The figure shows the positions and velocities | vedclass

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(B) The particles are moving under mutual attraction,which means the net external force on the system is zero $(F_{ext} = 0)$.Therefore,the velocity o

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$x = 6\,m$

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(B) The particles are moving under mutual attraction,which means the net external force on the system is zero $(F_{ext} = 0)$.Therefore,the velocity of the centre of mass $(v_{CM})$ remains constant.Given: $m_1 = 1\,kg$,$x_1 = 2\,m$,$v_1 = 5\,m/s$; $m_2 = 1\,kg$,$x_2 = 8\,m$,$v_2 = -3\,m/s$.The initial position of the centre of mass is $X_{CM}(0) = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} = \frac{(1)(2) + (1)(8)}{1 + 1} = \frac{10}{2} = 5\,m$.The velocity of the centre of mass is $v_{CM} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} = \frac{(1)(5) + (1)(-3)}{1 + 1} = \frac{2}{2} = 1\,m/s$.The position of the centre of mass at time $t$ is given by $X_{CM}(t) = X_{CM}(0) + v_{CM} \cdot t$.At $t = 1\,s$,$X_{CM}(1) = 5 + (1)(1) = 6\,m$.

The figure shows the positions and velocities of two particles. If the particles move under their mutual attraction,then the position of the centre of mass at $t = 1\,s$ is

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