Two masses m_1 and m_2 are connected by a mas | vedclass

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(D) The initial positions of the masses are $x_1 = 0$ and $x_2 = l$.The initial position of the centre of mass $(x_{CM})$ at $t=0$ is given by:$x_{CM}

Vedclass · Accepted Answer

$x=\frac{m_2 l}{m_1+m_2}+\frac{m_1 v_0 t}{m_1+m_2}$

Vedclass · Answer

(D) The initial positions of the masses are $x_1 = 0$ and $x_2 = l$.The initial position of the centre of mass $(x_{CM})$ at $t=0$ is given by:$x_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{m_1(0) + m_2(l)}{m_1 + m_2} = \frac{m_2 l}{m_1 + m_2}$The initial velocities of the masses are $v_1 = v_0$ and $v_2 = 0$.The velocity of the centre of mass $(v_{CM})$ is constant because there is no external force acting on the system:$v_{CM} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{m_1 v_0 + m_2(0)}{m_1 + m_2} = \frac{m_1 v_0}{m_1 + m_2}$At a later time $t$,the position of the centre of mass $(x'_{CM})$ is given by:$x'_{CM} = x_{CM} + v_{CM} t$$x'_{CM} = \frac{m_2 l}{m_1 + m_2} + \left( \frac{m_1 v_0}{m_1 + m_2} \right) t$$x'_{CM} = \frac{m_2 l}{m_1 + m_2} + \frac{m_1 v_0 t}{m_1 + m_2}$Thus,the correct option is $D$.

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