Two identical bodies A and B of equal masses | vedclass

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(C) The acceleration of the centre of mass is given by $a_{cm} = \frac{m_A a_1 + m_B a_2}{m_A + m_B}$.Since $m_A = m_B = m$, we have $a_{cm} = \frac{m

Vedclass · Accepted Answer

straight line

Vedclass · Answer

(C) The acceleration of the centre of mass is given by $a_{cm} = \frac{m_A a_1 + m_B a_2}{m_A + m_B}$.Since $m_A = m_B = m$, we have $a_{cm} = \frac{m(6\hat{i} + 6\hat{j}) + m(0)}{2m} = 3\hat{i} + 3\hat{j} 	ext{ m/s}^2$.The initial velocity of the centre of mass is $v_{cm} = \frac{m_A v_1 + m_B v_2}{m_A + m_B} = \frac{m(4\hat{i}) + m(4\hat{j})}{2m} = 2\hat{i} + 2\hat{j} 	ext{ m/s}$.Since the acceleration vector $a_{cm} = 3\hat{i} + 3\hat{j}$ and the initial velocity vector $v_{cm} = 2\hat{i} + 2\hat{j}$ are parallel (specifically, $a_{cm} = 1.5 v_{cm}$), the motion of the centre of mass is along a straight line.

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