}} = \frac{3\hat{i} + 1\hat{j} + 2\hat{k}}{6} \]
- This simplifies to:
\[ \mathbf{a_{CM}} = \left(\frac{3}{6}\hat{i} + \frac{1}{6}\hat{j} + \frac{2}{6}\hat{k}\right) = \left(\frac{1}{2}\hat{i} + \frac{1}{6}\hat{j} + \frac{1}{3}\hat{k}\right) \, \text{m/s}^2 \]

5. **Calculate the Magnitude of the Acceleration:**
- The magnitude of the acceleration \( |\mathbf{a_{CM}}| \) is given by:
\[ |\mathbf{a_{CM}}| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{1}{6}\right)^2 + \left(\frac{1}{3}\right)^2} \]
- Calculating each term:
\[ = \sqrt{\frac{1}{4} + \frac{1}{36} + \frac{1}{9}} \]
- Finding a common denominator (36):
\[ = \sqrt{\frac{9}{36} + \frac{1}{36} + \frac{4}{36}} = \sqrt{\frac{14}{36}} = \sqrt{\frac{7}{18}} \]
- Therefore, the magnitude of the acceleration of the center of mass is:
\[ |\mathbf{a_{CM}}| = \frac{\sqrt{14}}{6} \, \text{m/s}^2 \]

### Final Answer:
The magnitude of acceleration of the center of mass of the system is \( \frac{\sqrt{14}}{6} \, \text{m/s}^2 \).