A 500,kg boat is 9,m long and is floating wit | vedclass

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(C) Since there is no external horizontal force acting on the system (man + boat),the center of mass of the system remains stationary.Let $M = 500\,kg

Vedclass · Accepted Answer

$1.5\,m$ in the direction opposite to the displacement of the man

Vedclass · Answer

(C) Since there is no external horizontal force acting on the system (man + boat),the center of mass of the system remains stationary.Let $M = 500\,kg$ be the mass of the boat and $m = 100\,kg$ be the mass of the man.Let $L = 9\,m$ be the length of the boat.Let $x$ be the displacement of the boat relative to the water.When the man moves from one end to the other,his displacement relative to the boat is $L = 9\,m$.His displacement relative to the water is $(L - x) = (9 - x)$.Using the principle of conservation of the center of mass: $m(L - x) - Mx = 0$.Substituting the values: $100(9 - x) - 500x = 0$.$900 - 100x - 500x = 0$.$900 = 600x$.$x = \frac{900}{600} = 1.5\,m$.Since the boat moves in the direction opposite to the man's motion to keep the center of mass fixed,the displacement is $1.5\,m$ in the direction opposite to the displacement of the man.

$A$ $500\,kg$ boat is $9\,m$ long and is floating without motion on still water. $A$ man of mass $100\,kg$ is at one end and if he runs to the other end of the boat and stops,the displacement of the boat is

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