Two identical bodies A and B of equal masses have initial velocities v1 = 4i m/s and v2 = 4j m/s respectively. The body A has acceleration a1 = 6i + 6j m/s^2 while the acceleration of the other body B is zero. The centre of mass of the two bodies moves in _______ path.

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q32 – JEE-MAIN 2026

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Home Physics Rotational MotionQuestion #436

Two identical bodies A and B of equal masses have initial velocities

\vec{v_{1}} = 4 \hat{i}

m/s and

\vec{v_{2}} = 4 \hat{j}

m/s respectively. The body A has acceleration

\vec{a_{1}} = 6 \hat{i} + 6 \hat{j}

m/s

^{2}

while the acceleration of the other body B is zero. The centre of mass of the two bodies moves in _______ path.

Clue 1 of 3: Center of Mass Motion

The motion of the center of mass of a system of particles is governed by the net external force acting on the system.

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Step 1: Calculate the acceleration of the center of mass✦ Active

Given that bodies A and B are identical and have equal masses ( $m_{A} = m_{B} = m$ ), the acceleration of the center of mass (CM) is:

\vec{A_{C M}} = \frac{m_{A} \vec{a_{A}} + m_{B} \vec{a_{B}}}{m_{A} + m_{B}} = \frac{m (6 \hat{i} + 6 \hat{j}) + m (\vec{0})}{m + m} = \frac{6 \hat{i} + 6 \hat{j}}{2} = 3 \hat{i} + 3 \hat{j} {m/s}^{2}

The acceleration of the center of mass is constant.

Step 2: Calculate the initial velocity of the center of mass○ Expand

The initial velocity of the center of mass is:

\vec{V_{C M, 0}} = \frac{m_{A} \vec{v_{A}} + m_{B} \vec{v_{B}}}{m_{A} + m_{B}} = \frac{m (4 \hat{i}) + m (4 \hat{j})}{m + m} = \frac{4 \hat{i} + 4 \hat{j}}{2} = 2 \hat{i} + 2 \hat{j} m/s

Step 3: Determine the path of the center of mass○ Expand

The center of mass moves with a constant acceleration $\vec{A_{C M}} = 3 \hat{i} + 3 \hat{j}$ and has an initial velocity $\vec{V_{C M, 0}} = 2 \hat{i} + 2 \hat{j}$ . We observe the relationship between these two vectors:

\vec{A_{C M}} = \frac{3}{2} (2 \hat{i} + 2 \hat{j}) = \frac{3}{2} \vec{V_{C M, 0}}

Since the acceleration vector is parallel to the initial velocity vector, the velocity of the center of mass will always remain parallel to its initial velocity. Therefore, the direction of motion is constant, and the center of mass moves in a straight line.

💡 Teacher's Secret Hint

Remember that a parabolic path occurs when the initial velocity and constant acceleration are not parallel.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 32 – JEE-MAIN 2026

Clue 1 of 3: Center of Mass Motion

🧠 Full Solution Path