Two particles of mass m and 2m have their position vectors as a function of time as `r_(1) (t) = hat I - t^(3) hat(j) +2t^(2)hat(k) and r_(2) (t) = t hat (i) -t^(3)hat(j) -t^(2) hatk` respectively (where t is the time). Which one of the following graphs represents the path of the centre of mass ?

Options

A.
B.
C.
D.

Answer

To find the path of the center of mass of the two particles with given position vectors, we can follow these steps:

### Step 1: Identify the position vectors
The position vectors of the two particles are given as:
- For particle 1 (mass m):
\[ \mathbf{r}_1(t) = \hat{i} - t^3 \hat{j} + 2t^2 \hat{k} \]
- For particle 2 (mass 2m):
\[ \mathbf{r}_2(t) = t \hat{i} - t^3 \hat{j} - t^2 \hat{k} \]

### Step 2: Write the formula for the center of mass
The position vector of the center of mass \(\mathbf{r}_{\text{com}}\) is given by:
\[ \mathbf{r}_{\text{com}} = \frac{m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2}{m_1 + m_2} \]
where \(m_1 = m\) and \(m_2 = 2m\).

### Step 3: Substitute the values into the formula
Substituting the values of \(m_1\), \(m_2\), \(\mathbf{r}_1\), and \(\mathbf{r}_2\):
\[ \mathbf{r}_{\text{com}} = \frac{m \mathbf{r}_1 + 2m \mathbf{r}_2}{m + 2m} \]
This simplifies to:
\[ \mathbf{r}_{\text{com}} = \frac{m (\hat{i} - t^3 \hat{j} + 2t^2 \hat{k}) + 2m (t \hat{i} - t^3 \hat{j} - t^2 \hat{k})}{3m} \]

### Step 4: Simplify the expression
Now, we can simplify the numerator:
\[ = \frac{m \hat{i} - mt^3 \hat{j} + 2m \cdot 2t^2 \hat{k} + 2mt \hat{i} - 2mt^3 \hat{j} - 2mt^2 \hat{k}}{3m} \]
Combining like terms:
\[ = \frac{(m + 2mt) \hat{i} + (-mt^3 - 2mt^3) \hat{j} + (2mt^2 - 2mt^2) \hat{k}}{3m} \]
This simplifies to:
\[ = \frac{(1 + 2t) \hat{i} - 3t^3 \hat{j}}{3} \]

### Step 5: Write the components of the center of mass
From the above expression, we can identify the components:
- \(x = \frac{1 + 2t}{3}\)
- \(y = -t^3\)

### Step 6: Analyze the path
To analyze the path, we can express \(y\) in terms of \(x\):
From \(x = \frac{1 + 2t}{3}\), we can express \(t\) as:
\[ t = \frac{3x - 1}{2} \]
Substituting this into the equation for \(y\):
\[ y = -\left(\frac{3x - 1}{2}\right)^3 \]
This indicates that the path is a cubic curve in the fourth quadrant since \(y\) will be negative for positive \(x\).

### Conclusion
The graph representing the path of the center of mass is a curve in the fourth quadrant, which corresponds to option D.

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