Q57 / 75 · JEE Main 2026 (06 April Shift 2) · Physics
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.
- Acircular
- Bparabolic
- Correct answerCstraight line
- Delliptical
Solution
Velocity of the centre of mass is given by:
$\vec{v}_{cm} = \dfrac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2}$
Since $m_1 = m_2 = m$, we get:
$\vec{v}_{cm} = \dfrac{m(4\hat{i}) + m(4\hat{j})}{2m} = 2\hat{i} + 2\hat{j}$ m/s
Acceleration of the centre of mass is given by:
$\vec{a}_{cm} = \dfrac{m_1\vec{a}_1 + m_2\vec{a}_2}{m_1 + m_2}$
$\vec{a}_{cm} = \dfrac{m(6\hat{i} + 6\hat{j}) + 0}{2m} = 3\hat{i} + 3\hat{j}$ m/s$^2$
Here, $\vec{v}_{cm} = 2(\hat{i} + \hat{j})$ and $\vec{a}_{cm} = 3(\hat{i} + \hat{j})$.
Since the initial velocity and acceleration of the centre of mass are parallel to each other, the centre of mass moves in a straight line path.
Answer: straight line
$\vec{v}_{cm} = \dfrac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2}$
Since $m_1 = m_2 = m$, we get:
$\vec{v}_{cm} = \dfrac{m(4\hat{i}) + m(4\hat{j})}{2m} = 2\hat{i} + 2\hat{j}$ m/s
Acceleration of the centre of mass is given by:
$\vec{a}_{cm} = \dfrac{m_1\vec{a}_1 + m_2\vec{a}_2}{m_1 + m_2}$
$\vec{a}_{cm} = \dfrac{m(6\hat{i} + 6\hat{j}) + 0}{2m} = 3\hat{i} + 3\hat{j}$ m/s$^2$
Here, $\vec{v}_{cm} = 2(\hat{i} + \hat{j})$ and $\vec{a}_{cm} = 3(\hat{i} + \hat{j})$.
Since the initial velocity and acceleration of the centre of mass are parallel to each other, the centre of mass moves in a straight line path.
Answer: straight line
Hence, the correct answer is option: 'C'.