A man of 50 ,kg is standing at one end of a b | vedclass

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(A) Let $v_b$ be the velocity of the boat with respect to the water and $v_m$ be the velocity of the man with respect to the water.Given, the velocity

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(A) Let $v_b$ be the velocity of the boat with respect to the water and $v_m$ be the velocity of the man with respect to the water.Given, the velocity of the man with respect to the boat is $v_{mb} = 2 \,ms^{-1}$.By definition, $v_{mb} = v_m - v_b$, so $v_m = v_b + 2$.Since there is no external horizontal force acting on the system (man + boat), the linear momentum of the system is conserved.Initial momentum $P_i = 0$.Final momentum $P_f = m_{boat} v_b + m_{man} v_m = 200 v_b + 50(v_b + 2)$.Equating $P_i = P_f$:$200 v_b + 50 v_b + 100 = 0$$250 v_b = -100$$v_b = -\frac{100}{250} = -0.4 \,ms^{-1}$.The magnitude of the velocity of the boat is $0.4 \,ms^{-1}$ (in the opposite direction of the man's motion).

$A$ man of $50 \,kg$ is standing at one end of a boat of length $25 \,m$ and mass $200 \,kg$. If he starts running and when he reaches the other end, he has a velocity $2 \,ms^{-1}$ with respect to the boat. The final velocity of the boat is: (in $ms^{-1}$)

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