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 is the distance moved by the boat if they interchange their positions?

Answer 1

Hint:In order to calculate the movement of the boat, a vertical Reference Line is considered through that end of the boat where the person $A$, whose mass is 50 kg, is standing at the initial position of the boat. The length of the boat is perpendicular to the reference line because the length of the boat is horizontal and the reference line is vertical.

Complete step by step answer:
It is given that the boat is floating at rest in the still water. So, according to the law of conservation of momentum its centre of mass should be at rest.Now, in order to locate the position of the Centre of mass, let us consider that it is situated at a distance of $x\,metre$ from the line of reference.The following quantities are given to us in the question:
mass of boat ${m_b} = 100\,kg$
mass of the person$A$ ${m_A} = 50\,kg$
mass of the person $B$ ${m_B} = 100\,kg$$250d = 250$
length of boat $L = 5\,m$

Distance of centre of mass of boat from reference line $\dfrac{L}{2} = 2.5m$
$x = \dfrac{{{m_A} \times 0 + {m_b} \times \dfrac{L}{2} + {m_B} \times L}}{{{m_b} + {m_B} + {m_A}}}m$
On putting the given values,
$x = \dfrac{{50 \times 0 + 100 \times 2.5 + 100 \times 5}}{{100 + 100 + 50}}m$
When the two persons interchange their positions on the boat, then let the boat be shifted $d$ metre from its initial position. But this shift will not change the distance of the centre of mass of the system from the reference line.For the shifted position of the boat,
$x = \dfrac{{{m_B} \times d + {m_b} \times \left( {d + \dfrac{L}{2}} \right) + {m_A} \times \left( {d + L} \right)}}{{{m_b} + {m_B} + {m_A}}}m$

On putting the required values, we get,
$3 = \dfrac{{100 \times d + 100 \times \left( {d + 2.5} \right) + 50 \times \left( {d + 5} \right)}}{{100 + 100 + 50}}m$
$\Rightarrow 3 = \dfrac{{100 \times d + 100 \times \left( {d + 2.5} \right) + 50 \times \left( {d + 5} \right)}}{{250}}m$
On cross multiplying,
$3 \times 250 = 100d + 100d + 250 + 50d + 250$
$\Rightarrow 250d = 750 - 500$
On further solving, we get,
$250d = 250$
$\therefore d = 1\,m$

So, the distance moved by the boat when the two persons interchange their positions is $d = 1\,m$.

Note: In this question, the calculation is made keeping in mind that the distance of centre of mass in respect of a reference line remains unchanged even if there is a rearrangement of masses within the system, provided that no external force is exerted on the system.