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

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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 of the man with respect to the boat is $v_{m/b} = 2 \,ms^{-1}$.By relative velocity definition, $v_{m/b} = v_m - v_b$, so $v_m = v_b + 2$.Since there is no external horizontal force on the system (man + boat), the momentum of the system is conserved.Initial momentum $P_i = 0$.Final momentum $P_f = m_{boat} v_b + m_{man} v_m = 0$.Substituting the values: $200 v_b + 50(v_b + 2) = 0$.$200 v_b + 50 v_b + 100 = 0$.$250 v_b = -100$.$v_b = -100 / 250 = -0.4 \,ms^{-1}$.The magnitude of the velocity is $0.4 \,ms^{-1}$ or $2/5 \,ms^{-1}$.

$A$ 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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