A girl of mass 50, kg jumps out of a moving b | vedclass

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(A) According to the law of conservation of momentum,the total momentum before the jump is equal to the total momentum after the jump.Initially,the bo

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(A) According to the law of conservation of momentum,the total momentum before the jump is equal to the total momentum after the jump.Initially,the boat and the girl are moving together,so the initial momentum is $0$ (assuming the system starts from rest or considering the relative change).Let $m_1 = 50\, kg$ be the mass of the girl and $v_1 = 3\, m s^{-1}$ be her velocity.Let $m_2 = 300\, kg$ be the mass of the boat and $v_2$ be the velocity of the boat.Using the principle of conservation of linear momentum: $m_1 v_1 + m_2 v_2 = 0$.Substituting the values: $(50 	imes 3) + (300 	imes v_2) = 0$.$150 + 300 v_2 = 0$.$300 v_2 = -150$.$v_2 = -\frac{150}{300} = -0.5\, m s^{-1}$.The negative sign indicates that the boat moves in the opposite direction (backwards) with a velocity of $0.5\, m s^{-1}$.

$A$ girl of mass $50\, kg$ jumps out of a moving boat of mass $300\, kg$ on to the bank with a horizontal velocity of $3\, m s^{-1}$. With what velocity does the boat begin to move backwards (in $, m s^{-1}$)?

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