Question

Two particles of the same masses moving initially with velocity \( 4 \hat{i} \) and \( 4 \hat{j} \) m/s respectively.
Acceleration of the first particle is \( 6 \hat{i} + 6 \hat{j} \) and of the second is zero. Find the path of the center of mass of the system.

• A. straight line
• B. circle
• C. parabola
• D. ellipse

Hint:

When two particles of equal mass move with constant velocities and accelerations, the center of mass moves along a straight line.

Answer 1

The Correct Option is A

Let the masses of both particles be \( m \), and their initial velocities be: - \( \vec{v_1} = 4 \hat{i} + 4 \hat{j} \) - \( \vec{v_2} = 4 \hat{i} + 4 \hat{j} \) The position vectors of the two particles at any time \( t \) are given by: \[ \vec{r_1} = \vec{v_1} t + \frac{1}{2} \vec{a_1} t^2 = (4 \hat{i} + 4 \hat{j}) t + \frac{1}{2} (6 \hat{i} + 6 \hat{j}) t^2 \] \[ \vec{r_2} = \vec{v_2} t + \frac{1}{2} \vec{a_2} t^2 = (4 \hat{i} + 4 \hat{j}) t + 0 \] Now, the position of the center of mass is given by: \[ \vec{R} = \frac{m \vec{r_1} + m \vec{r_2}}{m + m} = \frac{\vec{r_1} + \vec{r_2}}{2} \] Substituting the expressions for \( \vec{r_1} \) and \( \vec{r_2} \): \[ \vec{R} = \frac{(4 \hat{i} + 4 \hat{j}) t + \frac{1}{2} (6 \hat{i} + 6 \hat{j}) t^2 + (4 \hat{i} + 4 \hat{j}) t}{2} \] \[ \vec{R} = \frac{(8 \hat{i} + 8 \hat{j}) t + 3 ( \hat{i} + \hat{j}) t^2}{2} \] \[ \vec{R} = \left( 4 \hat{i} + 4 \hat{j} \right) t + \frac{3}{2} (\hat{i} + \hat{j}) t^2 \] Thus, the center of mass follows a straight line, because there is no \( t^3 \) or higher power term. Final Answer: (A) straight line