On a horizontal frictionless frozen lake,a gi | vedclass

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(D) Since there is no external horizontal force acting on the system,the position of the centre of mass of the system remains unchanged.Let the initia

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$4 \,m$

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(D) Since there is no external horizontal force acting on the system,the position of the centre of mass of the system remains unchanged.Let the initial position of the girl be at the origin $(x = 0)$ and the box be at $x = 20 \,m$.The mass of the girl is $m_1 = 36 \,kg$ and the mass of the box is $m_2 = 9 \,kg$.The initial position of the centre of mass $(X_{CM})$ is:$X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{36 	imes 0 + 9 	imes 20}{36 + 9} = \frac{180}{45} = 4 \,m$.Let the girl move a distance $x$ towards the box and the box move a distance $(20 - x)$ towards the girl. When they meet,they will both be at the position of the centre of mass,$X_{CM} = 4 \,m$.Therefore,the girl has travelled $4 \,m$ from her initial position to reach the centre of mass,and the box has travelled $20 - 4 = 16 \,m$ to reach the same point.Thus,the girl has travelled $4 \,m$.

On a horizontal frictionless frozen lake,a girl of mass $36 \,kg$ and a box of mass $9 \,kg$ are connected to each other by means of a rope. Initially,they are $20 \,m$ apart. The girl exerts a horizontal force on the box,pulling it towards her. How far has the girl travelled when she meets the box?

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