30 ,kg boy stands at the far edge of a floati | vedclass

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(C) In the absence of any external horizontal force, the position of the Centre of Mass $(COM)$ of the system remains unchanged.Let the shore be the o

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$7.5$

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(C) In the absence of any external horizontal force, the position of the Centre of Mass $(COM)$ of the system remains unchanged.Let the shore be the origin $(x = 0)$.The mass of the plank is $M = 10 \,kg$ and its length is $L = 10 \,m$. Its $COM$ is at $x_p = 5 \,m$.The mass of the boy is $m = 30 \,kg$. Initially, he is at the far edge, so his position is $x_b = 10 \,m$.The initial position of the system's $COM$ is:$X_{COM} = \frac{M x_p + m x_b}{M + m} = \frac{10 	imes 5 + 30 	imes 10}{10 + 30} = \frac{50 + 300}{40} = \frac{350}{40} = 8.75 \,m$.When the boy walks to the near edge, the plank moves by a distance $d$ away from the shore. The new position of the plank's $COM$ is $x_p' = 5 + d$, and the new position of the boy is $x_b' = d$.Since the $COM$ remains at the same position:$X_{COM} = \frac{M x_p' + m x_b'}{M + m}$$8.75 = \frac{10(5 + d) + 30(d)}{40}$$8.75 	imes 40 = 50 + 10d + 30d$$350 = 50 + 40d$$300 = 40d$$d = \frac{300}{40} = 7.5 \,m$.

$30 \,kg$ boy stands at the far edge of a floating plank, whose near edge is against the shore of a river. The plank is $10 \,m$ long and weighs $10 \,kg$. If the boy walks to the near edge of the plank, how far from the shore does the plank move (in $\,m$)?

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