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Q. A man of $80 \, kg$ attempts to jump from a small boat of mass $40 \, kg$ on to the shore. He can generate a relative velocity of $6 \, m \, s^{- 1}$ between himself and the boat. His velocity towards the shore is

NTA AbhyasNTA Abhyas 2020

Solution:

Let the velocity of boat be $V$ opposite to the man. By momentum conservation, $-40V+80\left(\right.6-V\left.\right)=0$
$\therefore \, V=4ms^{- 1}$
$\therefore \, $ Velocity of the man $=6-4=2ms^{- 1}$