Question

A $ 500 \,kg $ boat is $ 9\,m $ long and is floating without motion on still water. A man of mass $ 100 \,kg $ is at one end and if he runs to the other end of the boat and stops, the displacement of the boat is

• A. 1.5 m in the direction of displacement of the man
• B. 0.75 m in the direction of displacement of the man
• C. 1.5 m in the direction opposite to the displacement of the man
• D. 0.75 m in the direction opposite to the displacement of the man

Answer 1

Answer: (C)

Given that the system is initially at rest,
i. e. , $\,\,v _{ CM }=0 $
$\therefore \,\,\,\frac{m_{1}\, v _{1}+m_{2} \,v _{2}}{m_{1}+m_{2}}=0 $
$m_{1} \,v _{1}+m_{2} \,v _{2}=0 $
$m_{1} \frac{\Delta \,r_{1}}{\Delta \,t}+m_{2} \frac{\Delta \,r_{2}}{\Delta\,, t}=0 $
$m_{1} \Delta r _{1}+m_{2} r _{2}=0$
Now here, in boat-man system, if the man moves towards right the boat moves toward left
$m_{1} \,\Delta \,r_{1} \pm \,m_{2} \,\Delta \,r_{2}$
If $\Delta r_{2}$ is the displacement of boat relative to shore; there the displacement of man relative to shore would be $\left(9-\Delta r_{2}\right)$
$\Delta r_{1} =9-\Delta r_{2} $
$m_{1}\left(9-\Delta r_{2}\right) =m_{2} \Delta r_{2} $
$100\left(9-\Delta r_{2}\right) =500\, \Delta r_{2} $
$\Delta r_{2} =\frac{100 \times 9}{600}=1.5 \,m $
i. e. , Boat moves $1.5 \,m$ relative to shore in the direction opposite to the displacement of man.