Question

Two particles of equal mass have initial velocities $2\hat{i}$ ms$^{-1}$ and $2\hat{j}$ ms$^{-1}$. First particle has a constant acceleration $(\hat{i}+\hat{j})$ ms$^{-2}$ while the acceleration of the second particle is always zero. The centre of mass of the two particles moves in

• A. Circle
• B. Parabola
• C. Ellipse
• D. Straight line

Answer 1

Answer: (D)

Straight line

Answer 2

The problem asks us to determine the path of the center of mass of two particles.

1. Given Information:

• Masses of the two particles are equal: $m_1 = m_2 = m$.
• Initial velocity of the first particle: $\vec{v}_1 = 2\hat{i}$ ms$^{-1}$.
• Initial velocity of the second particle: $\vec{v}_2 = 2\hat{j}$ ms$^{-1}$.
• Acceleration of the first particle: $\vec{a}_1 = (\hat{i}+\hat{j})$ ms$^{-2}$.
• Acceleration of the second particle: $\vec{a}_2 = 0$ ms$^{-2}$.

2. Calculate the initial velocity of the center of mass ($\vec{V}_{CM}$): The formula for the velocity of the center of mass is: $\vec{V}_{CM} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1 + m_2}$ Since $m_1 = m_2 = m$: $\vec{V}_{CM} = \frac{m(2\hat{i}) + m(2\hat{j})}{m + m} = \frac{2m(\hat{i} + \hat{j})}{2m} = (\hat{i} + \hat{j})$ ms$^{-1}$

3. Calculate the acceleration of the center of mass ($\vec{A}_{CM}$): The formula for the acceleration of the center of mass is: $\vec{A}_{CM} = \frac{m_1 \vec{a}_1 + m_2 \vec{a}_2}{m_1 + m_2}$ Since $m_1 = m_2 = m$: $\vec{A}_{CM} = \frac{m(\hat{i}+\hat{j}) + m(0)}{m + m} = \frac{m(\hat{i}+\hat{j})}{2m} = \frac{1}{2}(\hat{i}+\hat{j})$ ms$^{-2}$

4. Determine the path of the center of mass: We have the initial velocity of the center of mass $\vec{V}_{CM} = (\hat{i} + \hat{j})$ and the constant acceleration of the center of mass $\vec{A}_{CM} = \frac{1}{2}(\hat{i}+\hat{j})$.

Notice that $\vec{A}_{CM} = \frac{1}{2} \vec{V}_{CM}$. This means the acceleration of the center of mass is always parallel to its initial velocity. When the acceleration of a body is constant and in the same direction as its initial velocity (or opposite to it), the body moves in a straight line. The velocity vector will always maintain the same direction, only its magnitude will change.

Let $\vec{R}_{CM}(t)$ be the position vector of the center of mass at time $t$. Assuming the initial position of the center of mass is at the origin ($\vec{R}_{CM}(0) = 0$), its position at time $t$ is given by: $\vec{R}_{CM}(t) = \vec{V}_{CM}t + \frac{1}{2}\vec{A}_{CM}t^2$ $\vec{R}_{CM}(t) = (\hat{i} + \hat{j})t + \frac{1}{2}\left(\frac{1}{2}(\hat{i} + \hat{j})\right)t^2$ $\vec{R}_{CM}(t) = \left(t + \frac{t^2}{4}\right)(\hat{i} + \hat{j})$

Let $\vec{R}_{CM}(t) = x(t)\hat{i} + y(t)\hat{j}$. From the equation above, we have: $x(t) = t + \frac{t^2}{4}$ $y(t) = t + \frac{t^2}{4}$ Thus, $y(t) = x(t)$. This is the equation of a straight line passing through the origin with a slope of 1.

The path of the center of mass is a straight line.