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 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 \( m \)):
\[ \vec{r}_1(t) = \hat{i} - t^3 \hat{j} + 2t^2 \hat{k} \]
- For particle 2 (mass \( 2m \)):
\[ \vec{r}_2(t) = t \hat{i} - t^3 \hat{j} - t^2 \hat{k} \]

### Step 2: Write the formula for the center of mass
The position vector of the center of mass \( \vec{r}_{cm} \) for two particles can be calculated using the formula:
\[ \vec{r}_{cm} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2} \]
Where:
- \( m_1 = m \) (mass of particle 1)
- \( m_2 = 2m \) (mass of particle 2)

### Step 3: Substitute the values into the formula
Substituting the masses and position vectors into the formula:
\[ \vec{r}_{cm} = \frac{m \vec{r}_1 + 2m \vec{r}_2}{m + 2m} \]
This simplifies to:
\[ \vec{r}_{cm} = \frac{m \vec{r}_1 + 2m \vec{r}_2}{3m} \]
\[ \vec{r}_{cm} = \frac{1}{3} \left( \vec{r}_1 + 2 \vec{r}_2 \right) \]

### Step 4: Calculate \( \vec{r}_{cm} \)
Now, we will calculate \( \vec{r}_{cm} \):
\[ \vec{r}_{cm} = \frac{1}{3} \left( \left( \hat{i} - t^3 \hat{j} + 2t^2 \hat{k} \right) + 2 \left( t \hat{i} - t^3 \hat{j} - t^2 \hat{k} \right) \right) \]
Expanding this:
\[ = \frac{1}{3} \left( \hat{i} - t^3 \hat{j} + 2t^2 \hat{k} + 2t \hat{i} - 2t^3 \hat{j} - 2t^2 \hat{k} \right) \]
Combining like terms:
\[ = \frac{1}{3} \left( (1 + 2t) \hat{i} + (-1 - 2)t^3 \hat{j} + (2 - 2)t^2 \hat{k} \right) \]
This simplifies to:
\[ = \frac{1}{3} \left( (1 + 2t) \hat{i} - 3t^3 \hat{j} \right) \]

### Step 5: Write the final expression for \( \vec{r}_{cm} \)
Thus, the position vector of the center of mass is:
\[ \vec{r}_{cm} = \frac{1 + 2t}{3} \hat{i} - t^3 \hat{j} \]

### Step 6: Analyze the components
From the expression:
- The x-component is \( \frac{1 + 2t}{3} \) which is always positive for \( t \geq 0 \).
- The y-component is \( -t^3 \) which is always negative for \( t > 0 \).

### Step 7: Determine the path
The path of the center of mass will be a curve where:
- As \( t \) increases, the x-coordinate increases linearly, while the y-coordinate decreases cubically.

### Conclusion
Based on the analysis, the graph representing the path of the center of mass will show a positive x-coordinate and a negative y-coordinate, indicating a curve that moves upwards to the right and downwards in the y-direction.