A man of mass 60 ,kg is standing on a boat of | vedclass

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Question

(A) Since there is no external horizontal force acting on the system (man + boat),the center of mass of the system remains stationary.Let $m_m = 60 \,

Vedclass · Accepted Answer

$15.8$

Vedclass · Answer

(A) Since there is no external horizontal force acting on the system (man + boat),the center of mass of the system remains stationary.Let $m_m = 60 \,kg$ be the mass of the man and $m_b = 140 \,kg$ be the mass of the boat.The man walks a distance $d = v 	imes t = 1.5 \,m/s 	imes 4 \,s = 6.0 \,m$ relative to the boat.Let the boat move away from the shore by a distance $x$ due to the man walking towards the shore.By the principle of conservation of the center of mass,the displacement of the center of mass is zero:$m_m \Delta x_m + m_b \Delta x_b = 0$Taking the direction towards the shore as negative,the displacement of the man relative to the ground is $-(6.0 - x)$ and the displacement of the boat is $+x$.$60 	imes -(6.0 - x) + 140 	imes x = 0$$-360 + 60x + 140x = 0$$200x = 360$$x = 1.8 \,m$The man's displacement relative to the ground is $-(6.0 - 1.8) = -4.2 \,m$.The initial distance from the shore was $20 \,m$.Final distance $= 20 - 4.2 = 15.8 \,m$.

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