Question

Two identical bodies \(A\) and \(B\) of equal masses have initial velocities \(\vec v_1 = 4\hat{i}\,\text{m/s}\) and \(\vec v_2 = 4\hat{j}\,\text{m/s}\) respectively. The body \(A\) has acceleration \(\vec a_1 = 6\hat{i} + 6\hat{j}\,\text{m/s}^2\) while the acceleration of body \(B\) is zero. The centre of mass of the two bodies moves in ______ path.

• A. circular
• B. parabolic
• C. straight line
• D. elliptical

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to determine the trajectory of the center of mass (CM) of a system consisting of two equal masses with given initial velocities and accelerations.
Step 2: Key Formula or Approach:
For two equal masses \(m\), the velocity and acceleration of the center of mass are:
\[ \vec{v}_{cm} = \frac{\vec{v}_1 + \vec{v}_2}{2} \]
\[ \vec{a}_{cm} = \frac{\vec{a}_1 + \vec{a}_2}{2} \]
If the acceleration vector of the center of mass is parallel (or anti-parallel) to its initial velocity vector, the path will be a straight line.
Step 3: Detailed Explanation:
Initial velocity of the center of mass:
\[ \vec{v}_{cm} = \frac{4\hat{i} + 4\hat{j}}{2} = (2\hat{i} + 2\hat{j}) \text{ m/s} \]
Acceleration of the center of mass (since \(\vec{a}_2 = 0\)):
\[ \vec{a}_{cm} = \frac{(6\hat{i} + 6\hat{j}) + 0}{2} = (3\hat{i} + 3\hat{j}) \text{ m/s}^2 \]
Note the relationship between \(\vec{a}_{cm}\) and \(\vec{v}_{cm}\):
\[ \vec{a}_{cm} = 1.5 (2\hat{i} + 2\hat{j}) = 1.5 \vec{v}_{cm} \]
Since \(\vec{a}_{cm}\) is proportional to \(\vec{v}_{cm}\), the acceleration vector is parallel to the initial velocity vector.
Step 4: Final Answer:
Because the acceleration is in the same direction as the initial velocity, the direction of motion never changes, and the center of mass follows a straight line.