Two particles of mass $m$ and $2m$ have their position vectors as a function of time as $\vec{r}_1(t)=t\hat{i}-t^3\hat{j}+2t^2\hat{k}$ and $\vec{r}_2(t)=t\hat{i}-t^3\hat{j}-t^2\hat{k}$ respectively (where $t$ is the time). Which one of the following graphs represents the path of the centre of mass
Show Hint
Always check if any components (like $\hat{k}$ in this problem) cancel out. If the centre of mass remains in one plane, the 3D problem simplifies significantly to a 2D trajectory analysis.
Concept:
The position vector of the centre of mass ($\vec{r}_{cm}$) for a system of particles is given by:
\[ \vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2} \]
To find the path in the $xy$-plane, we need to determine the $x$ and $y$ coordinates of the centre of mass as functions of time and then eliminate $t$ to find the relationship between $y$ and $x$.
Step 1: Calculate the position vector of the centre of mass.
Given $m_1 = m$, $m_2 = 2m$.
\[ \vec{r}_{cm}(t) = \frac{m(t\hat{i} - t^3\hat{j} + 2t^2\hat{k}) + 2m(t\hat{i} - t^3\hat{j} - t^2\hat{k})}{m + 2m} \]
\[ \vec{r}_{cm}(t) = \frac{(mt + 2mt)\hat{i} + (-mt^3 - 2mt^3)\hat{j} + (2mt^2 - 2mt^2)\hat{k}}{3m} \]
\[ \vec{r}_{cm}(t) = \frac{3mt\hat{i} - 3mt^3\hat{j} + 0\hat{k}}{3m} = t\hat{i} - t^3\hat{j} \]
Step 2: Determine the equation of the path in the $xy$-plane.
From $\vec{r}_{cm}(t) = x\hat{i} + y\hat{j}$, we have:
\[ x = t \]
\[ y = -t^3 \]
Eliminating $t$ by substituting $t = x$ into the equation for $y$:
\[ y = -x^3 \]
Step 3: Identify the correct graph.
The equation $y = -x^3$ represents a cubic curve. For $t > 0$ (positive time), $x$ is positive and $y$ is negative. The curve starts at the origin $(0,0)$ and curves downwards into the fourth quadrant increasingly steeply. This matches the curve shown in graph (D).
