Question

Two persons A and B of mass 50 kg and 100 kg are standing at extremities of a boat of mass 100 kg with length 5 m floating at rest in still water.What the distance moved by the boat if they interchange their positions ?

Answer 1

The above figure represents the phenomenon described in the problem.
To calculate the movement of the boat a vertical Reference Line is imagined though that end of the boat where the person A of mass 50 kg is standing at initial position of the boat. Length of the boat is perpendicular to the reference line.

Given that the boat is floating at rest in still water . So by the law of conservation of momentum its center of mass must be at rest.

Hence calculation is made using the condition that the distance of center of mass of a system of masses in respect of a reference line remains unaltered even if there occurs rearrangement of masses within the system ,provided no external force exerted on the system .

Now to locate the position of CM let us consider that it is situated at a distance of `x` m from the line of reference.

Given that

• the mass of the boat `m_b=100kg`,
  mass of person A `m_A=50` kg
  mass of person B`m_B=100` kg
  length of the boat `L=5`m
  distance of CM of boat from reference line `=L/2=2.5`m

So

`x=(m_A*0+m_b*L/2+m_B*L)/(m_b+m_B+m_A)` m

`=>x=(50*0+100*2.5+100*5)/(100+100+50)=3` m

When the two persons exchange their positions . Let the boat be shifted `d` m from its initial position. But this shift will not change the distance of the CM of the system from reference line,
For the shifted position of the boat we get

`x=(m_B*d+m_b*(d+L/2)+m_A*(d+L))/(m_b+m_B+m_A)`

`=>3=(100*d+100*(d+2.5)+50*(d+5))/(100+100+50)`

`=>3*250=100d+100d+250+50d+250`

`=>250d=750-500=250`

`=>d=1` m

So the distance moved by the boat if the two persons interchange their positions will be `d=1`m