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

Concept: The acceleration of the centre of mass is given by \[ \vec a_{cm} = \frac{\sum m_i \vec a_i}{\sum m_i} \] Similarly, velocity of centre of mass: \[ \vec v_{cm} = \frac{\sum m_i \vec v_i}{\sum m_i} \] Step 1: {Find velocity of centre of mass.} Since masses are equal: \[ \vec v_{cm}=\frac{\vec v_1+\vec v_2}{2} \] \[ =\frac{4\hat{i}+4\hat{j}}{2} \] \[ =2\hat{i}+2\hat{j} \] Step 2: {Find acceleration of centre of mass.} \[ \vec a_{cm}=\frac{\vec a_1+\vec a_2}{2} \] \[ =\frac{(6\hat{i}+6\hat{j})+0}{2} \] \[ =3\hat{i}+3\hat{j} \] Step 3: {Analyze the motion.} Initial velocity: \[ (2,2) \] Acceleration: \[ (3,3) \] Thus both components vary linearly with time but acceleration is constant. Hence trajectory satisfies a quadratic relation between \(x\) and \(y\), which represents a parabola.