Comprehension # 5
One particle of mass `1 kg` is moving along positive x-axis with velocity `3 m//s`. Another particle of mass `2 kg` is moving along y-axis with `6 m//s`. At time `t = 0, 1 kg` mass is at `(3m, 0)` and `2 kg` at `(0, 9m), x - y` plane is the horizontal plane. (Surface is smooth for question 1 and rough for question 2 and 3)
Co-ordinates of centre of mass where it will stop finally are :-

Options

A. `(2.0 m, 14.25 m)`
B. `(2.25 m, 10 m)`
C. `(3.75 m, 9 m)`
D. `(1.75 m, 12 m)`

Explanation

To find the coordinates of the center of mass where it will stop finally, we will follow these steps:

### Step 1: Identify the initial positions and velocities of the particles
- The first particle (mass \( m_1 = 1 \, \text{kg} \)) is at position \( (3 \, \text{m}, 0) \) and moving along the x-axis with a velocity of \( 3 \, \text{m/s} \).
- The second particle (mass \( m_2 = 2 \, \text{kg} \)) is at position \( (0, 9 \, \text{m}) \) and moving along the y-axis with a velocity of \( 6 \, \text{m/s} \).

### Step 2: Calculate the initial position of the center of mass
The position of the center of mass \( R_{cm} \) is given by the formula:
\[ R_{cm} = \frac{m_1 R_1 + m_2 R_2}{m_1 + m_2} \]
Where:
- \( R_1 = (3 \, \text{m}, 0) \)
- \( R_2 = (0, 9 \, \text{m}) \)

Calculating \( R_{cm} \):
\[ R_{cm} = \frac{1 \cdot (3 \hat{i} + 0 \hat{j}) + 2 \cdot (0 \hat{i} + 9 \hat{j})}{1 + 2} = \frac{(3 \hat{i} + 18 \hat{j})}{3} = (1 \hat{i} + 6 \hat{j}) \, \text{m} \]

### Step 3: Calculate the velocity of the center of mass
The velocity of the center of mass \( V_{cm} \) is given by:
\[ V_{cm} = \frac{m_1 V_1 + m_2 V_2}{m_1 + m_2} \]
Where:
- \( V_1 = (3 \hat{i}, 0) \)
- \( V_2 = (0, 6 \hat{j}) \)

Calculating \( V_{cm} \):
\[ V_{cm} = \frac{1 \cdot (3 \hat{i} + 0 \hat{j}) + 2 \cdot (0 \hat{i} + 6 \hat{j})}{1 + 2} = \frac{(3 \hat{i} + 12 \hat{j})}{3} = (1 \hat{i} + 4 \hat{j}) \, \text{m/s} \]

### Step 4: Determine the time until both particles stop
Using the frictional force to find the deceleration:
- For the first particle:
\[ a_1 = -\frac{\mu m_1 g}{m_1} = -\mu g = -0.2 \cdot 10 = -2 \, \text{m/s}^2 \]
- For the second particle:
\[ a_2 = -\frac{\mu m_2 g}{m_2} = -\mu g = -0.2 \cdot 10 = -2 \, \text{m/s}^2 \]

Using the equation \( V = U + at \):
- For the first particle:
\[ 0 = 3 - 2t_1 \implies t_1 = 1.5 \, \text{s} \]
- For the second particle:
\[ 0 = 6 - 2t_2 \implies t_2 = 3 \, \text{s} \]

### Step 5: Find the distance traveled by each particle until they stop
- For the first particle:
\[ x_1 = U t + \frac{1}{2} a t^2 = 3 \cdot 1.5 + \frac{1}{2} \cdot (-2) \cdot (1.5)^2 = 4.5 - 1.5 = 3 \, \text{m} \]
Final position:
\[ R_1' = 3 + 3 = 6 \, \text{m} \]

- For the second particle:
\[ y_2 = U t + \frac{1}{2} a t^2 = 6 \cdot 3 + \frac{1}{2} \cdot (-2) \cdot (3)^2 = 18 - 9 = 9 \, \text{m} \]
Final position:
\[ R_2' = 9 + 9 = 18 \, \text{m} \]

### Step 6: Calculate the final position of the center of mass
Now we can calculate the final position of the center of mass when both particles stop:
\[ R_{cm}' = \frac{m_1 R_1' + m_2 R_2'}{m_1 + m_2} = \frac{1 \cdot (6 \hat{i}) + 2 \cdot (0 \hat{i} + 18 \hat{j})}{3} \]
Calculating:
\[ R_{cm}' = \frac{(6 \hat{i} + 36 \hat{j})}{3} = (2 \hat{i} + 12 \hat{j}) \, \text{m} \]

### Final Answer
The coordinates of the center of mass where it will stop finally are \( (2 \, \text{m}, 12 \, \text{m}) \).

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