Question

A man weighing 72.5 kg stands in a boat so that he is 4.58 m from the shore. He walks 2.44 m in the boat towards the shore and then stops. How far from the shore will he be at the end of this time? The boat weighs 90.7 kg, and there is assumed to be no friction between it and the water.

• A. 2.14 m
• B. 2.41 m
• C. 3.22 m
• D. 3.5 m

Answer 1

The correct option is C 3.22 m



Since there is no external force on the system (boat + man). The centre of mass of the system remains at rest.
Let, x = the distance moved by man on boat together = 2.44 m

Distance move, ${\overrightarrow{x}}_{mg}={\overrightarrow{x}}_{md}+{\overrightarrow{x}}_{bg}$

$x_{mg}=$ distance moved by man wrt ground in x
$x_{md}=$ distance moved by man wrt boat = 2.44 m
$x_{bg}=$ distance moved by boat wrt ground

Let, $x_{bg}=x$

${\overrightarrow{X}}_{mg}=2.44-x[\because$ boat moves in backward direction]

m = mass of man, M = mass of boat

$m(2.44-x)-Mx=0$

$72.5\times (2.44-x)-90.7x=0$

$\frac{72.5\times 2.44}{(72.5+90.7}=x$

$\Rightarrow x=1.08$

Hence boat moves 1.08 m in backward direction
Thus distance of man from shore

$=4.58-2.44+1.08=3.22m$