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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Laws of Motion

A $250 \;kg$ boat $4.5\;m$ long is floating with out motion on still water, with a boy of mass $50\; kg$ at one end. If he runs to the other end of the boat and stops, by how much distance the boat moves back?

\[(a)\;1\; m \quad (b)\;0.75\; m \quad (c)\;0.50\; m \quad(d)\;0.25\;m \]
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1 Answer

Since momentum is conserved in the system
Forward momentum of boy = backward momentum of boat
$L$ - Length of boat
$M$ - mass of boat
$t$ - time
$m$ - mass of boy
$x$ - distance moved by boat
$ M(L/t)=(M+m)x/t$
$50 \times 4.5 =(250 +50)x$
$x=0.75\; m$
Hence b is the correct answer.
answered Jul 5, 2013 by meena.p
edited May 26, 2014 by lmohan717

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